RD Sharma Volume 1 - Algebra Of Matrices Solution For Class 12 Mathematics

Class 12^th Mathematics RD Sharma Volume 1 Solution

Exercise 5.1

If a matrix has 8 elements, what are the possible orders it can have? What if it…
If a = [a_jj] = [ccc 2&3&-5 1&4&9 0&7&-2] and b = [b_ij] = [rr 2&-1 -3&4 1&2]…
If a = [a_jj] = [ccc 2&3&-5 1&4&9 0&7&-2] and b = [b_ij] = [rr 2&-1 -3&4 1&2]…
Let A be a matrix of order 3 × 4. If R1 denotes the first row of A and C2…
aij = i × j Construct a 2 ×3 matrix A = [ajj] whose elements ajj are given by :…
aij = 2i - j Construct a 2 ×3 matrix A = [ajj] whose elements ajj are given by…
aij = i + j Construct a 2 ×3 matrix A = [ajj] whose elements ajj are given by :…
aij = (i+j)^2/2 Construct a 2 ×3 matrix A = [ajj] whose elements ajj are given…
(i+j)^2/2 Construct a 2 × 2 matrix A = [ajj] whose elements ajj are given by :…
a_jj = (i-j)^2/2 Construct a 2 × 2 matrix A = [ajj] whose elements ajj are…
a_jj = (i-2j)^2/2 Construct a 2 × 2 matrix A = [ajj] whose elements ajj are…
a_jj = (2i+j)^2/2 Construct a 2 × 2 matrix A = [ajj] whose elements ajj are…
a_jj = |2i-3 j|/2 Construct a 2 × 2 matrix A = [ajj] whose elements ajj are…
a_jj = |-3i + j|/2 Construct a 2 × 2 matrix A = [ajj] whose elements ajj are…
aij = e2ix sin xj Construct a 2 × 2 matrix A = [ajj] whose elements ajj are…
ajj = i + j Construct a 3×4 matrix A = [ajj] whose elements ajj are given by :…
ajj = i - j Construct a 3×4 matrix A = [ajj] whose elements ajj are given by :…
ajj = 2i Construct a 3×4 matrix A = [ajj] whose elements ajj are given by :…
ajj = j Construct a 3×4 matrix A = [ajj] whose elements ajj are given by :…
a_jj = 1/2 |-3i+j| Construct a 3×4 matrix A = [ajj] whose elements ajj are…
a_jj = 2i + i/j Construct a 4 × 3 matrix A = [ajj] whose elements ajj are given…
a_ij = i-j/i+j Construct a 4 × 3 matrix A = [ajj] whose elements ajj are given…
ajj = i Construct a 4 × 3 matrix A = [ajj] whose elements ajj are given by :…
[ccc 3x+4y&2 a+b&2a-b&-1] = [ccc 2&2&4 5&-5&-1] Find x, y, a and b if…
[cc 2a+b 5c-d&4c+3d] = [cc 4&-3 11&24] Find x, y, a and b if
Find the values of a, b, c and d from the following equations: [cc 2a+b…
Find x, y and z so that A = B, where a = [ccc x-2&3&2z 18z+2&6z] , b = [ccc y&6…
If [cc x&3x-y 2x+z&3y-6] = [ll 3&2 4&7] find x, y, z, ω.
If [cc x&3x-y 2x+z&3y-6] = [ll 3&2 4&7] find x, y, z,ω.
If [ccc x+3+4&2y-7 4x+6&0 b-3&3b+2c] = [ccc 0&6&3y-2 2x&-3&2c+2 2b+4&-21&0]…
If [cc 2x+1&5x 0& y^2 + 1] = [cc x+3&10 0&26] find the value of (x + y).…
If [cc xy&4 z+6+y] = [ll 8&0 0&6] then find the values of x, y, z and ω.…
a row matrix which is also a column matrix Give an example of
a diagonal matrix which is not scalar Give an example of
a triangular matrix. Give an example of
The sales figure of two car dealers during January 2013 showed that dealer A…
For what value of x and y are the following matrices equal? a = [cc 2x+1&2y 0&…
Find the values of x and y if [cc x+10& y^2 + 2y 0&-4] = [cc 3x+4&3 0& y^2 -…
Find the values of a and b if A = B, where a = [cc a+4&3b 8&-6] , b = [cc 2a+2&…

Exercise 5.2

[rr 3&-2 1&4] + [cc -2&4 1&3] Compute the following sums:
[ccc 2&1&3 0&3&5 -1&2&5] + [rrr 1&-2&3 2&6&1 0&-3&1] Compute the following…
Let a = [ll 2&4 3&2] , b = [rr 1&3 -2&5] and c = [rr -2&5 3&4] Find each of the…
If a = [ll 2&3 5&7] , b = [ccc -1&0&2 3&4&1] , c = [ccc -1&2&3 2&1&0] find i. A…
Let a = [ccc -1&0&2 3&1&4] , b = [ccc 0&-2&5 1&-3&1] and c = [rrr 1&-5&2 6&0&-4]…
If A = diag (2, -5, 9), B = diag (1, 1, -4) and C = diag (-6, 3, 4), find i. A -…
Given the matrices a = [rrr 2&1&1 3&-1&0 0&2&4] , b = [rrr 9&7&-1 3&5&4 2&1&6]…
Find matrices X and Y, if X + Y = [ll 5&2 0&9] and x-y = [rr 3&6 0&-1]…
Find X, if y = [ll 3&2 1&4] and 2x-y [rr 1&0 -3&2]
Find matrices X and Y, if 2x-y = [ccc 6&-6&0 -4&2&1] and x+2y = [ccc 3&2&5…
If x-y = [lll 1&1&1 1&1&0 1&0&0] and x+y = [ccc 3&5&1 -1&1&4 11&8&0] find X and…
Find matrix A, if [rrr 1&2&-1 0&4&9]+a = [rrr 9&-1&4 -2&1&3]
If a = [ll 9&1 7&8] , b = [cc 1&5 7&12] find matrix C such that 5A + 3B + 2C is…
If a = [rr 2&-2 4&2 -5&1] , b = [rr 8&0 4&-2 3&6] find matrix X such that 2A +…
If a = [rrr 1&-3&2 2&0&2] and b = [ccc 2&-1&-1 1&0&-1] find the matrix C such…
[ccc x-y&2&-2 4&6] + [ccc 3&-2&2 1&0&-1] = [ccc 6&0&0 5&2x+y&5] Find x, y…
Find x, y satisfying the matrix equations. [x y + 2 z - 3] + [y 4 5] = [4 9,…
x [2 1]+y [3 5] + [-8 -11] = o Find x, y satisfying the matrix equations…
If 2 [ll 3&4 5] + [ll 1 0&1] = [cc 7&0 10&5] find x and y.
Find the value of λ, a non-zero scalar, if lambda [lll 1&0&2 3&4&5]+2 [ccc…
Find a matrix X such that 2A + B + X = O, where a = [rr -1&2 3&4] , b = [cc…
If a = [rr 8&0 4&-2 3&6] and b = [rr 2&-2 4&2 -5&1] then find the matrix X of…
3 [cc x z] = [cc x&6 -1&2t] + [cc 4+y z+t&3] Find x, y, z and t, if…
2 [cc x&5 7] + [cc 3&4 1&2] = [cc 7&14 15&14] Find x, y, z and t, if…
If X and Y are 2 × 2 matrices, then solve the following matrix equations for X…
In a certain city there are 30 colleges. Each college has 15 peons, 6 clerks, 1…
The monthly incomes of Aryan and Babban are in the ration 3: 4 and their…

Exercise 5.3

[cc a -b] [cc a&-b b] Compute the indicated products:
[rr 1&-2 2&3] [rrr 1&2&3 -3&2&-1] Compute the indicated products:…
[ccc 2&3&4 3&4&5 4&5&6] [ccc 1&-3&5 0&2&4 3&0&5] Compute the indicated…
a = [lr 5&-1 6&7] and b = [ll 2&1 3&4] Show that AB ≠ BA in each of the…
a = [rrr -1&1&0 0&-1&1 2&3&4] and b = [lll 1&2&3 0&1&0 1&1&0] Show that AB ≠ BA…
a = [lll 1&3&0 1&1&0 4&1&0] and b = [lll 0&1&0 1&0&0 0&5&1] Show that AB ≠ BA…
a = [cc 1&-2 2&3] and b = [lll 1&2&3 2&3&1] Compute the products AB and BA…
a = [rr 3&2 -1&0 -1&1] and b = [lll 4&5&6 0&1&2] Compute the products AB and BA…
a = [1-123] and b = [0 1 3 2] Compute the products AB and BA whichever exists…
[a , b] [c c d] + [lll a] [a b c d] Compute the products AB and BA whichever…
a = [ccc 1&3&-1 2&-1&-1 3&0&-1] and b = [rrr -2&3&-1 -1&2&-1 -6&9&-4] Show that…
a = [rrr 10&-4&-1 -11&5&0 9&-5&1] and b = [lll 1&2&1 3&4&2 1&3&2] Show that AB…
Evaluate the following:
[lll 1&2&3] [lll 1&0&2 2&0&1 0&1&2] [2 4 6] Evaluate the following:…
[cc 1&-1 0&2 2&3] ([ccc 1&0&2 2&0&1] - [ccc 0&1&2 1&0&2]) Evaluate the…
If a = [rr 1&0 0&-1] , b = [rr 1&0 0&-1] and c = [ll 0&1 1&0] then show that A^2…
If a = [rr 2&-1 3&2] and b = [rr 0&4 -1&7] find 3A^2 - 2B + I.
If a = [rr 4&2 -1&1] prove that (A - 2I) (A - 3I) = O.
If a = [ll 1&1 0&1] Show that a^2 = [ll 1&2 0&1] and a^3 = [ll 1&3 0&1]…
If a = [cc ab& b^2 - a^2 &-ab] show that A^2 = O
If a = [cc cos2theta -sin2theta] find A^2
If a = [rrr 2&-3&-5 -1&4&5 1&-3&-4] and b = [rrr -1&3&5 1&-3&-5 -1&3&5] show…
If a = [ccc 0&-b -c&0 b&-a&0] and b = [lll a^2 ab& b^2 ac& c^2] show that AB =…
If a = [rrr 2&-3&-5 -1&4&5 1&-3&-4] and b = [rrr 2&-2&-4 -1&3&4 1&-2&-3] show…
Let a = [rrr -1&1&-1 3&-3&3 5&5&5] and b = [rrr 0&4&3 1&-3&-3 -1&4&4] compute…
a = [ccc 1&2&0 -1&0&1] , b = [rr 1&0 -1&2 0&3] and c = [c 1 -1] For the…
a = [rrr 4&2&3 1&1&2 3&0&1] , b = [rrr 1&-1&1 0&1&2 2&-1&1] and c = [rrr…
a = [rr 1&-1 0&2] , b = [rr -1&0 2&1] and c = [rr 0&1 1&-1] For the following…
a = [rr 2&-1 1&1 -1&2] , b = [ll 0&1 1&1] and c = [rr 1&-1 0&1] For the…
If a = [ccc 1&0&-2 3&-1&0 -2&1&1] , b = [rrr 0&5&-4 -2&1&3 -1&0&2] and c = [rrr…
Compute the elements a43 and a22 of the matrix:
If a = [lll 0&1&0 0&0&1 p] and I is the identity matrix of order 3, show that…
If ω is a complex cube root of unity, show that
If a = [rrr 2&-3&-5 -1&4&5 1&-3&-4] show that A^2 = A.
Show that the matrix a = [rr 5&3 12&7] is a root of the equation A^2 - 12A - I…
If a = [rrr 4&-1&-4 3&0&-4 3&-1&-3] show that A^2 = I3
If [lll 1&1] [lll 1&0&2 0&2&1 2&1&0] [1 1 1] = 0 find x.
If [rr 2&3 5&7] [rr 1&-3 -2&4] = [cc -4&6 -9] find x.
If [ccc x&4&1] [ccc 2&1&2 1&0&2 0&2&-4] [c x 4 -1] = 0 find x
If [ccc 1&-1] [ccc 0&1&-1 2&1&3 1&1&1] [0 1 1] = 0 find x.
If a = [ll 3&-2 4&-2] and i = [ll 1&0 0&1] then prove that A^2 - A + 2I = 0.…
If a = [rr 3&1 -1&2] and i = [ll 1&0 0&1] then find λ so that A^2 = 5A + λI.…
If a = [rr 3&1 -1&2] show that A^2 - 5A + 7I2 = 0.
If a = [rr 2&3 -1&0] show that A^2 - 2A + 3I2 = 0.
Show that the matrix a = [ll 2&3 1&2] satisfies the equation A^3 - 4A^2 + A =…
If a = [rr 3&-5 -4&2] find A^2 - 5A - 14I.
If a = [rr 3&1 -1&2] show that A^2 -5A + 7I = 0. Use this to find A^4 .…
If a = [ll 3&-2 4&-2] find k such that A^2 = kA - 2I2.
If a = [rr 1&0 -1&7] find k such that A^2 - 8A + kI - 0.
If a = [ll 1&2 2&1] and f(x) = x^2 - 2x - 3, show that f(A) = 0.
If a = [ll 2&3 1&2] and i = [ll 1&0 0&1] then find λ, μ so that A^2 = λ A + μ I…
Find the value of x for which the matrix product [ccc 2&0&7 0&1&0 1&-2&1] [ccc…
Solve the matrix equations:
[ccc 1&2&1] [lll 1&2&0 2&0&1 1&0&2] [0 2 x] = 0 Solve the matrix equations:…
[x-5-1] [lll 1&0&2 0&2&1 2&0&3] [x 4 1] = 0 Solve the matrix equations:…
[rr 2x&3] [rr 1&2 -3&0] [x 8] = 0 Solve the matrix equations:
If a = [rrr 1&2&0 3&-4&5 0&-1&3] compute A^2 - 4A + 3I3.
If f(x) = x^2 - 2x, find f(A), where a = [lll 0&1&2 4&5&0 0&2&3]
If f(x) = x^3 + 4x^2 - x, find f(A), where a = [rrr 0&1&2 2&-3&0 1&-1&0]…
If a = [lll 1&0&2 0&2&1 2&0&3] then show that A is a root of the polynomial…
If a = [lll 1&2&2 2&1&2 2&2&1] then prove that A^2 - 4A - 5I = 0.…
If a = [lll 3&2&0 1&4&0 0&0&5] show that A^2 - 7A + 10I3 = 0.
Without using the concept of the inverse of a matrix, find the matrix [ll x z]…
[ll 1&1 0&1]a = [lll 3&3&5 1&0&1] Find the matrix A such that
a [lll 1&2&3 4&5&6] = [ccc -7&-8&-9 2&4&6] Find the matrix A such that…
[4 1 3]a = [lll -4&8&4 -1&2&1 -3&6&3] Find the matrix A such that…
[ccc 2&1&3] [ccc -1&0&-1 -1&1&0 0&1&1] [1 0 -1] = a Find the matrix A such…
[cc 2&-1 1&0 -3&4]a = [ccc -1&-8&-10 1&-2&-5 9&22&15] Find the matrix A such…
a = [lll 1&2&3 4&5&6] = [ccc -7&-8&-9 2&4&6 11&10&9] Find the matrix A such…
Find a 2 × 2 matrix A such that a [cc 1&-2 1&4] = 6i_2
If a = [ll 0&0 4&0] find A^16 .
If and x^2 = - 1, then show that (A + B)^2 = A^2 + B^2 .
If a = [ccc 1&0&-3 2&1&3 0&1&1] then verify that A^2 + A = (A + I), where I is…
If a = [rr 3&-5 -4&2] then find A^2 - 5A - 14I. Hence, obtain A^3 .…
If p (x) = [cc cosx -sinx] then show that P(x) P(y) = P(x + y) = P(y) P(x).…
If p = [lll x&0&0 0&0 0&0] and q = [lll a&0&0 0&0 0&0] prove that pq = [lll…
If a = [rrr 2&0&1 2&1&3 1&-1&0] find A^2 - 5A + 4I and hence find a matrix X…
If a = [ll 1&1 0&1] prove that a^n = [ll 1 0&1] for all positive integers n.…
If a = [ll a 0&1] prove that a^n = [cc a^n & b (a^n - 1/a-1) 0&1] for every…
If a = [cc costheta heta isintegrate heta] then prove by principle of…
If a = [cc cosalpha +sinalpha & root 2 sinalpha - root 2 sinalpha -sinalpha]…
If a = [lll 1&1&1 0&1&1 0&0&1] then use the principle of mathematical induction…
If B, C are n rowed matrices and if A = B + C, BC = CB, C^2 = O, then show that…
If A = diag(a b c), show that An = diag(an bn cn) for all positive integers n.…
If A is a square matrix, using mathematical induction prove that (AT)n = (An)T…
A matrix X has a + b rows and a + 2 columns while the matrix Y has b + 1 rows…
A and B such that AB ≠ BA. Give examples of matrices
A and B such that AB = O but A ≠ O, B ≠ O. Give examples of matrices…
A and B such that AB = O but BA ≠ O. Give examples of matrices
A, B and C such that AB = AC but B ≠ C, A ≠ O. Give examples of matrices…
Let A and B be square matrices of the same order. Does (A + B)^2 = A^2 + 2AB +…
If A and B are square matrices of the same order, explain, why in general (i)…
Let A and B be square matrices of the order 3 × 3. Is (AB)^2 = A^2 B^2 ? Give…
If A and B are square matrices of the same order such that AB = BA, then show…
Let a = [lll 1&1&1 3&3&3] , b = [rr 3&1 5&2 -2&4] and c = [rr 4&2 -3&5 5&0]…
Three shopkeepers, A, B and C go to a store to buy stationary. A purchases 12…
The cooperative stores of a particular school has 10 dozen physics books, 8…
In a legislative assembly election, a political group hired a public relations…
A trust fund has Rs 30000 that must be invested in two different types of…
75. To promote making of toilets for women, an organization tried to generate…
There are 2 families A and B. There are 4 men, 6 women and 2 children in family…
In a parliament election, a political party hired a public relations firm to…
The monthly incomes of Aryan and Babbar are in the ratio 3:4 and their monthly…
A trust invested some money in two types of bonds. The first bond pays 10%…

Exercise 5.4

Let a = [rr 2&-3 -7&5] and b = [rr 1&0 2&-4] verify that (2A)T = 2AT…
Let a = [rr 2&-3 -7&5] and b = [rr 1&0 2&-4] verify that (A + B)T = AT + BT…
Let a = [rr 2&-3 -7&5] and b = [rr 1&0 2&-4] verify that (A - B)T = AT - BT…
Let a = [rr 2&-3 -7&5] and b = [rr 1&0 2&-4] verify that (AB)T = BTAT…
If a = [3 5 2] and B = [1 0 4], verify that (AB)T = BTAT.
Let a = [ccc 1&-1&0 2&1&3 1&2&1] and b = [lll 1&2&3 2&1&3 0&1&1] Find AT, BT…
Let a = [ccc 1&-1&0 2&1&3 1&2&1] and b = [lll 1&2&3 2&1&3 0&1&1] Find AT, BT…
Let a = [ccc 1&-1&0 2&1&3 1&2&1] and b = [lll 1&2&3 2&1&3 0&1&1] Find AT, BT…
If a = [c -2 4 5] B= [1 3 -6], verify that (AB)T = BTAT.
If a = [rrr 2&4&-1 -1&0&2] , b = [rr 3&4 -1&2 2&1] find (AB)T .
For two matrices A and B, a = [lll 2&1&3 4&1&0] , b = [cc 1&-1 0&2 5&0] verify…
For the matrices, A and B, verify that (AB)T = BTAT, where a = [ll 1&3 2&4] , b…
If a^t = [cc 3&4 -1&2 0&1] and b = [rrr -1&2&1 1&2&3] find AT - BT.…
If a = [cc cosalpha -sinalpha] then verify that ATA = I2.
If a = [cc sinalpha -cosalpha] verify that ATA = I2.
If l_i , m_i , n_i i = 1 , 2 , 3 denote the direction cosines of three mutually…

Exercise 5.5

If a = [ll 2&3 4&5] prove that A - AT is a skew-symmetric matrix.…
If a = [ll 3&-4 1&-1] show that A - AT is a skew-symmetric matrix.…
If the matrix a = [rrr 5&2 y&-3 4&-7] is a symmetric matrix, find x, y, z and t.…
Let a = [ccc 3&2&7 1&4&3 -2&5&8] Find matrices X and Y such that X + Y = A,…
Express the matrix a = [ccc 4&2&-1 3&5&7 1&-2&1] as the sum of a symmetric and a…
Define a symmetric matrix. Prove that for a = [ll 2&4 5&6] , a+a^t is a…
Express the matrix a = [ll 3&-4 1&-1] as the sum of a symmetric and a…
Express the matrix [ccc 3&-2&-4 3&-2&-5 -1&1&2] as the sum of a symmetric and…

Very Short Answer

If A is an m × n matrix and B is n × p matrix does AB exist? If yes, write its order.…
If a = [ {lll} {2}&{1}&{4} {4}&{1}&{5} ] and b = [ {cc} {3}&{-1} {2}&{2} {1}&{3} ] .…
If a = [ {ll} {4}&{3} {1}&{2} ] and b = [ {c} {-4} {3} ] , write AB.…
If a = [ {1} {2} {3} ] , write AAT.
Give an example of two non-zero 2 × 2 matrices A and B such that AB = O.…
If a = [ {ll} {2}&{3} {5}&{7} ] , find A + AT.
If a = [ {ll} {2}&{3} {5}&{7} ] , write A2.
If a = [ {ll} {cosx}&{sinx} {-sinx}&{cosx} ] , find x satisfying 0 when A + AT = I.…
If a = [ {cc} {cosx}&{-sinx} {sinx}&{cosx} ] , find AAT.
If [ {cc} {1}&{0} {y}&{5} ]+2 [ {cc} {x}&{0} {1}&{-2} ] = i , where I is 2 × 2 unit…
If a = [ {cc} {1}&{-1} {-1}&{1} ] , satisfies the matrix equation A2 = kA, write the…
If a = [ {ll} {1}&{1} {1}&{1} ] satisfies A4 = λA, then write the value of λ.…
If a = [ {ccc} {-1}&{0}&{0} {0}&{-1}&{0} {0}&{0}&{-1} ] , find A2.…
If a = [ {ccc} {-1}&{0}&{0} {0}&{-1}&{0} {0}&{0}&{-1} ] , find A3.…
If a = [ {cc} {-3}&{0} {0}&{-3} ] , find A4.
If [x 2] [ {3} {4} ] = 2 , find x.
If A = [aij] is a 2 × 2 matrix such that aij = i + 2j, write A.
Write matrix A satisfying a + [ {cc} {2}&{3} {-1}&{4} ] = [ {cc} {3}&{-6} {-3}&{8} ]…
If A = [aij] is a square matrix such that aij = i2 – j2, then write whether A is…
For any square matrix write whether AAT is symmetric or skew-symmetric.…
If A = [aij] is a skew-symmetric matrix, then write the value of sum _{i}a_{ij} .…
If A = [aij] is a skew-symmetric matrix, then write the value of sum _{i}…
If A and B are symmetric matrices, then write the condition for which AB is also…
If B is a skew-symmetric matrix, write whether the matrix AB AT is symmetric or…
If B is a symmetric matrix, write whether the matrix AB AT is symmetric or…
If A is a skew-symmetric and n ∈ N such that (An)T = λAn, write the value of λ.…
If A is a symmetric matrix and n ∈ N, write whether An is symmetric or skew-symmetric…
If A is a skew-symmetric matrix and n is an even natural number, write whether An is…
If A is a skew-symmetric matrix and n is an odd natural number, write whether An is…
If A and B are symmetric matrices of the same order, write whether AB – BA is…
Write a square matrix which is both symmetric as well as skew-symmetric.…
Find the value of x and y, if 2 [ {ll} {1}&{3} {0}&{x} ] + [ {ll} {y}&{0} {1}&{2} ] =…
If [ {cc} {x+3}&{4} {y-4}&{x+y} ] = [ {cc} {5}&{4} {3}&{9} ] , find x and y.…
Find the value of x from the following: [ {cc} {2x-y}&{5} {3}&{y} ] = [ {cc} {6}&{5}…
Find the value of y, if [ {cc} {x-y}&{2} {x}&{5} ] = [ {ll} {2}&{2} {3}&{5} ] .…
Find the value of x, if [ {c} {3x+y}&{-y} {2y-x}&{3} ] = [ {cc} {1}&{2} {-5}&{3} ] .…
If matrix a = [ {lll} {1}&{2}&{3} ] , write AAT.
If [ {cc} {2x+y}&{3y} {0}&{4} ] = [ {ll} {6}&{0} {6}&{4} ] , then find x.…
If a = [ {ll} {1}&{2} {3}&{4} ] , find A + AT.
If [ {rr} {a+b}&{2} {5}&{b} ] = [ {ll} {6}&{5} {2}&{2} ] , then find a.…
If A is a matrix of order 3 × 4 and B is a matrix of order 4 × 3, find the order of…
If a = [ {cc} {cosalpha }&{-sinalpha} {sinalpha}&{cosalpha} ] is identity matrix,…
If [ {ll} {1}&{2} {3}&{4} ] [ {ll} {3}&{1} {2}&{5} ] = [ {ll} {7}&{11} {k}&{23} ] ,…
If I is the identity matrix and A is a square matrix such A2 = A, then what is the…
If a = [ {ll} {1}&{2} {0}&{3} ] is written as B + C, where B is a symmetric matrix…
If A is 2 × 3 matrix and B is a matrix such that ATB and BAT both are defined, then…
What is the total number of 2 × 2 matrices with each entry 0 or 1?…
If [ {cc} {x}&{x-y} {2x+y}&{7} ] = [ {ll} {3}&{1} {8}&{7} ] , then find the value of…
If a matrix has 5 elements, write all possible orders it can have.…
For a 2 × 2 matrix A = [aij] whose elements are given by a_{ij} = {i}/{j} , write…
If , find the value of x.
If [ {ccc} {9}&{-1}&{4} {-2}&{1}&{3} ] = a + [ {ccc} {1}&{2}&{-1} {0}&{4}&{9} ] ,…
If [ {cc} {a-b}&{2a+c} {2a-b}&{3c+d} ] = [ {cc} {-1}&{5} {0}&{13} ] , find the value…
For what value of x, is the matrix a = [ {ccc} {0}&{1}&{-2} {-1}&{0}&{3} {x}&{-3}&{0}…
If matrix a = [ {cc} {2}&{-2} {-2}&{2} ] and A2 = pA, then write the value of p.…
If A is a square matrix such that A2 = A, then write the value of 7A–(I + A)3, where I…
If 2 [ {ll} {3}&{4} {5}&{x} ] + [ {ll} {1}&{y} {0}&{1} ] = [ {ll} {7}&{0} {10}&{5} ]…
If [ {cc} {x}&{1} ] [ {cc} {1}&{0} {-2}&{0} ] = 0 , find x.
If [ {cc} {a+4}&{3b} {8}&{-6} ] = [ {cc} {2a+2}&{b+2} {8}&{a-8b} ] , write the value…
Write a 2 × 2 matrix which is both symmetric and skew-symmetric.
If [ {cc} {xy}&{4} {z+6}&{x+y} ] = [ {ll} {8}&{w} {0}&{6} ] , write the value of (x +…
Construct a 2 × 2 matrix A = [aij] whose elements aij are given by a_{ij} = { {ll} {…
If [ {x+y} {x-y} ] = [ {ll} {2}&{1} {4}&{3} ] [ {r} {1} {-2} ] , then write the value…
Matrix a = [ {ccc} {0}&{2b}&{-2} {3}&{1}&{3} {3a}&{3}&{-1} ] is given to be…
Write the number of all possible matrices of order 2 × 2 with each entry 1, 2 or 3.…
If [ {lll} {2}&{1}&{3} ] [ {ccc} {-1}&{0}&{-1} {-1}&{1}&{0} {0}&{1}&{1} ] [ {1} {0}…
If a = [ {ll} {3}&{5} {7}&{9} ] is written as A = P + Q, where as A = P + Q, where P…
Let A and B be matrices of orders 3 × 2 and 2 × 4 respectively. Write the order of…

Mcq

If a = [ {ccc} {1}&{0}&{0} {0}&{1}&{0} {a}&{b}&{-1} ] , then A2 is equal to…
If a = [ {ll} {0}&{i} {i}&{0} ] , n ∈ N, the A4n equals
If A and B are two matrices such that AB = A and BA = B, then B2 is equal to…
If AB = A and BA = B, where A and B are square matrices, then
If A and B are two matrices such that AB = B and BA = A, then A2 + B2 is equal to…
If [ {cc} { cos { 2 pi }/{7} } & { - sin frac { 2 pi }/{7} } { sin frac { 2 pi…
If the matrix AB is zero, then
Let a = [ {lll} {a}&{0}&{0} {0}&{a}&{0} {0}&{0}&{a} ] , then An is equal to…
If A, B and are square matrices or order 3, A is non-singular and AB = O, then B is a…
If a = [ {lll} {n}&{0}&{0} {0}&{n}&{0} {0}&{0}&{n} ] and b = [ {lll}…
If a = [ {ll} {1}&{a} {0}&{1} ] , then An (where n ∈ N) equals
If a = [ {lll} {1}&{2}&{x} {0}&{1}&{0} {0}&{0}&{1} ] and b = [ {ccc} {1}&{-2}&{y}…
If a = [ {ll} {1}&{-1} {2}&{-} ] , b = [ {cc} {a}&{1} {b}&{-1} ] and (A + B)2 = A2 +…
If a = [ {ll} { alpha } & { beta } { gamma } & { - alpha } ] is such that A2 =…
If S = [sij] is a scalar matrix such that sij = k and A is a square matrix of the same…
If A is a square matrix such that A2 = A, then (I + A)3 – 7A is equal to…
If a matrix A is both symmetric and skew-symmetric, then
The matrix [ {ccc} {0}&{5}&{-7} {-5}&{0}&{11} {7}&{-11}&{0} ] is…
If A is a square matrix, then AA is a
If A and B are symmetric matrices, then ABA is
If a = [ {ll} {5}&{x} {y}&{0} ] and A = AT, then
If A is 3 × 4 matrix and B is a matrix such that ATB and BAT are both defined. Then, B…
If A = [aij] is a square matrix of even order such that aij = i2 – j2, then…
If [ {cc} {costheta }&{-sintegrate heta} {sintheta}&{costheta} ] , then AT + A = I2,…
If a = [ {ccc} {2}&{0}&{-3} {4}&{3}&{1} {-5}&{7}&{2} ] is expressed as the sum of a…
Out of the following matrices, choose that matrix which is a scalar matrix:…
The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is…
Which of the given values of x and y make the following pairs of matrices equal? […
If a = [ {cc} {0}&{2} {3}&{-4} ] and ka = [ {ll} {0}&{3a} {2b}&{24} ] , then the…
If i = [ {ll} {1}&{0} {0}&{1} ] , j = [ {cc} {0}&{1} {-1}&{0} ] and b = [ {cc}…
The trace of the matrix a = [ {ccc} {1}&{-5}&{7} {0}&{7}&{9} {11}&{8}&{9} ] is…
If A = [aij] is scalar matrix of order n × n such that aij = k for all i, then trace…
The matrix a = [ {lll} {0}&{0}&{4} {0}&{4}&{0} {4}&{0}&{0} ] is a…
The number of possible matrices of order 3 × 3 with each entry 2 or 0 is…
If [ {cc} {2x+y}&{4x} {5x-7}&{4x} ] = [ {cc} {7}&{7y-13} {y}&{x+6} ] , then the value…
If A is a square matrix such that A2 = I, then (A – I)3 + (A + I)3 – 7A is equal to…
If A and B are two matrices of order 3 × m and 3 × n respectively and m = n, then the…
If A is a matrix of order m × n and B is a matrix such that ABT and BTA are both…
If A and B are matrices of the same order, then ABT – BTA is a
If matrix a = [a_{ij}]_ { 2 x 2 } , where a_{ij} = { {ll} { 1 , } & { i not equal…
If a = {1}/{ pi } [ {ccc} { sin^{-1} ( pix ) } & { tan^{-1} ( frac { pi }/{ pi } )…
If A and B are square matrices of the same order, then (A + B)(A – B) is equal to…
If a = [ {ccc} {2}&{-1}&{3} {-4}&{5}&{1} ] and b = [ {cc} {2}&{3} {4}&{-2} {1}&{5}…
The matrix a = [ {ccc} {0}&{-5}&{8} {5}&{0}&{12} {-8}&{-12}&{0} ] is a…
The matrix a = [ {lll} {1}&{0}&{0} {0}&{2}&{0} {0}&{0}&{4} ] is

Exercise 5.1

Question 1.

If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?

Answer:

If a matrix is of order m×n elements, it has mn elements. So, if the matrix has 8 elements, we will find the ordered pairs m and n.

mn = 8

Then, ordered pairs m and n can be

m×n be (8×1),(1×8),(4×2),(2×4)

Now, if it has 5 elements

Possible orders are (5×1), (1×5).

Question 2.

If and then find

a₂₂ + b₂₁

Answer:

A = [a_ij] = …......(1)

B = [b_ij] = …......(2)

Given, A = [a_ij] = B = [b_ij] =

Now, Comparing with equation (1) and (2)

a₂₂ = 4 and b₂₁ = – 3

a₂₂ + b₂₁ = 4 + ( – 3) = 1

Question 3.

If and then find

a₁₁ b₁₁ + a₂₂ b₂₂

Answer:

A = [a_ij] = …......(1)

B = [b_ij] = …......(2)

Given, A = [a_ij] = B = [b_ij] =

Now, Comparing with equation (1) and (2)

a₁₁ = 2, a₂₂ = 4, b₁₁ = 2, b₂₂ = 4

a₁₁ b₁₁ + a₂₂ b₂₂ = 2 × 2 + 4 × 4 = 4 + 16 = 20

Question 4.

Let A be a matrix of order 3 × 4. If R₁ denotes the first row of A and C₂ denotes its second column, then determine the orders of matrices R₁ and C₂.

Answer:

Let A be a matrix of order 3×4.

A = [a_ij]_3×4

R₁ = first row of A = [a₁₁,a₁₂,a₁₃,a₁₄]

So, order of matrix R₁ = 1×4

C₂ = second column of A =

Order of C₂ = 3×1

Question 5.

Construct a 2 ×3 matrix A = [a_jj] whose elements a_jj are given by :

a_ij = i × j

Answer:

Let A = [a_ij]_2×3

So, the elements in a 2×3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

A = ………………. (1)

a₁₁ = 1×1 = 1 a₁₂ = 1×2 = 2 a₁₃ = 1×3 = 3

a₂₁ = 2×1 = 2 a₂₂ = 2×2 = 4 a₂₃ = 2×3 = 6

So, from (1)

A =

Question 6.

Construct a 2 ×3 matrix A = [a_jj] whose elements a_jj are given by :

a_ij = 2i – j

Answer:

Let A = [a_ij]_2×3

So, the elements in a 2×3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

A = ………………. (1)

a₁₁ = 2×1 – 1 = 2 – 1 = 1 a₁₂ = 2×1 – 2 = 2 – 2 = 0 a₁₃ = 2×1 – 3 = 2 – 3 = – 1

a₂₁ = 2×2 – 1 = 4 – 1 = 3 a₂₂ = 2×2 – 2 = 4 – 2 = 2 a₂₃ = 2×2 – 3 = 4 – 3 = 1

So, from (1)

A =

Question 7.

Construct a 2 ×3 matrix A = [a_jj] whose elements a_jj are given by :

a_ij = i + j

Answer:

Let A = [a_ij]_2×3

So, the elements in a 2×3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

A = …… (1)

a₁₁ = 1 + 1 = 2 a₁₂ = 1 + 2 = 3 a₁₃ = 1 + 3 = 4

a₂₁ = 2 + 1 = 3 a₂₂ = 2 + 2 = 4 a₂₃ = 2 + 3 = 5

So, from (1)

A =

Question 8.

Construct a 2 ×3 matrix A = [a_jj] whose elements a_jj are given by :

a_ij =

Answer:

Let A = [a_ij]_2×3

So, the elements in a 2×3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

A = …… (1)

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

So, from (1)

A =

Question 9.

Construct a 2 × 2 matrix A = [a_jj] whose elements a_jj are given by :

Answer:

Let A = [a_ij]_2×2

So, the elements in a 2×2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂,

A = …… (1)

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

So, from (1)

A =

Question 10.

Construct a 2 × 2 matrix A = [a_jj] whose elements a_jj are given by :

Answer:

Let A = [a_ij]_2×2

So, the elements in a 2×2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂,

A = …… (1)

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

So, from (1)

A =

Question 11.

Construct a 2 × 2 matrix A = [a_jj] whose elements a_jj are given by :

Answer:

Let A = [a_ij]_2×2

So, the elements in a 2×2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂,

A = …… (1)

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

So, from (1)

A =

Question 12.

Construct a 2 × 2 matrix A = [a_jj] whose elements a_jj are given by :

Answer:

Let A = [a_ij]_2×2

So, the elements in a 2×2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂,

A = …… (1)

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

So, from (1)

A =

Question 13.

Construct a 2 × 2 matrix A = [a_jj] whose elements a_jj are given by :

Answer:

Let A = [a_ij]_2×2

So, the elements in a 2×2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂,

A = ……(1)

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

So, from (1)

A =

Question 14.

Construct a 2 × 2 matrix A = [a_jj] whose elements a_jj are given by :

Answer:

Let A = [a_ij]_2×2

So, the elements in a 2×2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂,

A = …… (1)

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

So, from (1)

A =

Question 15.

Construct a 2 × 2 matrix A = [a_jj] whose elements a_jj are given by :

a_ij = e^2ix sin xj

Answer:

Let A = [a_ij]_2×2

So, the elements in a 2×2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂,

A = …… (1)

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

So, from (1)

A =

Question 16.

Construct a 3×4 matrix A = [a_jj] whose elements a_jj are given by :

a_jj = i + j

Answer:

Let A = [a_ij]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃,a₂₄,a_31,a_32,a_33,a₃₄

A = ……(1)

a₁₁ = 1 + 1 = 2 a₁₂ = 1 + 2 = 3 a₁₃ = 1 + 3 = 4 a₁₄ = 1 + 4 = 5

a₂₁ = 2 + 1 = 3 a₂₂ = 2 + 2 = 4 a₂₃ = 2 + 3 = 5 a_{24 =} 2 + 4 = 6

a₃₁ = 3 + 1 = 4 a₃₂ = 3 + 2 = 5 a₃₃ = 3 + 3 = 6 a₃₄ = 3 + 4 = 7

So, from (1)

A =

Question 17.

Construct a 3×4 matrix A = [a_jj] whose elements a_jj are given by :

a_jj = i – j

Answer:

Let A = [a_ij]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃,a₂₄,a_31,a_32,a_33,a₃₄

A = …… (1)

a₁₁ = 1 – 1 = 0 a₁₂ = 1 – 2 = – 1 a₁₃ = 1 – 3 = – 2 a₁₄ = 1 – 4 = – 3

a₂₁ = 2 – 1 = 1 a₂₂ = 2 – 2 = 0 a₂₃ = 2 – 3 = – 1 a_{24 =} 2 – 4 = – 2

a₃₁ = 3 – 1 = 2 a₃₂ = 3 – 2 = 1 a₃₃ = 3 – 3 = 0 a₃₄ = 3 – 4 = – 1

So, from (1)

A =

Question 18.

Construct a 3×4 matrix A = [a_jj] whose elements a_jj are given by :

a_jj = 2i

Answer:

Let A = [a_ij]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃,a₂₄,a_31,a_32,a_33,a₃₄

A = ……(1)

a₁₁ = 2×1 = 2 a₁₂ = 2×1 = 2 a₁₃ = 2×1 = 2 a₁₄ = 2×1 = 2

a₂₁ = 2×2 = 4 a₂₂ = 2×2 = 4 a₂₃ = 2×2 = 4 a_{24 =} 2×2 = 4

a₃₁ = 2×3 = 6 a₃₂ = 2×3 = 6 a₃₃ = 2×3 = 6 a₃₄ = 2×3 = 6

So, from (1)

A =

Question 19.

Construct a 3×4 matrix A = [a_jj] whose elements a_jj are given by :

a_jj = j

Answer:

Let A = [a_ij]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃,a₂₄,a_31,a_32,a_33,a₃₄

A = ……(1)

a₁₁ = 1 a₁₂ = 2 a₁₃ = 3 a₁₄ = 4

a₂₁ = 1 a₂₂ = 2 a₂₃ = 3 a_{24 =} 4

a₃₁ = 1 a₃₂ = 2 a₃₃ = 3 a₃₄ = 4

So, from (1)

A =

Question 20.

Construct a 3×4 matrix A = [a_jj] whose elements a_jj are given by :

Answer:

Let A = [a_ij]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃,a₂₄,a_31,a_32,a_33,a₃₄

A = …… (1)

a₁₁ =

a₁₂ =

a₁₃ =

a₁₄ =

a₂₁ =

a₂₂ =

a₂₃ =

a_{24 =}

a₃₁ =

a₃₂ =

a₃₃ =

a₃₄ =

So, from (1)

A =

Question 21.

Construct a 4 × 3 matrix A = [a_jj] whose elements a_jj are given by :

Answer:

Let A = [a_ij]_4×3

So, the elements in a 4×3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃,a_31,a_32,a_33,a_41, a_42, a₄₃

A = ……(1)

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

a₃₁ =

a₃₂ =

a₃₃ =

a₄₁ =

a₄₂ =

a₄₃ =

So, from (1)

A =

Question 22.

Construct a 4 × 3 matrix A = [a_jj] whose elements a_jj are given by :

Answer:

Let A = [a_ij]_4×3

So, the elements in a 4×3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃,a_31,a_32,a_33,a_41, a_42, a₄₃

A = ……(1)

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

a₃₁ =

a₃₂ =

a₃₃ =

a₄₁ =

a₄₂ =

a₄₃ =

So, from (1)

A =

Question 23.

Construct a 4 × 3 matrix A = [a_jj] whose elements a_jj are given by :

a_jj = i

Answer:

Let A = [a_ij]_4×3

So, the elements in a 4×3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃,a_31,a_32,a_33,a_41, a_42, a₄₃

A = ……(1)

a₁₁ = 1

a₁₂ = 1

a₁₃ = 1

a₂₁ = 2

a₂₂ = 2

a₂₃ = 2

a₃₁ = 3

a₃₂ = 3

a₃₃ = 3

a₄₁ = 4

a₄₂ = 4

a₄₃ = 4

So, from (1)

A =

Question 24.

Find x, y, a and b if

Answer:

Given two matrices are equal.

We know that if two matrices are equal then the elements of each matrices are also equal.

∴3x + 4y = 2 …… (1)

and x – 2y = 4 …… (2)

and a + b = 5 ……(3)

2a – b = – 5 ……(4)

Multiplying equation (2) by 2 and adding to equation (1)

3x + 4y + 2x – 4y = 2 + 8

⇒ 5x = 10

⇒ x = 2

Now, Putting the value of x in equation (1)

3×2 + 4y = 2

⇒ 6 + 4y = 2

⇒ 4y = 2 – 6

⇒ 4y = – 4

⇒ y = – 1

Adding equation (3) and (4)

a + b + 2a – b = 5 + ( – 5)

⇒ 3a = 5 – 5 = 0

⇒ a = 0

Now, Putting the value of a in equation (3)

0 + b = 5

⇒ b = 5

∴ a = 0, b = 5, x = 2 and y = – 1

Question 25.

Find x, y, a and b if

Answer:

Given two matrices are equal.

We know that if two matrices are equal then the elements of each matrices are also equal.

∴2a + b = 4 ……(1)

And a – 2b = – 3 …… (2)

And 5c – d = 11 …… (3)

4c + 3d = 24 …… (4)

Multiplying equation (1) by 2 and adding to equation (2)

4a + 2b + a – 2b = 8 – 3

⇒ 5a = 5

⇒ a = 1

Now, Putting the value of a in equation (1)

2×1 + b = 4

⇒ 2 + b = 4

⇒ b = 4 – 2

⇒ b = 2

Multiplying equation (3) by 3 and adding to equation (4)

15c – 3d + 4c + 3d = 33 + 24

⇒ 19c = 57

⇒ c = 3

Now, Putting the value of c in equation (4)

4×3 + 3d = 24

⇒ 12 + 3d = 24

⇒ 3d = 24 – 12

⇒ 3d = 12

⇒ d = 4

∴ a = 1, b = 2, c = 3 and d = 4

Question 26.

Find the values of a, b, c and d from the following equations:

Answer:

Given two matrices are equal.

We know that if two matrices are equal then the elements of each matrices are also equal.

∴2a + b = 4 …… (1)

And a – 2b = – 3 ……(2)

And 5c – d = 11 …… (3)

4c + 3d = 24 …… (4)

Multiplying equation (1) by 2 and adding to equation (2)

4a + 2b + a – 2b = 8 – 3

⇒ 5a = 5

⇒ a = 1

Now, Putting the value of a in equation (1)

2×1 + b = 4

⇒ 2 + b = 4

⇒ b = 4 – 2

⇒ b = 2

Multiplying equation (3) by 3 and adding to equation (4)

15c – 3d + 4c + 3d = 33 + 24

⇒ 19c = 57

⇒ c = 3

Now, Putting the value of c in equation (4)

4×3 + 3d = 24

⇒ 12 + 3d = 24

⇒ 3d = 24 – 12

⇒ 3d = 12

⇒ d = 4

∴ a = 1, b = 2, c = 3 and d = 4

Question 27.

Find x, y and z so that A = B, where

Answer:

Given two matrices are equal as A = B.

We know that if two matrices are equal then the elements of each matrices are also equal.

∴x – 2 = y …… (1)

z = 3

And y + 2 = z …… (2)

2y = 6z

⇒ y = 3z ……(3)

Putting the value of z in equation (3)

∴ y = 3z = 3×3 = 9

Putting the value of y in equation (1)

x – 2 = 9

⇒ x – 2 = 9

⇒ x = 9 + 2

⇒ x = 11

∴ x = 11, y = 9, z = 3

Question 28.

If find x, y, z, ω.

Answer:

Given two matrices are equal.

We know that if two matrices are equal then the elements of each matrices are also equal.

∴x = 3 …… (1)

And 3x – y = 2 …… (2)

And 2x + z = 4 …… (3)

3y – ω = 7 …… (4)

Putting the value of x in equation (2)

3×3 – y = 2

⇒ 9 – y = 2

⇒ y = 9 – 2

⇒ y = 7

Now, putting the value of y in equation (4)

3×7 – ω = 7

⇒ 21 – ω = 7

⇒ ω = 21 – 7

⇒ ω = 14

Again, Putting the value of x in equation (3)

2×3 + z = 4

⇒ 6 + z = 4

⇒ z = 4 – 6

⇒ z = – 2

∴ x = 3, y = 7, z = – 2 and ω = 14

Question 29.

If find x, y, z,ω.

Answer:

Given two matrices are equal.

We know that if two matrices are equal then the elements of each matrices are also equal.

∴x = 3 ……(1)

And 3x – y = 2……(2)

And 2x + z = 4 …… (3)

3y – ω = 7 ……(4)

Putting the value of x in equation (2)

3×3 – y = 2

⇒ 9 – y = 2

⇒ y = 9 – 2

⇒ y = 7

Now, putting the value of y in equation (4)

3×7 – ω = 7

⇒ 21 – ω = 7

⇒ ω = 21 – 7

⇒ ω = 14

Again, Putting the value of x in equation (3)

2×3 + z = 4

⇒ 6 + z = 4

⇒ z = 4 – 6

⇒ z = – 2

∴ x = 3, y = 7, z = – 2 and ω = 14

Question 30.

If

Obtain the values of a, b, c, x, y and z.

Answer:

Given two matrices are equal.

We know that if two matrices are equal then the elements of each matrices are also equal.

∴x + 3 = 0

⇒ x = 0 – 3 = – 3 …… (1)

And z + 4 = 6

⇒ z = 6 – 4 = 2 …… (2)

And 2y – 7 = 3y – 2

⇒ 2y – 3y = – 2 + 7

⇒ – y = 5

⇒ y = – 5 …… (3)

4x + 6 = 2x …… (4)

a – 1 = – 3

⇒ a = – 3 + 1 = – 2 …… (5)

2c + 2 = 0

⇒ 2c = – 2

⇒ c = – 1 …… (6)

b – 3 = 2b + 4

⇒ b – 2b = 4 + 3

⇒ – b = 7

⇒ b = – 7 …… (7)

∴ x = – 3, y = – 5, z = 2 and a = – 2, b = – 7, c = – 1

Question 31.

If find the value of (x + y).

Answer:

Given two matrices are equal.

We know that if two matrices are equal then the elements of each matrices are also equal.

∴2x + 1 = x + 3 …… (1)

⇒ 2x – x = 3 – 1

⇒ x = 2

And 5x = 10 …… (2)

y² + 1 = 26…… (3)

⇒ y² = 26 – 1

⇒ y² = 25

⇒ y = 5 or – 5

∴ x = 2, y = 5 or – 5

∴ x + y = 2 + 5 = 7

Or x + y = 2 – 5 = – 3

Question 32.

If then find the values of x, y, z and ω.

Answer:

Given two matrices are equal.

We know that if two matrices are equal, then the elements of each matrix are also equal.

∴xy = 8 …… (1)

And ω = 4 ……(2)

And z + 6 = 0

⇒ z = – 6 ……(3)

x + y = 6 …… (4)

…… (from (1))

x = 2 or x = 4

∴ x = 2 or 4, z = – 6 and ω = 4

Question 33.

Give an example of

a row matrix which is also a column matrix

Answer:

As we know that order of a row matrix = 1× n

and order of a column matrix = m×1

So, order of a row as well as column matrix = 1×1

Therefore, required matrix A = [a_ij]_1×1

Question 34.

Give an example of

a diagonal matrix which is not scalar

Answer:

We know that a diagonal matrix has only a₁₁, a₂₂ and a₃₃ for a 3×3 matrix such that these elements are equal or different and all other entries 0 while scalar matrix has a₁₁ = a₂₂ = a₃₃ = k(say). So, a diagonal matrix which is not scalar must have a₁₁≠a₂₂≠a₃₃ for i≠j

Required matrix =

Question 35.

Give an example of

a triangular matrix.

Answer:

A triangular matrix is a square matrix,

A = [a_ij] such that a_ij = 0 for all i>j

Required matrix =

Question 36.

The sales figure of two car dealers during January 2013 showed that dealer A sold 5 deluxe, 3 premium and 4 standard cars, while dealer B sold 7 deluxe, 2 premium and 3 premium and 4 standard cars, while dealer B sold 7 deluxe, 2 premium and 3 standard cars. Total sales over the 2 month period of January – February revealed that dealer A sold 8 deluxe 7 premium and 6 standard cars. In the same 2 month period, dealer B sold 10 deluxe, 5 premium and 7 standard cars. Write 2 x 3 matrices summarizing sales data for January and 2 – month period for each dealer.

Answer:

By creating tables, we have

For January 2013

For January to February

Hence, we can form 2×3 matrices as

A = and B =

Question 37.

For what value of x and y are the following matrices equal?

Answer:

Given two matrices are equal .i.e, A = B.

We know that if two matrices are equal, then the elements of each matrices are also equal.

∴2x + 1 = x + 3

⇒ 2x – x = 3 – 1

⇒ x = 2 ……(1)

And 2y = y² + 2

⇒ y² – 2y + 2 = 0

⇒

⇒ (No real solutions) …… (2)

And y² – 5y = – 6

⇒ y² – 5y + 6 = 0

⇒ y² – 3y – 2y + 6 = 0

⇒ y(y – 3) – 2(y – 3) = 0

⇒ (y – 3)(y – 2) = 0

⇒ y = 3 or 2 …… (3)

∴ From the above equations we can say that A and B can’t be equal for any value of y.

Question 38.

Find the values of x and y if

Answer:

Given two matrices are equal .i.e, A = B.

We know that if two matrices are equal then the elements of each matrices are also equal.

∴x + 10 = 3x + 4

⇒ x – 3x = 4 – 10

⇒ – 2x = – 6

⇒ x = 3 …… (1)

And y² + 2y = 3

⇒ y² + 2y – 3 = 0

⇒ y² + 3y – y – 3 = 0

⇒ y(y + 3) – 1(y + 3) = 0

⇒ (y + 3)(y – 1) = 0

⇒ y = – 3 or 1 …… (2)

And y² – 5y = – 4

⇒ y² – 5y + 4 = 0

⇒ y² – 4y – y + 4 = 0

⇒ y(y – 4) – 1(y – 4) = 0

⇒ (y – 4)(y – 1) = 0

⇒ y = 4 or 1 ……(3)

∴ The common value is x = 3 and y = 1

Question 39.

Find the values of a and b if A = B, where

Answer:

Given two matrices are equal .i.e, A = B.

We know that if two matrices are equal then the elements of each matrices are also equal.

∴a + 4 = 2a + 2

⇒ a – 2a = 2 – 4

⇒ – a = – 2

⇒ a = 2 …… (1)

And 3b = b² + 2

⇒ b² – 3b + 2 = 0

⇒ b² – 2b – b + 2 = 0

⇒ b(b – 2) – 1(b – 2) = 0

⇒ (b – 2)(b – 1) = 0

⇒ b = 2 or 1 …… (2)

And – 6 = b² – 10

⇒ b² = – 10 + 6

⇒ b² = – 4

⇒ b = ±2i(No real solution) …… (3)

∴ a = 2, b = 2 or 1

Exercise 5.2

Question 1.

Compute the following sums:

Answer:

Hence, +=

Question 2.

Compute the following sums:

Answer:

Hence, +=

Question 3.

Let and Find each of the following :

i. 2A – 3B

ii. B – 4C

iii. 3A – C

iv. 3A – 2B + 3C

Answer:

(i) 2A==

= 3B==

= 2A-3B=-=

Hence, 2A-3B=

(ii) 4C==

B-4C=-

Hence, B-4C=

(iii) 3A=

= 3A-C=-

= =

Hence, 3A-C=

(iv) 3A==

= 2A==

= 3C==

= 3A-2B+3C=-+

= =

Hence, 3A-2B+3C=

Question 4.

If find

i. A + B and B + C

ii. 2B + 3A and 3C – 4B.

Answer:

i. A+B is not possible because matrix A is an order of 2x2 and Matrix B is an order of 2x3, So the Sum of the matrix is only possible when their order is same.

ii. B+C=+

Hence, B+C =

iii. 2B+3A also does not exist because the order of matrix B and matrix A is different , So we can not find the sum of these matrix.

iv. 3C-4B = -

=-=

Hence, 3C-4B =

Question 5.

Let and Compute 2A – 3B + 4C.

Answer:

2A==

= 3B==

= 4C==

= 2A-3B+4C

Hence, 2A-3B+4C=

Question 6.

If A = diag (2, -5, 9), B = diag (1, 1, -4) and C = diag (-6, 3, 4), find

i. A – 2B

ii. B + C – 2A

iii. 2A + 3B – 5C

Answer:

i. A-2B = diag -2diag

= diag-diag

= diag

Hence, A-2B=diag

ii. B+C-2A

=diag +diag-2diag

=diag+diag-diag

=diag

Hence, B+C-2A=diag

iii. 2A+3B-5C

=2diag+3diag-5diag

=diag+diag-diag

=diag

Hence, 2A+3B-5C=diag

Question 7.

Given the matrices

and

Verify that (A + B) + C = A + (B + C).

Answer:

L.H.S (A+B)+C

= (A+B)=+

= (A+B)+C =

= (A+B)+C

R.H.S A+(B+C)

= (B+C)+

= A+(B+C)+

Hence, L.H.S=R.H.S

Question 8.

Find matrices X and Y, if X + Y = and

Answer:

(X+Y)+(X-Y)=+

= 2X

= X

= X=

= (X+Y)-(X-Y)-

= 2Y

= Y

Hence, The value of X= and Y=

Question 9.

Find X, if and

Answer:

2X+Y=

= Put the Value of Y

= 2X+=

= 2X=-

= 2X=

= X=

Hence, The value of X=

Question 10.

Find matrices X and Y, if and

Answer:

2(2X-Y)+(X+2Y)=+

= 4X-2Y+X+2Y=

= 5X=

= X=

Now , We have to find Y , we will multiply the second equation by 2 and then Subtract from equation 1.

= (2X-Y)-2(X+2Y)=-

= 2X-Y-2X-4Y=

= -5Y=

= Y=

Hence, The Value of X= and Y=

Question 11.

If and find X and Y.

Answer:

(X-Y)+(X+Y)=

= X-Y+X+Y=

= 2X=

= X=

= (X-Y)-(X+Y)=

= X-Y-X-Y=

= -2Y=

= Y=

Hence, The Value of X=, and Y=

Question 12.

Find matrix A, if

Answer:

+A=

= A=-

= A=

Hence, A=

Question 13.

If find matrix C such that 5A + 3B + 2C is a null matrix.

Answer:

5A + 3B + 2C = 0

= +3+2=

= ++=

= …… (i)

= …… (ii)

= …… (iii)

= …… (iv)

= from equations (i),(ii),(iii),(iv), we get

Hence, C =

Question 14.

If find matrix X such that 2A + 3X = 5B.

Answer:

we have, 2A+3X=5B

= 3X= 5B-2A

= 3X=-

= 3X=

= X=

Hence, X

Question 15.

If and find the matrix C such that A + B + C is zero matrix.

Answer:

A+B+C=0.

= C=-A-B

= C=-

= C=

Hence, C=

Question 16.

Find x, y satisfying the matrix equations

Answer:

We know that, corresponding entries of equal matrices are equal.

= =

= -----(i) -----(ii)

-----(iii) -----(iv)

Now, Add the eq(iii) and eq(iv) and we get,

= X-Y+X+Y=3

= 2X=

= X=

Now, Put the Value of X in eq (iv) and we get,

= +Y=0

= Y=

Hence, Xand Y=

Question 17.

Find x, y satisfying the matrix equations.

[x y + 2 z - 3] + [y 4 5] = [4 9, 12]

Answer:

We know that , corresponding entries of equal matrices are equal.

= X+Y=4 ……(i)

= Y+6=9 ……(ii)

= Z+2=12 ……(iii)

On solving equation(i),(ii) and equation(iii) we get,

= Y=9-6

= Y=3

= Z=12-2

= Z=10

Put the value of Y in equation(i)…we get,

= X+3=4

= X=4-3

= X-1

Hence, X=1,Y=3 and Z=10

Question 18.

Find x, y satisfying the matrix equations

Answer:

To Find: Values of x and y

So,

2x + 3y -8 = 0

2x + 3y = 8…….(1)

x + 5y – 11 = 0

x + 5y = 11……(2)

Multiplying equation 2 by 2, we get

2x + 10 y = 22……..(3)

Subtracting equation 2 from 1, we get,

3y – 10 y = 11 – 22

-7y = -11

Putting this value in equation 1 we get,

Therefore, the values are,

Question 19.

If find x and y.

Answer:

= +=

We know that, corresponding entries of equal matrices are equal.

= Y+8=0 2X+1=5

= Y=-8 2X=5-1 = Y=-8 X=

= Y=-8 X=2

Hence, X=2 Y=-8

Question 20.

Find the value of λ, a non-zero scalar, if .

Answer:

We know that, corresponding entries of equal matrices are equal.

= +2=4

= =4-2

= =2

Since, -2=4

= =4+2

= +6

= =2

Hence, =2

Question 21.

Find a matrix X such that 2A + B + X = O, where

Answer:

we have 2A+B+X=0.

= X=-2A-B

= X=-

= X=

Hence, X=

Question 22.

If and then find the matrix X of order 3 x 2 such that 2A + 3X = 5B.

Answer:

2A+3X=5B.

= So, we can write as 3X=5B-2A

= 3X=-

= 3X=

= X=

Hence, X=

Question 23.

Find x, y, z and t, if

Answer:

= =

We know, Corresponding entries in equal matrices are equal.

= 3=+4 -----(i)

= 3-=4

= 2=4

= =2

Since, 3=6++------(ii)

= put the value of x in equation(ii)

= -=6+2

= 2 8

Therefore,

Since, +3-----(iii)

= =3

Therefore

Since, -1-----(iv)

= put the value of t in equation(iv)

= =3-1

= =2

Therefore,

Hence, ,

Question 24.

Find x, y, z and t, if

Answer:

= =7 =14

= =7-3 =14+4

= =4 =18

= =2 =9

Hence, =2,=9

Question 25.

If X and Y are 2 × 2 matrices, then solve the following matrix equations for X and Y.

Answer:

2X+3Y=-----(i)

= 3X+2Y-----(ii)

Multiply equation(i) by 3 and equation(ii) by 2, we get,

= 6X+9Y=

= 6X+4Y=

Subtract these equation then we get,

= 5Y=

= 5y=

= 5Y=

= Y=

Now, put the value of Y in equation (i)

= 2X+3

= 2X=-

= 2X=

= X=

Hence, Y= and X=

Question 26.

In a certain city there are 30 colleges. Each college has 15 peons, 6 clerks, 1 typist and 1 section officer. Express the given information as a column matrix. Using scalar multiplication, find the total number of posts of each kind in all the colleges.

Answer:

The Total number of post of each kind in 30 college. So,

= 30A=30

= 30A=

Question 27.

The monthly incomes of Aryan and Babban are in the ration 3: 4 and their monthly expenditures are in the ratio 5: 7. If each saves 15000 per month, find their monthly incomes using the matrix method. This problem reflects which value?

Answer:

Let us represent the situation through a matrix.

We will make two matrices: Income and Expenditure Matrices.

We know that Saving = Income – Expenditure.

Let the incomes of Aryan and Babban be 3x and 4x respectively and the expenditures be 5y and 7y respectively.

Income Matrix =

Expenditure Matrix =

Now, Saving =

Given: Saving = 15000 each

Therefore, we have,

So,

3 x – 5 y = 15000 ….(1)

4 x – 7 y = 15000 …..(2)

Solving equations 1 and 2, we get,

Multiplying eq(1) by 4 and eq(2) by 3 we get,

12 x – 20 y = 60000 ….(3)

12 x – 21 y = 45000 …..(4)

Eq(3) – Eq(4),

Y = 15000

Putting this value in eq(1) we get,

3 x – 4 × 15000 = 15000

3 x = 75000

X = 25000.

There monthly incomes are, 3 x = 3 × 15000 = 45000 and

4 x = 4 × 15000 = 60000.

Exercise 5.3

Question 1.

Compute the indicated products:

Answer:

Hence,

Question 2.

Compute the indicated products:

Answer:

Hence,

Question 3.

Compute the indicated products:

Answer:

Hence,

Question 4.

Show that AB ≠ BA in each of the following cases:

and

Answer:

given ,

......(1)

......(2)

From equation (1) and (2) we get

Question 5.

Show that AB ≠ BA in each of the following cases:

and

Answer:

given

......(1)

......(2)

From (1) and (2) AB BA

Question 6.

Show that AB ≠ BA in each of the following cases:

and

Answer:

Given

......(1)

......(2)

From equation (1) and (2) we get AB BA

Question 7.

Compute the products AB and BA whichever exists in each of the following cases:

and

Answer:

since the order of A is 22 and order of B is 23,

so AB is possible but BA is not the possible order of AB is 23.

Hence

And BA does not exits

Question 8.

Compute the products AB and BA whichever exists in each of the following cases:

and

Answer:

here,

since the order of A is 32 and order of B is 23,

AB and BA both exit and order of AB = 33 and order of BA = 22

Hence

Question 9.

Compute the products AB and BA whichever exists in each of the following cases:

and

Answer:

here

since the order of A is 14 and order of B is 4,

AB and BA both exit and order of AB = 1 and order of BA = 4

Hence

Question 10.

Compute the products AB and BA whichever exists in each of the following cases:

Answer:

Hence,

Question 11.

Show that AB ≠ BA in each of the following cases:

and

Answer:

......(1)

......(2)

From equation (1) and (2) ABBA

Question 12.

Show that AB ≠ BA in each of the following cases:

and

Answer:

......(1)

......(2)

From equation (1) and (2) ABBA

Question 13.

Evaluate the following:

Answer:

Hence,

Question 14.

Evaluate the following:

Answer:

Hence,

Question 15.

Evaluate the following:

Answer:

Hence,

Question 16.

If and then show that A² = B² = C² = I₂.

Answer:

given

......(1)

......(2)

......(3)

Hence,

From equation (1),(2) and (3),

Question 17.

If and find 3A² – 2B + I.

Answer:

given

Hence,

Question 18.

If prove that (A – 2I) (A – 3I) = O.

Answer:

given

(A-2I)(A-3I)

Hence,

Question 19.

If Show that and

Answer:

given

Hence,

Question 20.

If show that A² = O

Answer:

Given, A =

Hence,

Question 21.

If find A²

Answer:

Given,

|sin 22sin cos

Question 22.

If and show that AB = BA = O_{3 × 3}.

Answer:

Given, A B

AB ......(1)

BA ......(2)

From equation 1 and 2,

AB = BA

Question 23.

If and show that AB = BA = O_{3 × 3}.

Answer:

Given, A B

AB ......(1)

BA ......(2)

From equation 1 and 2,

AB = BA

Question 24.

If and show that AB = A and BA = B.

Answer:

Given, A B

AB = A

BA = B

Question 25.

Let and compute A² – B².

Answer:

Given, A B

......(1)

......(2)

Subtracting equation 2 from 1,

Hence,

Question 26.

For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A(BC).

and

Answer:

Given

......(1)

......(2)

From equation (1) and (2) we get,

Question 27.

For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A(BC).

and

Answer:

given

......(1)

......(2)

From equation (1) and (2)

Question 28.

For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A(B + C) = AB + AC.

and

Answer:

given

......(1)

......(2)

Using equation (1) and (2),

Question 29.

For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A(B + C) = AB + AC.

and

Answer:

given

......(1)

......(2)

From equation (1) and (2),

Question 30.

If and verify that A(B – C) = AB – AC

Answer:

Given,

A, B

A(B - C)

A(B - C) =

AB – AC

From equation (1) and (2),We get

Question 31.

Compute the elements a₄₃ and a₂₂ of the matrix:

Answer:

Given,

A =

Hence,

Question 32.

If and I is the identity matrix of order 3, show that A³ = pl + qA + rA².

Answer:

Given,

pqA + r

= p

Hence,

pqA + r

Hence proved.

Question 33.

If ω is a complex cube root of unity, show that

Answer:

Given, ω is a complex cube root of unity

Question 34.

If show that A² = A.

Answer:

Given:

Hence,

Question 35.

Show that the matrix is a root of the equation A² – 12A – I = 0.

Answer:

Given:

I is an identity matrix so

To show that

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for 12A, we get

So,

Substitute corresponding values from eqn(i) and (ii), we get

[as r_ij = a_ij + b_ij + c_ij]

Therefore,

Hence matrix A is the root of the given equation.

Question 36.

If show that A² = I₃

Answer:

Given,

Hence,

Question 37.

If find x.

Answer:

Given,

⇒

⇒ [2x + 1 + 2 + x + 3] = 0

⇒ 3x + 6 = 0

⇒ x = -2

Question 38.

If find x.

Answer:

Given,

By multiplication of matrices, we have

⇒ x = 13

⇒

Question 39.

If find x

Answer:

Given,

⇒

⇒ [(2x + 4)x + 4(x + 2)-1(2x + 4)] = 0

⇒ 2 + 4x + 4x + 8-2x-4 = 0

⇒ 2 + 6x + 4 = 0

⇒ 2 + 2x + 4x + 4 = 0

⇒ 2x(x + 1) + 4(x + 1) = 0

⇒ (x + 1)(2x + 4) = 0

⇒ x = -1 or x = -2

Hence, x = -1 or x = -2

Question 40.

If find x.

Answer:

Given:

We will multiply the 1×3 matrix with a 3×3 matrix, we get

, [as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now we will multiply these two matrices, we get

Therefore, the value of satisfying the given matrix condition is 2

Question 41.

If and then prove that A² – A + 2I = 0.

Answer:

Given: and

To prove A² – A + 2I = 0

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for 2I, we get

So,

Substitute corresponding values from eqn(i) and eqn(ii), we get

[as r_ij = a_ij + b_ij + c_ij]

Therefore,

Hence proved

Question 42.

If and then find λ so that A² = 5A + λI.

Answer:

Given: , and

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for 5A, we get

So,

Substitute corresponding values from eqn(i) and eqn(ii), we get

[as r_ij = a_ij + b_ij + c_ij]

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal,

Hence, and

So the value of λ so that is – 7

Question 43.

If show that A² – 5A + 7I₂ = 0.

Answer:

Given:

I₂ is an identity matrix of size 2, so

To show that

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for 5A, we get

Now,

So,

Substitute corresponding values from eqn(i), (ii) and (iii), we get

[as r_ij = a_ij + b_ij + c_ij]

Therefore,

Hence proved

Question 44.

If show that A² – 2A + 3I₂ = 0.

Answer:

Given:

I₂ is an identity matrix of size 2, so

To show that

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for 2A, we get

Now,

So,

Substitute corresponding values from eqn(i), (ii) and (iii), we get

[as r_ij = a_ij + b_ij + c_ij]

Therefore,

Hence proved

Question 45.

Show that the matrix satisfies the equation A³ – 4A² + A = 0.

Answer:

Given:

To show that

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for A³, we get

So,

Substitute corresponding values from eqn(i) and (ii), we get

[as r_ij = a_ij + b_ij + c_ij]

Therefore,

Hence matrix A satisfies the given equation.

Question 46.

If find A² – 5A – 14I.

Answer:

Given:

I is identity matrix so

To find

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for 5A, we get

So,

Substitute corresponding values from eqn(i) and (ii), we get

[as r_ij = a_ij + b_ij + c_ij]

Therefore,

Question 47.

If show that A² –5A + 7I = 0. Use this to find A⁴.

Answer:

Given:

I is identity matrix so

To show that

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for 5A, we get

So,

Substitute corresponding values from eqn(i) and (ii), we get

[as r_ij = a_ij + b_ij + c_ij]

Therefore,

Hence proved

We will find A⁴

Multiply both sides by A², we get

As multiplying by the identity matrix, I don’t change anything. Now will substitute the corresponding values we get

Hence this is the value of A⁴

Question 48.

If find k such that A² = kA – 2I₂.

Answer:

Given:

I₂ is an identity matrix of size 2, so

Also given,

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for kA, we get

So,

Substitute corresponding values from eqn(i) and (ii), we get

[as r_ij = a_ij + b_ij + c_ij],

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence,

Therefore, the value of k is 1

Question 49.

If find k such that A² – 8A + kI – 0.

Answer:

Given:

I is identity matrix, so

Also given,

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for 8A, we get

So,

Substitute corresponding values from eqn(i) and (ii), we get

[as r_ij = a_ij + b_ij + c_ij],

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence,

Therefore, the value of k is 7

Question 50.

If and f(x) = x² – 2x – 3, show that f(A) = 0.

Answer:

Given: and

To show that

Substitute in , we get

I is identity matrix, so

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for 2A, we get

Substitute corresponding values from eqn(ii) and (iii) in eqn(i), we get

[as r_ij = a_ij + b_ij + c_ij],

So,

Hence Proved

Question 51.

If and then find λ, μ so that A² = λ A + μ I

Answer:

Given: , and A² = λA + μI

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for λA, we get

But given, A² = λA + μI

Substitute corresponding values from eqn(i) and (ii), we get

[as r_ij = a_ij + b_ij + c_ij],

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence, λ + 0 = 4 ⇒ λ = 4

And also, 2λ + μ = 7

Substituting the obtained value of λ in the above equation, we get

2(4) + μ = 7 ⇒ 8 + μ = 7 ⇒ μ = – 1

Therefore, the value of λ and μ are 4 and – 1 respectively

Question 52.

Find the value of x for which the matrix product equal to an identity matrix.

Answer:

We know, is identity matrix of size 3.

So according to the given criteria

Now we will multiply the two matrices on LHS using the formula c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj, we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

So we get

So the value of x is

Question 53.

Solve the matrix equations:

Answer:

Now we will multiply the two first matrices on LHS using the formula c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj, we get

Again multiply these two LHS matrices, we get

⇒ x² – 2x – 15 = 0. This is form of quadratic equation, we will solve this by splitting the middle term, we get

⇒ x² – 5x + 3x – 15 = 0

⇒ x(x – 5) + 3(x – 5) = 0

⇒ (x – 5)(x + 3) = 0

⇒ x – 5 = 0 or x + 3 = 0

This gives, x = 5 or x = – 3 is the required solution of the matrices.

Question 54.

Solve the matrix equations:

Answer:

Now we will multiply the two first matrices on LHS using the formula c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj, we get

Again multiply these two LHS matrices, we get

⇒ 4 + 4x = 0 we will solve this linear equation, we get

⇒ 4x = – 4

This gives, x = – 1 is the required solution of the matrices.

Question 55.

Solve the matrix equations:

Answer:

Now we will multiply the two first matrices on LHS using the formula c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj, we get

Again multiply these two LHS matrices, we get

⇒ x² – 2x – 40 + 2x – 8 = 0

⇒ x² – 48 = 0. This is form of quadratic equation, we will solve this, we get

⇒ x² = 48 ⇒ x² = 16×3

This gives, x = ±4√3 is the required solution of the matrices.

Question 56.

Solve the matrix equations:

Answer:

Now we will multiply the two first matrices on LHS using the formula c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj, we get

Again multiply these two LHS matrices, we get

⇒ x² – 9x + 32x = 0

⇒ x² – 23x. This is form of quadratic equation, we will solve this, we get

⇒ x(x – 23) = 0

⇒ x = 0 or x – 23 = 0

This gives, x = 0 or x = 23 is the required solution of the matrices.

Question 57.

If compute A² – 4A + 3I₃.

Answer:

Given:

To find the value of A² – 4A + 3I₃

I₃ is an identity matrix of size 3, so

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for 4A, we get

So, Substitute corresponding values from eqn(i) and (ii)in equation A² – 4A + 3I₃, we get

[as r_ij = a_ij + b_ij + c_ij],

Hence the value of

Question 58.

If f(x) = x² – 2x, find f(A), where

Answer:

Given: and f(x) = x² – 2x

To find the value of f(A)

We will substitute x = A in the given equation we get

f(A) = A² – 2A……………..(i)

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for 2A, we get

So, Substitute corresponding values from eqn(i) and (ii) in equation f(A) = A² – 2A, we get

[as r_ij = a_ij + b_ij + c_ij],

Hence the value of

Question 59.

If f(x) = x³ + 4x² – x, find f(A), where

Answer:

Given: and f(x) = x³ + 4x² – x

To find the value of f(A)

We will substitute x = A in the given equation we get

f(A) = A³ + 4A² – A ……………..(i)

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for A³, we get

So, Substitute corresponding values from eqn(i) and (ii) in equation f(A) = A³ + 4A² – A, we get

[as r_ij = a_ij + b_ij + c_ij],

Hence the value of

Question 60.

If then show that A is a root of the polynomial f(x) = x³ – 6x² + 7x + 2.

Answer:

Given: and f(x) = x³ – 6x² + 7x + 2

To find the value of f(A)

We will substitute x = A in the given equation we get

f(A) = A³ – 6A² + 7A + 2I……………..(i)

Here I is identity matrix

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Now, we will find the matrix for A³, we get

So, Substitute corresponding values from eqn(i) and (ii) in equation f(A) = A³ – 6A² + 7A + 2I, we get

[as r_ij = a_ij + b_ij + c_ij],

Hence the A is the root of the given polynomial.

Question 61.

If then prove that A² – 4A – 5I = 0.

Answer:

Given:

To prove A² – 4A – 5I = 0

Here I is the identity matrix

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

So, Substitute corresponding values from eqn(i) in equation

A² – 4A – 5I, we get

[as r_ij = a_ij + b_ij + c_ij],

Hence the A² – 4A – 5I = 0 (Proved)

Question 62.

If show that A² – 7A + 10I₃ = 0.

Answer:

Given:

To prove A² –7A + 10I₃ = 0

Here I₃ is an identity matrix of size 3

Now, we will find the matrix for A², we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

So, Substitute corresponding values in equation

A² –7A + 10I₃, we get

[as r_ij = a_ij + b_ij + c_ij],

Hence the A² –7A + 10I₃ = 0 (Proved)

Question 63.

Without using the concept of the inverse of a matrix, find the matrix such that

Answer:

Given: =

Multiplying we get,

From above we can see that,

5 x – 7 z = – 16 …(1)

–2x + 3 z = 7 ….(2)

5 y – 7 u = – 6 …..(3)

–2y + 3 u = 2 ……(4)

Now we have to solve these equations to find values of x, y, z and u

Multiplying eq (1) by 2 and eq (2) by 5 and adding the equations we get,

10 x – 14 z + 10 x + 15 z = –32 + 35

Z = 3

Putting this value in eq(1) we get,

5 x – 21 = – 16

5 x = 5

X = 1

Now, multiplying eq(3) by 2 and eq(4) by 5 and adding we get,

10 y – 14 u + 10 y + 15 u = –12 + 10

u = –2

Putting value of u in equation (3) we get,

5 y + 14 = = – 6

5 y = – 20

Y = –4

Therefore now we have,

Question 64.

Find the matrix A such that

Answer:

We know that the two matrices are eligible for their product only when the number of columns of first matrix is equal to the number of rows of the second matrix.

So, is 2×2 matrix, and is 3×2 matrix

Now in order to get a 3×2 matrix as solution 2×2 matrix should be multiplied by 2×3 matrix. Hence matrix A is 2×3 matrix.

Let,

So the given question becomes,

Now we will multiply the two matrices on LHS, we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

To satisfy the above equality condition, corresponding entries of the matrices should be equal, i.e.,

d = 1, e = 0, f = 1

a + d = 3 ⇒ a + 1 = 3 ⇒ a = 2

b + e = 3 ⇒ b + 0 = 3 ⇒ b = 3

c + f = 5 ⇒ c + 1 = 5 ⇒ c = 4

Now substituting these values in matrix A, we get

is the matrix A.

Question 65.

Find the matrix A such that

Answer:

We know that the two matrices are eligible for their product only when the number of columns of first matrix is equal to the number of rows of the second matrix.

The matrix given on the RHS of the equation is a 2×3 matrix and the one given on the LHS of the equation is a 2×3 matrix.

Therefore, A has to be a 2×2 matrix.

Let,

So the given question becomes,

Now we will multiply the two matrices on LHS, we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

To satisfy the above equality condition, corresponding entries of the matrices should be equal, i.e.,

a + 4b = – 7, 2a + 5b = – 8, 3a + 6b = – 9

c + 4d = 2, 2c + 5d = 4, 3c + 6d = 6

Now, a + 4b = – 7 ⇒ a = – 7 – 4b…………(i)

∴ 2a + 5b = – 8

⇒ 2( – 7 – 4b) + 5b = – 8 (by substituting the value of a from eqn(i))

⇒ – 14 – 8b + 5b = – 8

⇒ 3b = – 14 + 8

⇒ b = – 2

Hence substitute the value of b in eqn(i), we get

a = – 7 – 4b

⇒ a = – 7 – 4( – 2) = – 7 + 8 = 1

⇒ a = 1

Now, c + 4d = 2 ⇒ c = 2 – 4d…………(ii)

∴ 2c + 5d = 4

⇒ 2(2 – 4d) + 5d = 4 by substituting the value of a from eqn(ii))

⇒ 4 – 8d + 5d = 4

⇒ 3d = 0

⇒ d = 0

Hence substitute the value of d in eqn(ii), we get

c = 2 – 4d

⇒ c = 2 – 4(0)

⇒ c = 2

Now substituting these values in matrix A, we get

is the matrix A.

Question 66.

Find the matrix A such that

Answer:

We know that the two matrices are eligible for their product only when the number of columns of first matrix is equal to the number of rows of the second matrix.

The matrix given on the RHS of the equation is a 3×3 matrix and the one given on the LHS of the equation is a 1×3 matrix.

Therefore, A has to be a 1×3 matrix.

Let,

So the given question becomes,

Now we will multiply the two matrices on LHS, we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

To satisfy the above equality condition, corresponding entries of the matrices should be equal, i.e.,

4a = – 4 ⇒ a = – 1

4b = 8 ⇒ b = 2

4c = 4⇒ c = 1

Now substituting these values in matrix A, we get

is the matrix A.

Question 67.

Find the matrix A such that

Answer:

We will multiply the first two matrices, on the LHS, we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

Again multiply the two matrices on the LHS, we get

is the matrix A.

Question 68.

Find the matrix A such that

Answer:

We know that the two matrices are eligible for their product only when the number of columns of first matrix is equal to the number of rows of the second matrix.

The matrix given on the RHS of the equation is a 3×3 matrix and the one given on the LHS of the equation is a 3×2 matrix.

Therefore, A has to be a 2×3 matrix.

Let,

So the given question becomes,

Now we will multiply the two matrices on LHS, we get

[as c_ij = a_i1b_1j + a_i2b_2j + … + a_inb_nj]

To satisfy the above equality condition, corresponding entries of the matrices should be equal, i.e.,

2a – d = – 1…(i),

2b – e = – 8….(ii),

2c – f = – 10…..(iii)

a = 1, b = – 2,c = – 5

– 3a + 4d = 9……..(iv),

– 3b + 4e = 22………(v),

– 3c + 4f = 15……..(vi)

Substitute the value of a in eqn(i), we get

2a – d = – 1 ⇒ 2(1) – d = – 1 ⇒ d = 3

Substitute the value of b in eqn(ii), we get

2b – e = – 8 ⇒ 2( – 2) – e = – 8 ⇒ – 4 – e = – 8 ⇒ e = 4

Substitute the value of c in eqn(iii), we get

2c – f = – 10 ⇒ 2( – 5) – f = – 10 ⇒ – 10 – f = – 10 ⇒ f = 0

Now substituting these values in matrix A, we get

is the matrix A.

Question 69.

Find the matrix A such that

Answer:

On multiplying A with 2 × 3 matrix we get 3 × 3 matrix

Therefore, A must be a matrix of order 3 × 2

Let A =

So we have,

a + 4 b = – 7 ….(1)

2 a + 5 b = – 8 …(2)

c + 4 d = 2 …….(3)

2 c + 5 d = 4 ..(4)

e + 4 f = 11 …(5)

2 e + 5 f = 10 …(6)

a = 1, b = –2, c = 2, d = 0, e = –5 and f = 4 on solving the above equations.

Hence, A =

Question 70.

Find a 2 × 2 matrix A such that

Answer:

Given A is a 2×2 matrix,

So let

Here I₂ is an identity matrix of size 2,

So the given equation becomes,

To satisfy the above equality condition, corresponding entries of the matrices should be equal, i.e.,

a + b = 6…….(i)

– 2a + 4b = 0 ⇒ 2a = 4b ⇒ a = 2b……..(ii)

c + d = 0 ⇒ c = – d……(iii)

– 2c + 4d = 6 ……(iv)

Substitute the values of eqn(ii) in eqn (i), we get

a + b = 6 ⇒ 2b + b = 6 ⇒ b = 2

So eqn(ii) becomes, a = 2b = 2(2) = 4⇒ a = 4

Substitute the values of eqn(iii) in eqn (iv), we get

– 2c + 4d = 6 ⇒ – 2( – d) + 4d = 6 ⇒ 2d + 4d = 6 ⇒ 6d = 6 ⇒ d = 1

So eqn(iii) becomes, c = – d⇒ c = – 1

Now substituting these values in matrix A, we get

is the matrix A.

Question 71.

If find A¹⁶.

Answer:

Given:

We will find A²,

Hence, A¹⁶ = (A²)⁸ = (0)⁸ = 0

Hence A¹⁶ is a nill matrix.

Question 72.

If and x² = – 1, then show that (A + B)² = A² + B².

Answer:

Given, and x² = –1.

We need to prove (A + B)² = A² + B².

Let us evaluate the LHS and the RHS one at a time.

To find the LHS, we will first calculate A + B.

We know (A + B)² = (A + B)(A + B).

(∵ x² = –1)

To find the RHS, we will first calculate A² and B².

We know A² = A × A.

(∵ x² = –1)

Similarly, we also have B² = B × B.

Now, the RHS is A² + B².

Thus, (A + B)² = A² + B².

Question 73.

If then verify that A² + A = (A + I), where I is the identity matrix.

Answer:

Given

We need to prove A² + A = A (A + I).

Let us evaluate the LHS and the RHS one at a time.

To find the LHS, we will first calculate A².

We know A² = A × A

Now, the LHS is A² + A.

To find the RHS, we will first calculate A + I.

Now, the RHS is A(A + I).

Thus, A² + A = A (A + I).

Question 74.

If then find A² – 5A – 14I. Hence, obtain A³.

Answer:

Given.

We need to find A² – 5A – 14I.

We know A² = A × A.

Now, we evaluate –5A = –5 × A.

Finally, matrix –14I = –14 × I.

The given expression is A² – 5A – 14I.

On multiplying both sides with matrix A, we get

A (A² – 5A – 14I) = O

⇒ A³ – 5A² – 14(A × I) = O

⇒ A³ – 5A² – 14A = O

⇒ A³ = 5A² + 14A

Thus, and .

Question 75.

If then show that P(x) P(y) = P(x + y) = P(y) P(x).

Answer:

Given.

We need to prove that P(x)P(y) = P(x + y) = P(y)P(x).

First, we will evaluate P(x)P(y).

Now, we will evaluate P(y)P(x).

Thus, P(x)P(y) = P(x + y) = P(y)P(x).

Question 76.

If and prove that

Answer:

Given and

We need to prove that.

First, we will evaluate PQ.

Now, we will evaluate QP.

Thus,

Question 77.

If find A² – 5A + 4I and hence find a matrix X such that A² – 5A + 4I + X = O.

Answer:

Given.

We need to find A² – 5A + 4I.

We know A² = A × A.

Now, we evaluate –5A = –5 × A.

Finally, matrix 4I = 4 × I.

The given expression is A² – 5A + 4I.

Given that A² – 5A + 4I + X = O

⇒ X = – (A² – 5A + 4I)

Thus, and

Question 78.

If prove that for all positive integers n.

Answer:

Given.

We need to prove that.

We will prove this result using the principle of mathematical induction.

Step 1: When n = 1, we have

Hence, the equation is true for n = 1.

Step 2: Let us assume the equation true for some n = k, where k is a positive integer.

To prove the given equation using mathematical induction, we have to show that.

We know A^k+1 = A^k × A.

Hence, the equation is true for n = k + 1 under the assumption that it is true for n = k.

Therefore, by the principle of mathematical induction, the equation is true for all positive integer values of n.

Thus, for all positive integers n.

Question 79.

If prove that for every positive integer n.

Answer:

Given.

We need to prove that.

We will prove this result using the principle of mathematical induction.

Step 1: When n = 1, we have

Hence, the equation is true for n = 1.

Step 2: Let us assume the equation true for some n = k, where k is a positive integer.

To prove the given equation using mathematical induction, we have to show that.

We know A^k+1 = A^k × A.

Hence, the equation is true for n = k + 1 under the assumption that it is true for n = k.

Therefore, by the principle of mathematical induction, the equation is true for all positive integer values of n.

Thus, for every positive integer n.

Question 80.

If then prove by principle of mathematical induction that for all n ∈ N.

Answer:

Given.

We need to prove that using the principle of mathematical induction.

Step 1: When n = 1, we have

Hence, the equation is true for n = 1.

Step 2: Let us assume the equation true for some n = k, where k is a positive integer.

To prove the given equation using mathematical induction, we have to show that.

We know A^k+1 = A^k × A.

However, we have i² = –1

Hence, the equation is true for n = k + 1 under the assumption that it is true for n = k.

Therefore, by the principle of mathematical induction, the equation is true for all positive integer values of n.

Thus, for all n ϵ N.

Question 81.

If prove that for all

Answer:

Given.

We need to prove that.

We will prove this result using the principle of mathematical induction.

Step 1: When n = 1, we have

Hence, the equation is true for n = 1.

Step 2: Let us assume the equation true for some n = k, where k is a positive integer.

To prove the given equation using mathematical induction, we have to show that.

We know A^k+1 = A^k × A.

We evaluate each value of this matrix independently.

(a) The value at index (1, 1)

(b) The value at index (1, 2)

(d) The value at index (2, 2)

So, the matrix A^k+1 is

Hence, the equation is true for n = k + 1 under the assumption that it is true for n = k.

Therefore, by the principle of mathematical induction, the equation is true for all positive integer values of n.

Thus, for all n ϵ N.

Question 82.

If then use the principle of mathematical induction to show that for every positive integer n.

Answer:

Given.

We need to prove that using the principle of mathematical induction.

Step 1: When n = 1, we have

Hence, the equation is true for n = 1.

Step 2: Let us assume the equation true for some n = k, where k is a positive integer.

To prove the given equation using mathematical induction, we have to show that.

We know A^k+1 = A^k × A.

Hence, the equation is true for n = k + 1 under the assumption that it is true for n = k.

Therefore, by the principle of mathematical induction, the equation is true for all positive integer values of n.

Thus, for every positive integer n.

Question 83.

If B, C are n rowed matrices and if A = B + C, BC = CB, C² = O, then show that for every n ϵ N, Aⁿ⁺¹ = Bⁿ(B + (n + 1)C).

Answer:

Given A = B + C, BC = CB and C² = O.

We need to prove that Aⁿ⁺¹ = Bⁿ(B + (n + 1)C).

We will prove this result using the principle of mathematical induction.

Step 1: When n = 1, we have Aⁿ⁺¹ = A¹⁺¹

⇒ Aⁿ⁺¹ = B¹(B + (1 + 1)C)

∴ Aⁿ⁺¹ = B(B + 2C)

For the given equation to be true for n = 1, Aⁿ⁺¹ must be equal to A².

It is given that A = B + C and we know A² = A × A.

⇒ A² = (B + C)(B + C)

⇒ A² = B(B + C) + C(B + C)

⇒ A² = B² + BC + CB + C²

However, BC = CB and C² = O.

⇒ A² = B² + CB + CB + O

⇒ A² = B² + 2CB

∴ A² = B(B + 2C)

Hence, Aⁿ⁺¹ = A²and the equation is true for n = 1.

Step 2: Let us assume the equation true for some n = k, where k is a positive integer.

⇒ A^k+1 = B^k(B + (k + 1)C)

To prove the given equation using mathematical induction, we have to show that A^k+2 = B^k+1(B + (k + 2)C).

We know A^k+2 = A^k+1 × A.

⇒ A^k+2 = [B^k(B + (k + 1)C)](B + C)

⇒ A^k+2 = [B^k+1 + (k + 1)B^kC)](B + C)

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^kC(B + C)

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^kCB + (k + 1)B^kC²

However, BC = CB and C² = O.

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^kBC + (k + 1)B^kO

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^k+1C + O

⇒ A^k+2 = B^k+1(B + C) + B^k+1[(k + 1)C]

⇒ A^k+2 = B^k+1[(B + C) + (k + 1)C]

⇒ A^k+2 = B^k+1[B + (1 + k + 1)C]

∴ A^k+2 = B^k+1[B + (k + 2)C]

Hence, the equation is true for n = k + 1 under the assumption that it is true for n = k.

Therefore, by the principle of mathematical induction, the equation is true for all positive integer values of n.

Thus, Aⁿ⁺¹ = Bⁿ(B + (n + 1)C) for every n ϵ N.

Question 84.

If A = diag(a b c), show that Aⁿ = diag(aⁿ bⁿ cⁿ) for all positive integers n.

Answer:

Given.

We need to prove that.

We will prove this result using the principle of mathematical induction.

Step 1: When n = 1, we have

Hence, the equation is true for n = 1.

Step 2: Let us assume the equation true for some n = k, where k is a positive integer.

To prove the given equation using mathematical induction, we have to show that.

We know A^k+1 = A^k × A.

∴ A^k+1 = diag(a^k+1 b^k+1 c^k+1)

Hence, the equation is true for n = k + 1 under the assumption that it is true for n = k.

Therefore, by the principle of mathematical induction, the equation is true for all positive integer values of n.

Thus, Aⁿ = diag(aⁿ bⁿ cⁿ) for all positive integers n.

Question 85.

If A is a square matrix, using mathematical induction prove that (A^T)ⁿ = (Aⁿ)^T for all n ϵ N.

Answer:

Given A is a square matrix.

We need to prove that (A^T)ⁿ = (Aⁿ)^T.

We will prove this result using the principle of mathematical induction.

Step 1: When n = 1, we have (A^T)¹ = A^T

∴ (A^T)¹ = (A¹)^T

Hence, the equation is true for n = 1.

Step 2: Let us assume the equation true for some n = k, where k is a positive integer.

⇒ (A^T)^k = (A^k)^T

To prove the given equation using mathematical induction, we have to show that (A^T)^k+1 = (A^k+1)^T.

We know (A^T)^k+1 = (A^T)^k × A^T.

⇒ (A^T)^k+1 = (A^k)^T × A^T

We have (AB)^T = B^TA^T.

⇒ (A^T)^k+1 = (AA^k)^T

⇒ (A^T)^k+1 = (A^1+k)^T

∴ (A^T)^k+1 = (A^k+1)^T

Hence, the equation is true for n = k + 1 under the assumption that it is true for n = k.

Therefore, by the principle of mathematical induction, the equation is true for all positive integer values of n.

Thus, (A^T)ⁿ = (Aⁿ)^T for all n ϵ N.

Question 86.

A matrix X has a + b rows and a + 2 columns while the matrix Y has b + 1 rows and a + 3 columns. Both matrices XY and YX exist. Find a and b. Can you say XY and YX are of the same type? Are they equal?

Answer:

X has a + b rows and a + 2 columns.

⇒ Order of X = (a + b) × (a + 2)

Y has b + 1 rows and a + 3 columns.

⇒ Order of Y = (b + 1) × (a + 3)

Recall that the product of two matrices A and B is defined only when the number of columns of A is equal to the number of rows of B.

It is given that the matrix XY exists.

⇒ Number of columns of X = Number of rows of Y

⇒ a + 2 = b + 1

∴ a = b – 1

The matrix YX also exists.

⇒ Number of columns of Y = Number of rows of X

⇒ a + 3 = a + b

∴ b = 3

We have a = b – 1

⇒ a = 3 – 1

∴ a = 2

Thus, a = 2 and b = 3.

Hence, order of X = 5 × 4 and order of Y = 4 × 5.

Order of XY = Number of rows of X × Number of columns of Y

⇒ Order of XY = 5 × 5

Order of YX = Number of rows of Y × Number of columns of X

⇒ Order of XY = 4 × 4

As the orders of the two matrices XY and YX are different, they are not of the same type and thus unequal.

Question 87.

Give examples of matrices

A and B such that AB ≠ BA.

Answer:

We need to find matrices A and B such that AB ≠ BA.

Let and.

First, we will find AB.

Now, we will find BA.

Thus, AB ≠ BA when and.

Question 88.

Give examples of matrices

A and B such that AB = O but A ≠ O, B ≠ O.

Answer:

We need to find matrices A and B such that AB = O but A ≠ O, B ≠ O.

Let and.

Now, we will find AB.

Thus, AB ≠ O when and.

Question 89.

Give examples of matrices

A and B such that AB = O but BA ≠ O.

Answer:

We need to find matrices A and B such that AB = O but BA ≠ O.

Let and.

First, we will find AB.

Now, we will find BA.

Thus, AB = O and BA ≠ O when and.

Question 90.

Give examples of matrices

A, B and C such that AB = AC but B ≠ C, A ≠ O.

Answer:

We need to find matrices A, B and C such that AB = AC but B ≠ C, A ≠ O.

Let, and.

First, we will find AB.

Now, we will find AC.

Thus, AB = AC but B ≠ C, A ≠ O when, and.

Question 91.

Let A and B be square matrices of the same order. Does (A + B)² = A² + 2AB + B² hold? If not, why?

Answer:

Given that A and B are square matrices of the same order.

We need to check if (A + B)² = A² + 2AB + B².

We know (A + B)² = (A + B)(A + B)

⇒ (A + B)² = A(A + B) + B(A + B)

∴ (A + B)² = A² + AB + BA + B²

For the equation (A + B)² = A² + 2AB + B² to hold, we need AB = BA that is the matrices A and B must satisfy the commutative property for multiplication.

However, here it is not mentioned that AB = BA.

Therefore, AB ≠ BA.

Thus, (A + B)² ≠ A² + 2AB + B².

Question 92.

If A and B are square matrices of the same order, explain, why in general

(i) (A + B)² ≠ A² + 2AB + B²

(ii) (A – B)² ≠ A² – 2AB + B²

(iii) (A + B)(A – B) = A² – B²

Answer:

(i) Given that A and B are square matrices of the same order.

We know (A + B)² = (A + B)(A + B)

⇒ (A + B)² = A(A + B) + B(A + B)

∴ (A + B)² = A² + AB + BA + B²

For the equation (A + B)² = A² + 2AB + B² to be valid, we need AB = BA.

As the multiplication of two matrices does not satisfy the commutative property in general, AB ≠ BA.

Thus, (A + B)² ≠ A² + 2AB + B².

(ii) Given that A and B are square matrices of the same order.

We know (A – B)² = (A – B)(A – B)

⇒ (A – B)² = A(A – B) – B(A – B)

∴ (A – B)² = A² – AB – BA + B²

For the equation (A – B)² = A² – 2AB + B² to be valid, we need AB = BA.

As the multiplication of two matrices does not satisfy the commutative property in general, AB ≠ BA.

Thus, (A – B)² ≠ A² – 2AB + B².

(iii) Given that A and B are square matrices of the same order.

We have (A + B)(A – B) = A(A – B) + B(A – B)

∴ (A + B)(A – B) = A² – AB + BA – B²

For the equation (A + B)(A – B) = A² – B² to be valid, we need AB = BA.

As the multiplication of two matrices does not satisfy the commutative property in general, AB ≠ BA.

Thus, (A + B)(A – B) ≠ A² – B².

Question 93.

Let A and B be square matrices of the order 3 × 3. Is (AB)² = A²B²? Give reasons.

Answer:

Given that A and B are square matrices of the order 3 × 3.

We know (AB)² = (AB)(AB)

⇒ (AB)² = A × B × A × B

⇒ (AB)² = A(BA)B

If the matrices A and B satisfy the commutative property for multiplication, then AB = BA.

We found (AB)² = A(BA)B.

Hence, when AB = BA, we have (AB)² = A(AB)B.

⇒ (AB)² = A × A × B × B

⇒ (AB)² = A²B²

Therefore, (AB)² = A²B² holds only when AB = BA.

Thus, (AB)² ≠ A²B² unless the matrices A and B satisfy the commutative property for multiplication.

Question 94.

If A and B are square matrices of the same order such that AB = BA, then show that (A + B)² = A² + 2AB + B².

Answer:

Given that A and B are square matrices of the same order such that AB = BA.

We need to prove that (A + B)² = A² + 2AB + B².

We know (A + B)² = (A + B)(A + B)

⇒ (A + B)² = A(A + B) + B(A + B)

⇒ (A + B)² = A² + AB + BA + B²

However, here it is mentioned that AB = BA.

⇒ (A + B)² = A² + AB + AB + B²

∴ (A + B)² = A² + 2AB + B²

Thus, (A + B)² = A² + 2AB + B² when AB = BA.

Question 95.

Let and

Verify that AB = AC though B ≠ C, A ≠ 0.

Answer:

Given, and.

We need to verify that AB = AC.

Let us evaluate the LHS and the RHS one at a time.

We will first calculate AB.

Now, the RHS is AC.

Thus, AB = AC even though B ≠ C and A ≠ O.

Question 96.

Three shopkeepers, A, B and C go to a store to buy stationary. A purchases 12 dozen notebooks, 5 dozen pens and 6 dozen pencils. B purchases 10 dozen notebooks, 6 dozen pens and 7 dozen pencils. C purchases 11 dozen notebooks, 13 dozen pens and 8 dozen pencils. A notebook costs 40 paise, a pen costs Rs 1.25 and a pencil costs 35 paise. Use matrix multiplication to calculate each individual’s bill.

Answer:

Given the purchase details of three shopkeepers A, B and C.

A: 12 dozen notebooks, 5 dozen pens and 6 dozen pencils

B: 10 dozen notebooks, 6 dozen pens and 7 dozen pencils

C: 11 dozen notebooks, 13 dozen pens and 8 dozen pencils

Hence, the items purchased by A, B and C can be represented in matrix form with rows denoting the shopkeepers and columns denoting the number of dozens of items as –

The price of each of the items is also given.

Cost of one notebook = 40 paise

⇒ Cost of one dozen notebooks = 12 × 40 paise

⇒ Cost of one dozen notebooks = 480 paise

∴ Cost of one dozen notebooks = Rs 4.80

Cost of one pen = Rs 1.25

⇒ Cost of one dozen pens = 12 × Rs 1.25

∴ Cost of one dozen pens = Rs 15

Cost of one pencil = 35 paise

⇒ Cost of one dozen notebooks = 12 × 35 paise

⇒ Cost of one dozen notebooks = 420 paise

∴ Cost of one dozen notebooks = Rs 4.20

Hence, the cost of purchasing one dozen of the items can be represented in matrix form with each row corresponding to an item as –

Now, the individual bill for each shopkeeper can be found by taking the product of the matrices X and Y.

Thus, the bills of shopkeepers A, B and C are Rs 157.80, Rs 167.40 and Rs 281.40 respectively.

Question 97.

The cooperative stores of a particular school has 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are Rs 8.30, Rs 3.45 and Rs 4.50 each respectively. Find the total amount the store will receive from selling all the items.

Answer:

Given the details of stock of various types of books.

Physics: 10 dozen books

Chemistry: 8 dozen books

Mathematics: 5 dozen books

Hence, the number of dozens of books available in the store can be represented in matrix form with each column corresponding to a different subject as -

The price of each of the items is also given.

Cost of one physics book = Rs 8.30

⇒ Cost of one dozen physics books = 12 × Rs 8.30

∴ Cost of one dozen physics books = Rs 99.60

Cost of one chemistry book = Rs 3.45

⇒ Cost of one dozen chemistry books = 12 × Rs 3.45

∴ Cost of one dozen chemistry books = Rs 41.40

Cost of one mathematics book = Rs 4.50

⇒ Cost of one dozen mathematics books = 12 × Rs 4.50

∴ Cost of one dozen mathematics books = Rs 54

Hence, the cost of purchasing a dozen books of each subject can be represented in matrix form with each row corresponding to a different subject as –

Now, the amount received by the store upon selling all the available books can be found by taking the product of the matrices X and Y.

⇒ XY = [10 × 99.60 + 8 × 41.40 + 5 × 54]

⇒ XY = [996 + 331.20 + 270]

∴ XY = [1597.20]

Thus, the total amount the store will receive from selling all the items is Rs 1597.20.

Question 98.

In a legislative assembly election, a political group hired a public relations firm to promote its candidates in three ways; telephone, house calls and letters. The cost per contact (in paise) is given matrix A as

The number of contacts of each type made in two cities X and Y is given in matrix B as

Find the total amount spent by the group in the two cities X and Y.

Answer:

Given matrix A containing the costs per contact in paisa for different types of contacting the public.

Matrix B contains the number of contacts of each type made in two cities X and Y.

Now, the total amount spent by the political group in the two cities for contacting the public can be obtained by taking the product of the matrices B and A.

Amount spent in City X = 340000 paisa = Rs 3400

Amount spent in City Y = 720000 paisa = Rs 7200

Total amount spent = Rs (3400 + 7200) = Rs 10600

Thus, the total amount spent by the party in both the cities is Rs 10600 with Rs 3400 spent in city X and Rs 7200 spent in city Y.

Question 99.

A trust fund has Rs 30000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and the second bond pays 7% per year. Using matrix multiplication, determine how to divide Rs 30000 among the two types of bonds if the trust fund must obtain an annual total interest of (i) Rs 1800 and (ii) Rs 2000.

Answer:

Given that Rs 30000 must be invested into two types of bonds with 5% and 7% interest rates.

Let Rs x be invested in bonds of the first type.

Thus, Rs (30000 – x) will be invested in the other type.

Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as -

(i) Annual interest obtained is Rs 1800.

We know the formula to calculate the interest on a principal of Rs P at a rate R% per annum for t years is given by,

Here, the time is one year and thus T = 1.

Hence, the interest obtained after one year can be expressed in matrix representation as -

⇒ 5x + 7(30000 – x) = 1800 × 100

⇒ 5x + 210000 – 7x = 180000

⇒ –2x = 180000 – 210000

⇒ –2x = –30000

∴ x = 15000

Amount invested in the first bond = x = Rs 15000

Amount invested in the second bond = 30000 – x

⇒ Amount invested in the second bond = 30000 – 15000

∴ Amount invested in the second bond = Rs 15000

Thus, the trust has to invest Rs 15000 each in both the bonds in order to obtain an annual interest of Rs 1800.

(ii) Annual interest obtained is Rs 2000.

As in the previous case, the interest obtained after one year can be expressed in matrix representation as -

⇒ 5x + 7(30000 – x) = 2000 × 100

⇒ 5x + 210000 – 7x = 200000

⇒ –2x = 200000 – 210000

⇒ –2x = –10000

∴ x = 5000

Amount invested in the first bond = x = Rs 5000

Amount invested in the second bond = 30000 – x

⇒ Amount invested in the second bond = 30000 – 5000

∴ Amount invested in the second bond = Rs 25000

Thus, the trust has to invest Rs 5000 in the first bond and Rs 25000 in the second bond in order to obtain an annual interest of Rs 2000.

Question 100.

75. To promote making of toilets for women, an organization tried to generate awareness through (i) house calls (ii) letters, and (iii) announcements. The cost for each mode per attempt is given below :

(i) ₹ 50

(ii) ₹ 20

(iii) ₹ 40

The number of attempts made in three villages X, Y and Z are given below :

Find the total cost incurred by the organization for three villages separately, using matrices.

Answer:

Given costs in Rs for making different types of attempts.

Cost for one house call = Rs 50

Cost for one letter = Rs 20

Cost for one announcement = Rs 40

Hence, the costs per contact in Rs for different types of contacting the people can be expressed in matrix form as -

Let matrix B contain the number of attempts of each type made in the three villages X, Y and Z.

From the given information, the number of attempts made can be expressed in the matrix form as -

Now, the total cost incurred by the organization in the three villages for creating awareness can be obtained by taking the product of the matrices B and A.

Cost incurred in village X = Rs 30000

Cost incurred in village Y = Rs 23000

Cost incurred in village Z = Rs 39000

Thus, the cost incurred by the organization in villages X, Y and Z is Rs 30000, Rs 23000 and Rs 39000 respectively.

Question 101.

There are 2 families A and B. There are 4 men, 6 women and 2 children in family A, and 2 men, 2 women and 4 children in family B. The recommended daily amount of calories is 2400 for men, 1990 for women, 1800 for children and 45 grams of proteins for men, 55 grams for women and 33 grams for children. Represent the above information using a matrix. Using matrix multiplication, calculate the total requirement of calories and proteins for each of the two families. What awareness can you create among people about the planned diet from this question?

Answer:

Given the details of two families A and B.

A: 4 men, 6 women and 2 children

B: 2 men, 2 women and 4 children

Hence, the number of people in both the families A and B can be represented in matrix form with rows denoting the family and columns denoting the number of people of each type as -

The calorie and protein requirements for different types of people are also given.

Men: 2400 calories, 45gm proteins

Women: 1900 calories, 55gm proteins

Children: 1800 calories, 33gm proteins

Hence, the required calories and proteins can be represented in matrix form with each row corresponding to different type of people as –

Now, the required number of calories and proteins for each of the two families can be obtained by taking the product of the matrices X and Y.

Thus, the requirement of calories and protein is as follows –

Family A: 24600 calories and 576 grams protein

Family B: 15800 calories and 332 grams protein

It can be said that a balanced diet with proper amounts of calories and protein must be consumed by the people of all ages in order to lead a healthy life.

Question 102.

In a parliament election, a political party hired a public relations firm to promote its candidates in three ways – telephone, house calls and letters. The cost per contact (in paisa) is given in matrix A as

The number of contacts of each type made in two cities X and Y is given in matrix B as

Find the total amount spent by the party in the two cities.

What should one consider before casting his/her vote – party’s promotional activity or their social activities?

Answer:

Given matrix A contains the costs per contact in paisa for different types of contacting the public.

Matrix B contains the number of contacts of each type made in two cities X and Y.

Now, the total amount spent by the political party in the two cities for contacting the public can be obtained by taking the product of the matrices B and A.

Amount spent in City X = 990000 paisa = Rs 9900

Amount spent in City Y = 2120000 paisa = Rs 21200

Total amount spent = Rs (9900 + 21200) = Rs 31110

Thus, the total amount spent by the party in both the cities is Rs 31110 with Rs 9990 spent in city X and Rs 21200 spent in city Y.

One must surely consider the party’s social activities instead of their promotional activities before casting his/her vote.

Question 103.

The monthly incomes of Aryan and Babbar are in the ratio 3:4 and their monthly expenditures are in the ratio 5:7. If each saves Rs 15000 per month, find their monthly incomes using matrix method. This problem reflects which value?

Answer:

Let the monthly incomes of Aryan and Babbar be 3x and 4x respectively.

Let their monthly expenditures be 5y and 7y respectively.

Given that both of them save Rs 15000 per month.

We know that the savings is the difference between the income and the expenditure.

Thus, we have two equations –

3x – 5y = 15000

4x – 7y = 15000

Recall that the solution to the system of equations that can be written in the form AX = B is given by X = A^-1B.

Here,

We know the inverse of a matrix is given by

|A| = (3)(–7) – (4)(–5) = –21 + 20 = –1

We have X = A^–1B.

Monthly income of Aryan = 3x = 3 × Rs 30000 = Rs 90000

Monthly income of Babbar = 4x = 4 × Rs 30000 = Rs 120000

Thus, the monthly incomes of Aryan and Babbar are Rs 90000 and Rs 120000 respectively.

This problem tells us that savings are important and our income must not be spent wastefully.

Question 104.

A trust invested some money in two types of bonds. The first bond pays 10% interest and the second bond pays 12%. The trust received Rs 2800 as interest. However, if trust had interchanged money in bonds, they would have got Rs 100 less as interest. Using matrix method, find the amount invested by the trust.

Answer:

Given that some amount is invested into two types of bonds with 10% and 12% interest rates.

Let the amount invested in bonds of the first type and the second type be Rs x and Rs y respectively.

Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as -

The annual interest obtained is Rs 2800.

We know the formula to calculate the interest on a principal of Rs P at a rate R% per annum for t years is given by,

Here, the time is one year and thus T = 1.

Hence, the interest obtained after one year can be expressed in matrix representation as -

⇒ 10x + 12y = 2800 × 100

⇒ 10x + 12y = 280000

∴ 5x + 6y = 140000 …… (1)

However, on reversing the invested amounts, the interest received is Rs 100 less than the earlier value (Rs 2800).

Now, the amount invested in the second bond is Rs x and that in the first bond is Rs y with the annual interest obtained being Rs 2700.

Hence, the interest obtained by exchanging the invested amount of the two bonds after one year can be expressed in matrix representation as –

⇒ 10y + 12x = 2700 × 100

⇒ 12x + 10y = 270000

∴ 6x + 5y = 135000 …… (2)

Recall that the solution to the system of equations that can be written in the form AX = B is given by X = A^–1B.

Here,

We know the inverse of a matrix is given by

|A| = (5)(5) – (6)(6) = 25 – 36 = –11

We have X = A^–1B.

Amount invested in the first bond = x = Rs 10000

Amount invested in the second bond = y = Rs 15000

Thus, the trust invested Rs 10000 in the first bond and Rs 15000 in the second bond.

Exercise 5.4

Question 1.

Let and verify that

(2A)^T = 2A^T

Answer:

Given,

and

Put the value of A

L.H.S = R.H.S

Hence verified.

Question 2.

Let and verify that

(A + B)^T = A^T + B^T

Answer:

and

L.H.S = R.H.S

Hence proved.

Question 3.

Let and verify that

(A – B)^T = A^T - B^T

Answer:

and

L.H.S = R.H.S

Question 4.

Let and verify that

(AB)^T = B^TA^T

Answer:

and

So,

Question 5.

If and B = [1 0 4], verify that (AB)^T = B^TA^T.

Answer:

Given,

L.H.S = R.H.S

So,

Question 6.

Let and Find A^T, B^T and verify that

(A + B)^T = A^T + B^T

Answer:

Given,

L.H.S = R.H.S

So,

Question 7.

Let and Find A^T, B^T and verify that

(AB)^T = B^T A^T

Answer:

Given,

L.H.S = R.H.S

So,

Question 8.

Let and Find A^T, B^T and verify that

(2A)^T = 2A^T

Answer:

Given,

L.H.S = R.H.S

So,

Question 9.

If B= [1 3 -6], verify that (AB)^T = B^TA^T.

Answer:

Given,

L.H.S = R.H.S

So,

Question 10.

If find (AB)^T .

Answer:

Given,

So,

Question 11.

For two matrices A and B, verify that (AB)^T = B^TA^T.

Answer:

Given,

L.H.S = R.H.S

So,

Question 12.

For the matrices, A and B, verify that (AB)^T = B^TA^T, where

Answer:

Given,

L.H.S = R.H.S

So,

Question 13.

If and find A^T – B^T.

Answer:

Given,

Transpose matrix of B,

Question 14.

If then verify that A^TA = I₂.

Answer:

Given,

Hence verified

Question 15.

If verify that A^TA = I₂.

Answer:

Given,

Hence,

Question 16.

If denote the direction cosines of three mutually perpendicular vectors in space, prove that AA^T = I, where

Answer:

Given,

are direction cosines of three mutually perpendicular vectors

And given,

Hence,

Exercise 5.5

Question 1.

If prove that A – A^T is a skew-symmetric matrix.

Answer:

Given,

…(i)

…(ii)

From (i) and (ii) we can see that

a skew-symmetric matrix is a square matrix whose transpose equal to its negative, that is,

X = – X^T

So, A – A^T is a skew-symmetric.

Question 2.

If show that A – A^T is a skew-symmetric matrix.

Answer:

Given,

…(i)

…(ii)

From (i) and (ii) we can see that

a skew-symmetric matrix is a square matrix whose transpose equals its negative, that is,

X = – X^T

So, A – A^T is a skew-symmetric matrix.

Question 3.

If the matrix is a symmetric matrix, find x, y, z and t.

Answer:

Given,

is a symmetric matrix.

We know that is a symmetric matrix if

So,

Hence,

X=4, y=2, t=-3 and z can have any value.

Question 4.

Let Find matrices X and Y such that X + Y = A, where X is a symmetric and Y is a skew symmetric matrix.

Answer:

Given, Then

Now,

X is a symmetric matrix.

Now,

Y is a skew symmetric matrix.

And,

Hence, X+Y=A

Question 5.

Express the matrix as the sum of a symmetric and a skew-symmetric matrix.

Answer:

Given, Then

X is a symmetric matrix.

Y is a skew symmetric matrix.

Now,

Hence, X + Y = A.

Question 6.

Define a symmetric matrix. Prove that for is a symmetric matrix where A^T is the transpose of A.

Answer:

A square matrix ‘A’ is called a symmetric matrix, if A = A^T.

Here,

From equation (i) and (ii),

So, A + A^T is a symmetric matrix.

Question 7.

Express the matrix as the sum of a symmetric and a skew-symmetric matrix.

Answer:

Given,

Let,

Now,

Hence, X is a symmetric matrix.

Now let,

Now,

Y is a skew symmetric.

Now,

X+Y = A

Question 8.

Express the matrix as the sum of a symmetric and skew-symmetric matrix and verify your result.

Answer:

Given, Then,

Let,

X is a symmetric matrix.

And,

Y is a skew symmetric matrix.

Very Short Answer

Question 1.

If A is an m × n matrix and B is n × p matrix does AB exist? If yes, write its order.

Answer:

Given: A = m × n matrix and B = n × p matrix

∴ the product AB is defined and the size of the product matrix AB is m × p.

Question 2.

If and . Write the order of AB and BA.

Answer:

Given:

In matrix A, there are 2 rows and 3 columns.

∴ A is a 2 × 3 matrix

In matrix B, there are 3 rows and 2 columns.

∴ B is a 3 × 2 matrix

So, the product matrix AB will be

∴ the order of AB is 2 × 2 matrix

and the order of product matrix BA will be

∴ the order of BA is 3 × 3 matrix.

Question 3.

If and , write AB.

Answer:

Given:

So, AB will be

Question 4.

If , write AA^T.

Answer:

Given:

Now, firstly we find the A^T

So, the product AA^T will be

Question 5.

Give an example of two non-zero 2 × 2 matrices A and B such that AB = O.

Answer:

Example 1:

Let and are the two non – zero matrices

Now, we will check that AB = 0 or not

Hence, and are the two non – zero matrices such that AB = 0

Example 2:

Let and are the two non – zero matrices

Now, we will check that AB = 0 or not

Hence, and are the two non – zero matrices such that AB = 0

Question 6.

If , find A + A^T.

Answer:

Given:

To find: A + A^T

Firstly, we find the A^T

So,

Question 7.

If , write A².

Answer:

Given:

To find: A²

Question 8.

If , find x satisfying when A + A^T = I.

Answer:

Given:

To find: x

Firstly, we find the A^T

So,

It is given that A + A^T = I when

So,

Comparing both the matrices, we get

2cosx = 1

Question 9.

If , find AA^T.

Answer:

Given:

To find: AA^T

Firstly, we find the A^T

So,

[∵ cos²x + sin²x = 1]

⇒ AA^T = I

Question 10.

If , where I is 2 × 2 unit matrix. Find x and y.

Answer:

Here, it is given that I is a 2 × 2 unit matrix

So,

So, given equation becomes

Comparing the matrices, we get

1 + 2x = 1 …(i)

and y + 2 = 0 …(ii)

Solving eq. (i), we get

1 + 2x = 1

⇒ 2x = 0

⇒ x = 0

Solving eq. (ii), we get

y + 2 = 0

⇒ y = -2

Hence, the value of x = 0 and y = -2

Question 11.

If , satisfies the matrix equation A² = kA, write the value of k.

Answer:

Given:

and it satisfies the matrix equation A² = kA

Firstly, we find the A²

= 2 A

∴ k = 2

Hence, the value of k = 2

Question 12.

If satisfies A⁴ = λA, then write the value of λ.

Answer:

Given:

and it satisfies the matrix equation A⁴ = λA

Firstly, we find the A⁴

= 2 A

So, A⁴ = A² × A²

⇒ A⁴ = 2A × 2A

⇒ A⁴ = 4A²

⇒ A⁴ = 4 × 2A [∵A² = 2A]

⇒ A⁴ = 8A

∴ λ = 8

Hence, the value of λ = 8

Question 13.

If , find A².

Answer:

Now, we have to find the A²

⇒ A² = I

Question 14.

If , find A³.

Answer:

Now, we have to find the A³

Now, we will find A³

⇒ A³ = A

Question 15.

If , find A⁴.

Answer:

Given:

To find: A⁴

A⁴ = A × A × A × A

⇒ A⁴ = A² × A²

So,

⇒ A² = 9I

∵ A⁴ = A² × A²

= 9I × 9I

= 81 I²

= 81 I [∵ I² = I]

Question 16.

If [x 2], find x.

Answer:

Given:

Here, we have to find the x

Solving the given matrix, we get

⇒ [3x + 8] = 2

⇒ 3x = 2 – 8

⇒ 3x = -6

⇒ x = -2

Question 17.

If A = [a_ij] is a 2 × 2 matrix such that a_ij = i + 2j, write A.

Answer:

Given: A = [a_ij] is a 2 × 2 matrix

…(i)

Given that a_ij = i + 2j

So, a₁₁ = 1 + 2 × 1 = 1 + 2 = 3

a₁₂ = 1 + 2 × 2 = 1 + 4 = 5

a₂₁ = 2 + 2 × 1 = 2 + 2 = 4

a₂₂ = 2 + 2 × 2 = 2 + 4 = 6

Putting the values in eq. (i), we get

Question 18.

Write matrix A satisfying .

Answer:

Given:

To find: A

Let

Solving for matrix A, we get

Comparing the values, we get

a + 2 = 3 …(i)

b + 3 = -6 …(ii)

c – 1 = -3 …(iii)

and d + 4 = 8 …(iv)

Solving eq. (i), we get

a + 2 = 3

⇒ a = 1

Solving eq. (ii), we get

b + 3 = -6

⇒ b = -6 – 3

⇒ b = -9

Solving eq. (iii), we get

c – 1 = -3

⇒ c = -3 + 1

⇒ c = -2

Solving eq. (iv), we get

d + 4 = 8

⇒ d = 8 – 4

⇒ d = 4

Putting the value of a, b , c and d to get the matrix A, we get

Question 19.

If A = [a_ij] is a square matrix such that a_ij = i² – j², then write whether A is symmetric or skew-symmetric.

Answer:

Given: A = [a_ij] is a square matrix such that a_ij = i² – j²

Suppose A is a 2 × 2 square matrix i.e.

Here,

a_ij = i² – j²

So, a₁₂ = (1)² – (2)² = 1 – 4 = - 3

and a₂₁ = (2)² – (1)² = 4 – 1 = 3

For diagonal elements, i = j, we have

a₁₁ = (1)² – (1)² = 0

and a₂₂ = (2)² – (2)² = 0

So, Matrix A becomes

Now, we have to check A is symmetric or skew – symmetric.

We know that, if a matrix is symmetric then A^T = A

and if a matrix is skew – symmetric then A^T = -A

So, firstly we find the A^T

So,

⇒ A^T = - A

∴ A is a skew – symmetric matrix.

Question 20.

For any square matrix write whether AA^T is symmetric or skew-symmetric.

Answer:

Here, we have any square matrix

To Find: AA^T is symmetric or skew – symmetric

Proof: Firstly, we take the transpose of AA^T, so we get

(AA^T)^T = (A^T)^T A^T [∵ (AB)^T = B^TA^T]

⇒ (AA^T)^T = AA^T [∵ (A^T)^T = A]

∴ AA^T is a symmetric matrix

Question 21.

If A = [a_ij] is a skew-symmetric matrix, then write the value of .

Answer:

Given: A = [a_ij] is a skew – symmetric matrix

⇒ a_ij = - a_ji …(i)

[for all values of i, j]

For diagonal elements,

⇒ a_ii = - a_ii [for all values of i]

⇒ a_ii + a_ii = 0

⇒ 2a_ii = 0

⇒ a_ii = 0 …(ii)

Now,

= 0 + a₁₂ + a₁₃ + … + (-a₁₂) + 0 + a₂₃ + …+ (-a₁₃)+ (- a₂₃)+ 0 + …

[from (i) and (ii)]

= 0

Thus,

Hence Proved.

Question 22.

If A = [a_ij] is a skew-symmetric matrix, then write the value of .

Answer:

Given: A = [a_ij] is a skew – symmetric matrix

⇒ a_ij = - a_ji …(i)

[for all values of i, j]

For diagonal elements,

⇒ a_ii = - a_ii [for all values of i]

⇒ a_ii + a_ii = 0

⇒ 2a_ii = 0

⇒ a_ii = 0 …(ii)

Now,

= 0 + a₁₂ + a₁₃ + … + (-a₁₂) + 0 + a₂₃ + …+ (-a₁₃)+ (- a₂₃)+ 0 + …

[from (i) and (ii)]

= 0

Thus,

Hence Proved.

Question 23.

If A and B are symmetric matrices, then write the condition for which AB is also symmetric.

Answer:

If A and B are symmetric matrices, then AB is symmetric if and only if A and B commute .i.e.

AB = (AB)^T = B^TA^T = BA

[∵ B^T = B and A^T = A]

Question 24.

If B is a skew-symmetric matrix, write whether the matrix AB A^T is symmetric or skew-symmetric.

Answer:

B is a skew – symmetric matrix, then

B^T = -B …(i)

Consider

(ABA^T)^T = (A^T)^T B^T A^T

[∵ (AB)^T = B^TA^T]

⇒ (ABA^T)^T = AB^TA^T [∵ (A^T)^T = A]

⇒ (ABA^T)^T = A(- B)A^T [from (i)]

⇒ (ABA^T)^T = - ABA^T

∴ ABA^T is a skew – symmetric matrix

Question 25.

If B is a symmetric matrix, write whether the matrix AB A^T is symmetric or skew-symmetric.

Answer:

B is a symmetric matrix, then

B^T = B …(i)

Consider

(ABA^T)^T = (A^T)^T B^T A^T

[∵ (AB)^T = B^TA^T]

⇒ (ABA^T)^T = AB^TA^T [∵ (A^T)^T = A]

⇒ (ABA^T)^T = A(B)A^T [from (i)]

⇒ (ABA^T)^T = ABA^T

∴ ABA^T is a symmetric matrix

Question 26.

If A is a skew-symmetric and n ∈ N such that (Aⁿ)^T = λAⁿ, write the value of λ.

Answer:

Let A is a skew – symmetric matrix, then

A^T = - A …(i)

Consider

⇒ (Aⁿ)^T = λAⁿ [given]

⇒ (A^T)ⁿ = λAⁿ

⇒ (-A)ⁿ = λAⁿ [from (i)]

⇒ (-1)ⁿ(A)ⁿ = λAⁿ

Comparing both the sides, we get

λ = (-1)ⁿ

Question 27.

If A is a symmetric matrix and n ∈ N, write whether Aⁿ is symmetric or skew-symmetric or neither of these two.

Answer:

Given that A is a symmetric matrix

∴ A = A^T …(i)

Now, we have to check Aⁿ is symmetric or skew – symmetric

(Aⁿ)^T = (A×A×A×A…A)^T [for all n Є N]

⇒ (Aⁿ)^T = (A^T × A^T … A^T)

[∵ (AB)^T = B^TA^T]

= A × A … A [from (i)]

= Aⁿ

⇒ (Aⁿ)^T = Aⁿ

Case 1: If n is an even natural number, then

(Aⁿ)^T = Aⁿ

So, Aⁿ is a symmetric matrix

Case 2: If n is odd natural number, then

(Aⁿ)^T = Aⁿ

So, Aⁿ is a symmetric matrix

Question 28.

If A is a skew-symmetric matrix and n is an even natural number, write whether Aⁿ is symmetric or skew-symmetric or neither of these two.

Answer:

Let A is a skew – symmetric matrix, then

A^T = - A …(i)

Now, we have to check Aⁿ is symmetric or skew – symmetric

(Aⁿ)^T = (A^T)ⁿ [for all n Є N]

⇒ (Aⁿ)^T = (- A)ⁿ [from (i)]

⇒ (Aⁿ)^T = (-1)ⁿ (A)ⁿ

Given that n is an even natural number, then

(Aⁿ)^T = Aⁿ

[∵ (-1)² = 1, (-1)⁴ =1,… (-1)ⁿ = 1]

So, Aⁿ is a symmetric matrix

Question 29.

If A is a skew-symmetric matrix and n is an odd natural number, write whether Aⁿ is symmetric or skew-symmetric or neither of the two.

Answer:

Let A is a skew – symmetric matrix, then

A^T = - A …(i)

Now, we have to check Aⁿ is symmetric or skew – symmetric

(Aⁿ)^T = (A^T)ⁿ [for all n Є N]

⇒ (Aⁿ)^T = (- A)ⁿ [from (i)]

⇒ (Aⁿ)^T = (-1)ⁿ (A)ⁿ

Given that n is odd natural number, then

(Aⁿ)^T = - Aⁿ

[∵ (-1)³ = - 1, (-1)⁵ = - 1,… (-1)ⁿ = - 1]

So, Aⁿ is a skew - symmetric matrix

Question 30.

If A and B are symmetric matrices of the same order, write whether AB – BA is symmetric or skew-symmetric or neither of the two.

Answer:

A and B are symmetric matrices,

∴ A’ = A and B’ = B …(i)

Consider (AB – BA)’ = (AB)’ – (BA)’ [(a – b)’ = a’ – b’]

= B’A’ – A’B’ [(AB)’ = B’A’]

= BA – AB [from (i)]

= - (AB – BA)

∴ (AB – BA)’ = - (AB – BA)

Hence, (AB – BA) is a skew symmetric matrix.

Question 31.

Write a square matrix which is both symmetric as well as skew-symmetric.

Answer:

We must understand what symmetric matrix is.

A symmetric matrix is a square matrix that is equal to its transpose.

A symmetric matrix ⬄ A = A^T

Now, let us understand what skew-symmetric matrix is.

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that, it satisfies the condition

A skew symmetric matrix ⬄ A^T = -A

And,

A square matrix is a matrix with the same number of rows and columns. An n-by-n matrix is known as a square matrix of order n.

We need to find a square matrix which is both symmetric as well as skew symmetric.

Take a 2 × 2 null matrix.

Say,

Let us take transpose of the matrix A.

We know that, the transpose of a matrix is a new matrix whose rows are the columns of the original.

So,

Since, A = A^T.

∴, A is symmetric.

Take the same matrix and multiply it with -1.

Let us take transpose of the matrix –A.

So,

Since,

A^T = -A

∴, A is skew-symmetric.

Thus, A (a null matrix) is both symmetric as well as skew-symmetric.

Question 32.

Find the value of x and y, if .

Answer:

We are given that,

We need to find the value of x and y.

Taking Left Hand Side (LHS) matrix of the equation,

Multiplying the scalar, 2 by each element of the matrix ,

Adding the corresponding elements,

Equate LHS to Right Hand Side (RHS) equation,

We know that if we have,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

Similarly, the corresponding elements of two matrices are equal,

2 + y = 5 …(i)

6 = 6

1 = 1

2x + 2 = 8 …(ii)

We have equations (i) and (ii) to solve for x and y.

From equation (i),

2 + y = 5

⇒ y = 5 – 2

⇒ y = 3

From equation (ii),

2x + 2 = 8

⇒ 2x = 8 – 2

⇒ 2x = 6

⇒ x = 3

Thus, we have x = 3 and y = 3.

Question 33.

If , find x and y.

Answer:

We are given with,

We need to find the values of x and y.

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

So, if we have

Corresponding elements of two matrices are equal.

That is,

x + 3 = 5 …(i)

4 = 4

y – 4 = 3 …(ii)

x + y = 9 …(iii)

To solve for x and y, we have three equations (i), (ii) and (iii).

From equation (i),

x + 3 = 5

⇒ x = 5 – 3

⇒ x = 2

From equation (ii),

y – 4 = 3

⇒ y = 3 + 4

⇒ y = 7

We need not solve equation (iii) as we have got the values of x and y.

Thus, the values of x = 2 and y = 7.

Question 34.

Find the value of x from the following: .

Answer:

We are given with matrix equation,

We need to find the values of x and y.

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

So, if we have

Corresponding elements of two matrices are equal.

That is,

2x – y = 6 …(i)

5 = 5

3 = 3

y = -2 …(ii)

To solve for x and y, we have equations (i) and (ii).

From equation (ii),

y = -2

Substituting y = -2 in equation (i), we get

2x – y = 6

⇒ 2x – (-2) = 6

⇒ 2x + 2 = 6

⇒ 2x = 6 – 2

⇒ 2x = 4

⇒ x = 2

Thus, we get x = 2 and y = -2.

Question 35.

Find the value of y, if .

Answer:

We are given that,

We need to find the values of x and y.

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

So, if we have

Corresponding elements of two matrices are equal.

That is,

x – y = 2 …(i)

2 = 2

x = 3 …(ii)

5 = 5

To solve for x and y, we have equations (i) and (ii).

From equation (ii),

x = 3

Substituting x = 3 in equation (i), we get

3 – y = 2

⇒ y = 3 – 2

⇒ y = 1

Thus, we get x = 3 and y = 1.

Question 36.

Find the value of x, if .

Answer:

We are given that,

We need to find the values of x and y.

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

So, if we have

Corresponding elements of two matrices are equal.

That is,

3x + y = 1 …(i)

-y = 2 …(ii)

2y – x = -5 …(iii)

3 = 3

To solve for x and y, we have equations (i), (ii) and (iii).

From equation (ii),

-y = 2

Multiplying both sides by -1,

-1 × -y = -1 × 2

⇒ y = -2

Substituting y = -2 in either of the equations (i) or (iii), say (i)

3x + y = 1

⇒ 3x + (-2) = 1

⇒ 3x – 2 = 1

⇒ 3x = 1 + 2

⇒ 3x = 3

⇒ x = 1

Thus, we get x = 1 and y = -2.

Question 37.

If matrix , write AA^T.

Answer:

We are given that,

We need to compute AA^T.

We know that the transpose of a matrix is a new matrix whose rows are the columns of the original.

So, transpose of matrix A will be given as

Multiplying A by A^T,

In multiplication of matrices,

Dot multiply the matching members of 1^st row of first matrix and 1^st column of second matrix and then sum up.

(a₁₁ a₁₂ a₁₃)(b₁₁ b₂₁ b₃₁) = a₁₁ × b₁₁ + a₁₂ × b₂₁ + a₁₃ × b₃₁

So,

(1 2 3)(1 2 3) = 1 × 1 + 2 × 2 + 3 × 3

⇒ (1 2 3)(1 2 3) = 1 + 4 + 9

⇒ (1 2 3)(1 2 3) = 14

Thus,

Question 38.

If , then find x.

Answer:

We are given that,

We need to find the value of x.

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

So, if we have

Corresponding elements of two elements are equal.

That is,

2x + y = 6 …(i)

3y = 0 …(ii)

To solve for x, we have equations (i) and (ii).

We can’t solve for x using only equation (i) as equation (i) contains x as well as y. We need to find the value of y from equation (ii) first.

From equation (ii),

3y = 0

⇒ y = 0

Substituting y = 0 in equation (i),

2x + y = 6

⇒ 2x + (0) = 6

⇒ 2x = 6 – 0

⇒ 2x = 6

⇒ x = 3

Thus, we get x = 3.

Question 39.

If , find A + A^T.

Answer:

We are given that,

We need to find the value of A + A^T.

We know that the transpose of a matrix is a new matrix whose rows are the columns of the original.

We have,

Here,

1^st row of A = (1 2)

2^nd row of A = (3 4)

Transpose of this matrix A, A^T will be given as

1^st column of A^T = 1^st row of A = (1 2)

2^nd column of A^T = 2^nd row of A = (3 4)

Then,

For addition of two matrices, say X and Y, where

Add the corresponding elements of matrices X and Y.

Similarly, we need to add these two matrices, A and A^T.

Adding the corresponding elements of the matrices A and A^T,

Thus, we get the matrix .

Question 40.

If , then find a.

Answer:

We are given that,

We need to find the value of x.

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

So, if we have

Corresponding elements of two elements are equal.

That is,

a + b = 6 …(i)

b = 4 …(ii)

To solve for a, we have equations (i) and (ii).

We can’t solve for a using only equation (i) as equation (i) contains a as well as b. We need to find the value of b from equation (ii) first.

From equation (ii),

b = 4

Substituting the value of b = 4 in equation (i),

a + b = 6

⇒ a + 4 = 6

⇒ a = 6 – 4

⇒ a = 2

Thus, we get a = 2.

Question 41.

If A is a matrix of order 3 × 4 and B is a matrix of order 4 × 3, find the order of the matrix of AB.

Answer:

We are given that,

Order of matrix A = 3 × 4

Order of matrix B = 4 × 3

We need to find the order of the matrix of AB.

We know that,

For matrices X and Y such that,

Order of X = m × n

Order of Y = r × s

In order to multiply the two matrices X and Y, the number of columns in X must be equal to the number of rows in Y. That is,

n = r

And order of the resulting matrix, XY is given as

Order of XY = m × s

Provided n = r.

So, we know

Order of A = 3 × 4

Here,

Number of rows = 3

Number of columns = 4

Order of B = 4 × 3

Here,

Number of rows = 4

Number of columns = 3

Note that,

Number of columns in A = Number of rows in B = 4

So,

Order of the resulting matrix, AB is given as

Order of AB = 3 × 3

Thus, order of AB = 3 × 3

Question 42.

If is identity matrix, then write the value of α.

Answer:

We are given that,

is an identity matrix.

We need to find the value of α.

We must understand what an identity matrix is.

An identity matrix is a square matrix in which all the elements of the principal diagonal are ones and all other elements are zeroes.

An identity matrix is denoted by

According to the question,

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

So, if we have

The corresponding elements of matrices are equal.

That is,

cos α = 1

-sin α = 0

sin α = 0

cos α = 1

Since, the equations are repetitive, take

cos α = 1

⇒ α = cos^-1 1

⇒ α = 0°

Thus, the value of α = 0°.

Question 43.

If , then write the value of k.

Answer:

We are given with

We need to find the value of k.

Take Left Hand Side (LHS) of the matrix equation.

In multiplication of matrices,

For c₁₁: dot multiply the matching members of 1^st row of first matrix and 1^st column of second matrix and then sum up.

(a₁₁ a₁₂)(b₁₁ b₂₁) = a₁₁ × b₁₁ + a₁₂ × b₂₁

Thus,

(1 2)(3 2) = 1 × 3 + 2 × 2

⇒ (1 2)(3 2) = 3 + 4

⇒ (1 2)(3 2) = 7

For c₁₂: dot multiply the matching members of 1^st row of first matrix and 2^nd column of second matrix and then sum up.

(a₁₁ a₁₂)(b₁₂ b₂₂) = a �₁₁ × b₁₂ + a₁₂ × b₂₂

Thus,

(1 2)(1 5) = 1 × 1 + 2 × 5

⇒ (1 2)(1 5) = 1 + 10

⇒ (1 2)(1 5) = 11

Similarly, do the same for other elements.

Since,

Substituting the value of LHS,

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

Thus,

7 = 7

11 = 11

17 = k

23 = 23

Hence, k = 17.

Question 44.

If I is the identity matrix and A is a square matrix such A² = A, then what is the value of (I + A)² – 3A?

Answer:

We are given that,

I is the identity matrix.

A is a square matrix such that A² = A.

We need to find the value of (I + A)² – 3A.

We must understand what an identity matrix is.

An identity matrix is a square matrix in which all the elements of the principal diagonal are ones and all other elements are zeroes.

Take,

(I + A)² – 3A = (I)²+ (A)² + 2(I)(A) – 3A

[∵, by algebraic identity,

(x + y)² = x² + y² + 2xy]

⇒ (I + A)² – 3A = (I)(I) + A² + 2(IA) – 3A

By property of matrix,

(I)(I) = I

IA = A

⇒ (I + A)² – 3A = I + A² + 2A – 3A

⇒ (I + A)² – 3A = I + A + 2A – 3A [∵, given in question, A² = A]

⇒ (I + A)² – 3A = I + 3A – 3A

⇒ (I + A)² – 3A = I + 0

⇒ (I + A)² – 3A = I

Thus, the value of (I + A)² – 3A = I.

Question 45.

If is written as B + C, where B is a symmetric matrix and C is a skew-symmetric matrix, then find B.

Answer:

We are given that,

Where,

B = symmetric matrix

C = skew-symmetric matrix

We need to find B.

A symmetric matrix is a square matrix that is equal to its transpose.

A symmetric matrix ⬄ A = A^T

Now, let us understand what skew-symmetric matrix is.

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that, it satisfies the condition

A skew symmetric matrix ⬄ A^T = -A

So, let the matrix B be

Let us calculate A^T.

We know that the transpose of a matrix is a new matrix whose rows are the columns of the original.

We have,

Here,

1^st row of A = (1 2)

2^nd row of A = (0 3)

Transpose of this matrix A, A^T will be given as

1^st column of A^T = 1^st row of A = (1 2)

2^nd column of A^T = 2^nd row of A = (0 3)

Then,

Substituting the matrix A and A^T in B,

Taking transpose of B,

1^st row of B = (1 1)

2^nd row of B = (1 3)

Transpose of this matrix B, B^T will be given as

1^st column of B^T = 1^st row of B = (1 1)

2^nd column of A^T = 2^nd row of A = (1 3)

Then,

Since, B = B^T. Thus, B is symmetric.

Now, let the matrix C be

Substituting the matrix A and A^T in C,

Multiplying -1 on both sides,

Taking transpose of C,

1^st row of C = (0 1)

2^nd row of C = (-1 0)

Transpose of this matrix C, C^T will be given as

1^st column of C^T = 1^st row of C = (0 1)

2^nd column of C^T = 2^nd row of C = (-1 0)

Then,

Since, C^T = -C. Thus, C is skew-symmetric.

Check:

Put the value of matrices B and C.

Matrices B and C satisfies the equation.

Hence, .

Question 46.

If A is 2 × 3 matrix and B is a matrix such that A^TB and BA^T both are defined, then what is the order of B?

Answer:

We are given that,

Order of matrix A = 2 × 3

A^TB and BA^T are defined matrices.

We need to find the order of matrix B.

We know that the transpose of a matrix is a new matrix whose rows are the columns of the original.

So, if the number of rows in matrix A = 2

And, number of columns in matrix A = 3

Then, the number of rows in matrix A^T = number of columns in matrix A = 3

Number of columns in matrix A^T = number of rows in matrix A = 2

So,

Order of matrix A^T can be written as

Order of matrix A^T = 3 × 2

Thus, we have

Number of rows of A^T = 3 …(i)

Number of columns of A^T = 2 …(ii)

If A^TB is defined, that is, it exists, then

Number of columns in A^T = Number of rows in B

⇒ 2 = Number of rows in B [from (ii)]

Or,

Number of rows in B = 2 …(iii)

If BA^T is defined, that is, it exists, then

Number of columns in B = Number of rows in A^T

Substituting value of number of rows in A^T from (i),

⇒ Number of columns in B = 3 …(iv)

From (iii) and (iv),

Order of B = Number of rows × Number of columns

⇒ Order of B = 2 × 3

Thus, order of B is 2 × 3.

Question 47.

What is the total number of 2 × 2 matrices with each entry 0 or 1?

Answer:

We are given with the information that,

Each element of the 2 × 2 matrix can be filled in 2 ways, either 0 or 1.

We need to find the number of total 2 × 2 matrices with each entry 0 or 1.

Let A be 2 × 2 matrix such that,

Note that, there are 4 elements in the matrix.

So, if 1 element can be filled in 2 ways, either 0 or 1.

That is,

Number of ways in which 1 element can be filled = 2¹

Then,

Number of ways in which 4 elements can be filled = 2⁴

⇒ Number of ways in which 4 elements can be filled = 16

Thus, total number of 2 × 2 matrices with each entry 0 or 1 is 16.

Question 48.

If , then find the value of y.

Answer:

We are given that,

We need to find the value of y.

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

So, if we have

Corresponding elements of two elements are equal.

That is,

x = 3 …(i)

x – y = 1 …(ii)

2x + y = 8 …(iii)

7 = 7

To solve for y, we have equations (i), (ii) and (iii).

From equation (i),

x = 3

Substituting the value of x = 3 in equation (ii),

x – y = 1

⇒ 3 – y = 1

⇒ y = 3 – 1

⇒ y = 2

Thus, we get y = 2.

Question 49.

If a matrix has 5 elements, write all possible orders it can have.

Answer:

We are given that,

A matrix has 5 elements.

We need to find all the possible orders.

We know that if there is a matrix A, of order m × n.

Then, there are mn elements.

Or,

If a matrix has mn elements, then

The order of the matrix = m × n or n × m

For example,

If a matrix is of order 1 × 2, then

There are 2 elements in the matrix.

Or,

If a matrix is of order 2 × 1, then

There are 2 elements in the matrix.

Similarly,

If a matrix has 5 elements, then

The order of this matrix are 1 × 5 or 5 × 1.

Thus, possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

Question 50.

For a 2 × 2 matrix A = [a_ij] whose elements are given by , write the value of a₁₂.

Answer:

We are given with,

A matrix of order 2 × 2, A = [a_ij].

We need to find the value of a₁₂.

Here, if A is of the order 2 × 2 then,

Number of rows of A = 2

Number of columns of A = 2

We can easily find the elements using the representation of element, .

Compare a_ij with a₁₂.

We get,

i = 1

j = 2

Putting these values in ,

Thus, the value of .

Question 51.

If , find the value of x.

Answer:

We are given with,

We need to find the value of x.

By property of matrices,

Similarly,

…(i)

And,

…(ii)

Adding equations (i) and (ii),

Now,

Since

Adding the two matrices on the right hand side by simply adding the corresponding elements,

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

So, this means that we can get two equations,

10 = 2x – y …(iii)

5 = 3x + y …(iv)

We have two equations and two variables.

Solving equations (iii) and (iv),

⇒ 5x = 15

⇒ x = 3

Thus, we get x = 3.

Question 52.

If , then find matrix A.

Answer:

We are given that,

We need to find the matrix A.

In order to find A, shift the matrix in addition with A to left hand side of the equation.

Just like in algebraic property,

X = A + Y

⇒ A = X – Y

Similarly,

Subtraction in matrices is done by subtraction of corresponding elements in the matrices.

Thus, we get .

Question 53.

If , find the value of b.

Answer:

We are given that,

We need to find the value of b.

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

Similarly,

a – b = -1 …(i)

2a + c = 5 …(ii)

2a – b = 0 …(iii)

3c + d = 13 …(iv)

We have the equations (i), (ii), (iii) and (iv).

We need not solve equations (ii) and (iv). We will be able to solve for b from equations (i) and (iii).

Multiply equation (i) by 2.

a – b = -1 [× 2

⇒ 2a – 2b = -2 …(v)

Subtracting equation (iii) from equation (v),

⇒ -b = -2

⇒ b = 2

Thus, the value of b = 2.

Question 54.

For what value of x, is the matrix a skew-symmetric matrix?

Answer:

We are given that,

is a skew-symmetric matrix.

We need to find the value of x.

Let us understand what skew-symmetric matrix is.

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that, it satisfies the condition

A skew symmetric matrix ⬄ A^T = -A

First, let us find –A.

Let us find the transpose of A.

We know that the transpose of a matrix is a new matrix whose rows are the columns of the original.

In matrix A,

1^st row of A = (0 1 -2)

2^nd row of A = (-1 0 3)

3^rd row of A = (x -3 0)

In the formation of matrix A^T,

1^st column of A^T = 1^st row of A = (0 1 -2)

2^nd column of A^T = 2^nd row of A = (-1 0 3)

3^rd column of A^T = 3^rd row of A = (x -3 0)

So,

Substituting the matrices –A and A^T, we get

-A = A^T

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

By comparing the corresponding elements of the two matrices,

x = 2

Thus, the value of x = 2.

Question 55.

If matrix and A² = pA, then write the value of p.

Answer:

We are given that,

A² = pA

We need to find the value of p.

First, let us find A².

We know that, A² = A.A

In multiplication of matrices A and A, such that A² = Z(say):

For the calculation of z₁₁: Dot multiply the 1^st row of first matrix and the 1^st column of second matrix and then sum up.

(2 -2)(2 -2) = 2 × 2 + (-2) × (-2)

⇒ (2 -2)(2 -2) = 4 + 4

⇒ (2 -2)(2 -2) = 8

So,

For the calculation of z₁₂: Dot multiply the 1^st row of first matrix and the 2^nd column of second matrix and then sum up.

(2 -2)(-2 2) = 2 × -2 + (-2) × 2

⇒ (2 -2)(-2 2) = -4 – 4

⇒ (2 -2)(-2 2) = -8

So,

Similarly,

So,

…(i)

Now, let us find pA.

Multiply p by matrix A,

…(ii)

Substituting value of A² and pA from (i) and (ii) in

A² = pA

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

So,

2p = 8

-2p = -8

2p = 8

Take equation,

2p = 8

⇒ p = 4

Thus, the value of p = 4.

Question 56.

If A is a square matrix such that A² = A, then write the value of 7A–(I + A)³, where I is the identity matrix.

Answer:

We are given that,

A is a square matrix such that,

A² = A

I is an identity matrix.

We need to find the value of 7A – (I + A)³.

Take,

7A – (I + A)³ = 7A – (I³ + A³ + 3I²A + 3IA²)

[∵, by algebraic identity, (x + y)³ = x³ + y³ + 3x²y + 3xy²]

⇒ 7A – (I + A)³ = 7A – I³ – A³ – 3I²A – 3IA²

⇒ 7A – (I + A)³ = 7A – I – A³ – 3I²A – 3IA²

⇒ 7A – (I + A)³ = 7A – I – A.A² – 3I²A – 3IA²

⇒ 7A – (I + A)³ = 7A – I – A.A² – 3A – 3A²

[∵, by property of identity matrix,

I²A = A & IA² = A²]

⇒ 7A – (I + A)³ = 7A – I – A.A – 3A – 3A

[∵, it is given that, A² = A]

⇒ 7A – (I + A)³ = 7A – I – A² – 6A

[∵, A.A = A²]

⇒ 7A – (I + A)³ = 7A – I – A – 6A

[∵, it is given that, A² = A]

⇒ 7A – (I + A)³ = 7A – I – 7A

⇒ 7A – (I + A)³ = -I

Thus, the value of 7A – (I + A)³ is –I.

Question 57.

If , find x – y.

Answer:

We are given that,

We need to find the value of (x – y).

Take,

Multiplying 2 by each element of the matrix,

In addition of matrices, we need to add the corresponding elements of the matrices.

So,

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

So,

7 = 7

8 + y = 0 …(i)

10 = 10

2x + 1 = 5 …(ii)

Let us find x and y using the equations (i) and (ii).

From equation (i),

8 + y = 0

⇒ y = -8

From equation (ii),

2x + 1 = 5

⇒ 2x = 5 – 1

⇒ 2x = 4

⇒ x = 2

So, x = 2 and y = -8.

Then,

x – y = 2 – (-8)

⇒ x – y = 2 + 8

⇒ x – y = 10

Thus, the value of (x – y) is 10.

Question 58.

If , find x.

Answer:

We are given that,

We need to find the value of x.

Let matrices be,

Then,

Order of A = 1 × 2 [∵, Matrix A has 1 row and 2 columns]

Order of B = 2 × 2 [∵, Matrix B has 2 rows and 2 columns]

Since,

Number of columns in A = Number of rows in B = 2

∴, Order of resulting matrix AB will be 1 × 2.

Resulting matrix = O

O is zero-matrix, where every element of the matrix is zero.

Order of O = 1 × 2

That is,

So,

…(i)

Let,

Let us solve the left hand side of the matrix equation.

In multiplication of matrices,

For z₁₁: Dot multiply 1^st row of first matrix and 1^st column of second matrix, and then sum up.

(x 1)(1 -2) = x × 1 + 1 × -2

⇒ (x 1)(1 -2) = x – 2

So,

For z₁₂: Dot multiply 1^st row of first matrix and 2^nd column of second matrix, and then sum up.

(x 1)(0 0) = x × 0 + 1 × 0

⇒ (x 1)(0 0) = 0 + 0

⇒ (x 1)(0 0) = 0

So,

Substituting the resulting matrix in left hand side of (i),

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

Therefore,

x – 2 = 0

⇒ x = 2

Thus, the value of x is 2.

Question 59.

If , write the value of a – 2b.

Answer:

We are given that,

We need to find the value of a – 2b.

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

Therefore,

a + 4 = 2a + 2 …(i)

3b = b + 2 …(ii)

8 = 8

-6 = a – 8b …(iii)

We have the equations (i), (ii) and (iii).

From equation (i),

a + 4 = 2a + 2

⇒ 2a – a = 4 – 2

⇒ a = 2

From equation (ii),

3b = b + 2

⇒ 3b – b = 2

⇒ 2b = 2

⇒ b = 1

We have a = 2 and b = 1.

Substituting the values of a and b in

a – 2b = 2 – 2(1)

⇒ a – 2b = 2 – 2

⇒ a – 2b = 0

Thus, the value of a – 2b is 0.

Question 60.

Write a 2 × 2 matrix which is both symmetric and skew-symmetric.

Answer:

We need to find a matrix of order 2 × 2 which is both symmetric and skew-symmetric.

We must understand what symmetric matrix is.

A symmetric matrix is a square matrix that is equal to its transpose.

A symmetric matrix ⬄ A = A^T

Now, let us understand what skew-symmetric matrix is.

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that, it satisfies the condition

A skew symmetric matrix ⬄ A^T = -A

And,

A square matrix is a matrix with the same number of rows and columns. An n-by-n matrix is known as a square matrix of order n.

Take a 2 × 2 null matrix.

Say,

Let us take transpose of the matrix A.

We know that, the transpose of a matrix is a new matrix whose rows are the columns of the original.

So,

Since, A = A^T.

∴, A is symmetric.

Take the same matrix and multiply it with -1.

Let us take transpose of the matrix –A.

So,

Since,

A^T = -A

∴, A is skew-symmetric.

Thus, A (a null matrix) of order 2 × 2 is both symmetric as well as skew-symmetric.

Question 61.

If , write the value of (x + y + z).

Answer:

We are given that,

We need to find the value of (x + y + z).

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

We have,

Therefore,

xy = 8 …(i)

4 = w …(ii)

z + 6 = 0 …(iii)

x + y = 6 …(iv)

We have the equations (i), (ii), (iii) and (iv).

We just need to find the values of x, y and z. So,

From equation (iii),

z + 6 = 0

⇒ z = -6

Now, let us find (x + y + z).

Substituting z = -6 and the value of (x + y) from equation (iv),

x + y + z = (x + y) + z

⇒ x + y + z = 6 + (-6)

⇒ x + y + z = 6 – 6

⇒ x + y + z = 0

Thus, the value of (x + y + z) is 0.

Question 62.

Construct a 2 × 2 matrix A = [a_ij] whose elements a_ij are given by .

Answer:

We are given that a_ij is given as

We need to construct a 2 × 2 matrix A defined as A = [a_ij].

Since, this is a 2 × 2 matrix where A = [a_ij], we know

Number of rows = 2

Number of columns = 2

∴, i = 1, 2

& j = 1, 2

First, put i = 1 and j = 1 in a_ij, here i = j.

For i = j,

a_ij = (i + j)²

So,

a₁₁ = (1 + 1)²

⇒ a₁₁ = 2²

⇒ a �₁₁ = 4

Put i = 1 and j = 2 in a_ij, here i ≠ j.

For i ≠ j,

So,

Put i = 2 and j = 1 in a_ij, here i ≠ j.

For i ≠ j,

So,

Put i = 2 and j = 2 in a_ij, here i = j.

For i = j,

a_ij = (i + j)²

So,

a₂₂ = (2 + 2)²

⇒ a₂₂ = 4²

⇒ a₂₂ = 16

Thus, we get

Question 63.

If , then write the value of (x, y).

Answer:

We are given that,

We need to find the value of (x, y).

Multiply the matrices on the right hand side of the equation,

For z₁₁: Dot multiply the 1^st row of first matrix and 1^st column of second matrix, then sum up.

(2 1)(1 -2) = 2 × 1 + 1 × -2

⇒ (2 1)(1 -2) = 2 – 2

⇒ (2 1)(1 -2) = 0

So,

For z₂₁: Dot multiply the 2^nd row of first matrix and 1^st column of second matrix, then sum up.

(4 3)(1 -2) = 4 × 1 + 3 × (-2)

⇒ (4 3)(1 -2) = 4 – 6

⇒ (4 3)(1 -2) = -2

So,

Equate the resulting matrix to the given matrix equation.

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

Therefore,

x + y = 0

x – y = -2

Adding these two equations, we get

(x + y) + (x – y) = 0 + (-2)

⇒ x + y + x – y = -2

⇒ x + x + y – y = -2

⇒ 2x + 0 = -2

⇒ 2x = -2

⇒ x = -1

Putting x = -1 in

x + y = 0

⇒ (-1) + y = 0

⇒ -1 + y = 0

⇒ y = 1

So, putting values of x and y from above in (x, y), we get

(x, y) = (-1, 1)

Thus, (x, y) is (-1, 1).

Question 64.

Matrix is given to be symmetric, find the values of a and b.

Answer:

We are given that,

is symmetric matrix.

We need to find the values of a and b.

We must understand what symmetric matrix is.

A symmetric matrix is a square matrix that is equal to its transpose.

A symmetric matrix ⬄ A = A^T

This means, we need to find the transpose of matrix A.

Let us take transpose of the matrix A.

We know that, the transpose of a matrix is a new matrix whose rows are the columns of the original.

We have,

1^st row of matrix A = (0 2b -2)

2^nd row of matrix A = (3 1 3)

3^rd row of matrix A = (3a 3 -1)

For matrix A^T, it will become

1^st column of A^T = 1^st row of A = (0 2b -2)

2^nd column of A^T = 2^nd row of A = (3 1 3)

3^rd column of A^T = 3^rd row of A = (3a 3 -1)

Now, as A = A^T.

Substituting the matrices A and A^T, we get

We know by the property of matrices,

This implies,

a₁₁ = b₁₁, a₁₂ = b₁₂, a₂₁ = b₂₁ and a₂₂ = b₂₂

Applying this property, we can write

2b = 3 …(i)

-2 = 3a …(ii)

3 = 2b

3a = -2

We can find a and b from equations (i) and (ii).

From equation (i),

2b = 3

From equation (ii),

-2 = 3a

Thus, we get and .

Question 65.

Write the number of all possible matrices of order 2 × 2 with each entry 1, 2 or 3.

Answer:

We are given with the information that,

Each element of the 2 × 2 matrix can be filled in 3 ways, either 1, 2 or 3.

We need to find the number of total 2 × 2 matrices with each entry 1, 2 or 3.

Let A be 2 × 2 matrix such that,

Note that, there are 4 elements in the matrix.

So, if 1 element can be filled in 3 ways, either 1, 2 or 3.

That is,

Number of ways in which 1 element can be filled = 3¹

Then,

Number of ways in which 4 elements can be filled = 3⁴

⇒ Number of ways in which 4 elements can be filled = 81

Thus, total number of 2 × 2 matrices with each entry 1, 2 or 3 is 81.

Question 66.

If , then write the order of matrix A.

Answer:

We are given that,

We need to find the order of the matrix A.

Let the matrices be,

Let us find the order of X.

Number of rows of matrix X = 1

Number of columns of matrix X = 3

So, order of matrix X = 1 × 3 …(i)

Now, let us find the order of Y.

Number of rows of matrix Y = 3

Number of columns of matrix Y = 3

So, order of matrix Y = 3 × 3 …(ii)

From (i) and (ii),

Order of resulting XY = 1 × 3 [∵, Number of columns of X = Number of rows of Y] …(iii)

Let us find the order of Z.

Number of rows of matrix Z = 3

Number of columns of matrix Z = 1

So, order of matrix Z = 3 × 1 …(iv)

Order of resulting XYZ = 1 × 1 [∵, Number of columns of XY = Number of rows of Z]

Thus, the order of matrix A = 1 × 1

Question 67.

If is written as A = P + Q, where as A = P + Q, where P is symmetric and Q is skew-symmetric matrix, then write the matrix P.

Answer:

We are given that,

Where,

P = symmetric matrix

Q = skew-symmetric matrix

We need to find P.

A symmetric matrix is a square matrix that is equal to its transpose.

A symmetric matrix ⬄ P = P^T

Now, let us understand what skew-symmetric matrix is.

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that, it satisfies the condition

A skew symmetric matrix ⬄ Q^T = -Q

So, let the matrix P be

Let us calculate A^T.

We know that the transpose of a matrix is a new matrix whose rows are the columns of the original.

We have,

Here,

1^st row of A = (3 5)

2^nd row of A = (7 9)

Transpose of this matrix A, A^T will be given as

1^st column of A^T = 1^st row of A = (3 5)

2^nd column of A^T = 2^nd row of A = (7 9)

Then,

Substituting the matrix A and A^T in P,

Taking transpose of P,

1^st row of P = (3 6)

2^nd row of P = (6 9)

Transpose of this matrix P, P^T will be given as

1^st column of P^T = 1^st row of P = (3 6)

2^nd column of P^T = 2^nd row of P = (6 9)

Then,

Since, P = P^T. Thus, P is symmetric.

Now, let the matrix Q be

Substituting the matrix A and A^T in Q,

Multiplying -1 on both sides,

Taking transpose of Q,

1^st row of Q = (0 -1)

2^nd row of Q = (1 0)

Transpose of this matrix Q, Q^T will be given as

1^st column of Q^T = 1^st row of Q = (0 -1)

2^nd column of Q^T = 2^nd row of Q = (1 0)

Then,

Since, Q^T = -Q. Thus, Q is skew-symmetric.

Check:

Put the value of matrices P and Q.

Matrices P and Q satisfies the equation.

Hence, .

Question 68.

Let A and B be matrices of orders 3 × 2 and 2 × 4 respectively. Write the order of matrix AB.

Answer:

We are given that,

Order of matrix A = 3 × 2

Order of matrix B = 2 × 4

We need to find the order of matrix AB.

We know that,

Matrix A × Matrix B = Matrix AB

If order of matrix A is (m × n) and order of matrix B is (r × s), then matrices A and B can be multiplied if and only if n = r.

That is,

Number of columns in A = Number of rows in B

Also, the order of resulting matrix AB comes out to be m × s.

Applying it,

Number of columns in A = 2

Number of rows in B = 2

This means,

Matrices A and B can be multiplied, and its order will be given as:

Order of matrix AB = 3 × 4

Thus, order of matrix AB = 3 × 4

Mcq

Question 1.

If , then A² is equal to
A. a null matrix

B. a unit matrix

C. –A

D. A

Answer:

Hence, the Option (B) is correct, as the main diagonal elements are 1 except other which are 0.

Question 2.

If , n ∈ N, the A⁴ⁿ equals
A.

B.

C.

D.

Answer:

{n = 1, so the exponent comes out to be 4 and if n = 2, which will turn the exponent to 8, and the same cycle will repeat.}

Option (C) is the answer.

Question 3.

If A and B are two matrices such that AB = A and BA = B, then B² is equal to
A. B

B. A

C. 1

D. 0

Answer:

AB = A ----- (i)

BA = B ------(ii)

From equation (ii)

B × (AB) = B

B²A= B

From equation (ii)

B²A = BA

B² = B

Option (A) is the answer.

Question 4.

If AB = A and BA = B, where A and B are square matrices, then
A. B² = B and A² = A

B. B² ≠ B and A² = A

C. A²≠ A, B² = B

D. A² ≠ A, B² ≠ B

Answer:

AB = A ----- (i)

BA = B ------(ii)

From equation (ii)

B × (AB) = B

B²A= B

From equation (ii)

B²A = BA

B² = B

From equation (i)

A × (BA) = A

A²B= A

From equation (i)

A²B = AB

A² = A

Hence, A² = A & B² = B.

Option (A) is the correct answer.

Question 5.

If A and B are two matrices such that AB = B and BA = A, then A² + B² is equal to
A. 2AB

B. 2BA

C. A + B

D. AB

Answer:

AB = A ----- (i)

BA = B ------(ii)

From equation (ii)

B × (AB) = B

B²A= B

From equation (ii)

B²A = BA

B² = B

From equation (i)

A × (BA) = A

A²B= A

From equation (i)

A²B = AB

A² = A

Hence, A² + B² = A + B.

Option (C) is the correct answer.

Question 6.

If , then the least positive integral value of k is
A. 3

B. 4

C. 6

D. 7

Answer:

|A| =

I^k = I {K can be anything}

Let

Similarly,

Hence,

Multiplying Cos & Sin, to LHS & RHS,

So, k = 7

A⁷ = I

Hence, k = 7.

Option (D) is the correct answer.

Question 7.

If the matrix AB is zero, then
A. It is not necessary that either A = O or, B = O

B. A = O or B = O

C. A = O and B = O

D. all the above statements are wrong

Answer:

If the matrix AB is zero, then, it is not necessary that either A = 0 or, B = 0

Option (A) is the correct answer

Question 8.

Let , then Aⁿ is equal to
A.

B.

C.

D.

Answer:

{n times, (where n ∈ N)}

Option (C) is the answer.

Question 9.

If A, B and are square matrices or order 3, A is non-singular and AB = O, then B is a
A. null matrix

B. singular matrix

C. unit matrix

D. non-singular matrix

Answer:

As AB = 0

And Order of the matrices A & B is 3,

Matrix B has to be a null matrix.

Option (A) is the answer.

Question 10.

If and , then AB is equal to
A. B

B. nB

C. Bⁿ

D. A + B

Answer:

Option (B) is the answer.

Question 11.

If , then Aⁿ (where n ∈ N) equals
A.

B.

C.

D.

Answer:

{n times, (where n ∈ N)}

Option (A) is the answer.

Question 12.

If and and AB = I₃, then x + y equals
A. 0

B. –1

C. 2

D. none of these

Answer:

Hence x+y=0

Option (A) is the answer.

Question 13.

If and (A + B)² = A² + B², then values of a and b are
A. a = 4, b = 1

B. a = 1, b = 4

C. a = 0, b = 4

D. a = 2, b = 4

Answer:

As, A² + B² = (A+B)²

a² = 1 & b = 4

a = ± 1

Hence Option (B) is the correct answer.

Question 14.

If is such that A² = I, then
A. 1 + α² + βγ = 0

B. 1 – α² + βγ = 0

C. 1 – α² – βγ = 0

D. 1 + α² – βγ = 0

Answer:

As A² = I,

So, α² + βγ = 1

α² + βγ – 1 = 0

1 – α² – βγ.

Hence, Option (C) is the correct answer.

Question 15.

If S = [s_ij] is a scalar matrix such that s_ij = k and A is a square matrix of the same order, then AS = SA = ?
A. A^k

B. k + A

C. kA

D. kS

Answer:

As, = k

Let {Square Matrix}

Hence, AS = SA = kA

Option (C) is the answer.

Question 16.

If A is a square matrix such that A² = A, then (I + A)³ – 7A is equal to
A. A

B. I – A

C. I

D. 3A

Answer:

(I + A)³ = I³ + A³ + 3A²I + 3AI² (Using the identity of (a + b)³ = a³ + b³ + 3ab (a + b))

(I + A)³ = I + A²(A) + 3AI + 3A [I stands for Identity Matrix]

(I + A)³ = I + A² + 3A + 3A

(I + A)³ = 7A + I

(I + A)³ – 7A

7A + I – 7A

= I

Option (C) is the answer.

Question 17.

If a matrix A is both symmetric and skew-symmetric, then
A. A is a diagonal matrix

B. A is a zero matrix

C. A is a scalar matrix

D. A is a square matrix

Answer:

If a matrix A is both symmetric and skew-symmetric,

A’ = A & A’ = -A

Comparing both the equations,

A = -A

A + A = 0

2A = 0

A = 0

then A is a zero matrix.

Option (B) is the answer.

Question 18.

The matrix is
A. a skew-symmetric matrix

B. a symmetric matrix

C. a diagonal matrix

D. an upper triangular matrix

Answer:

A^T = -A

Then, the given matrix is a skew – symmetric matrix.

Option (A) is the answer.

Question 19.

If A is a square matrix, then AA is a
A. skew-symmetric matrix

B. symmetric matrix

C. diagonal matrix

D. none of these

Answer:

If A is a square matrix,

Let

then AA is neither of the matrices given in the options of the question.

Option (D) is the answer.

Question 20.

If A and B are symmetric matrices, then ABA is
A. symmetric matrix

B. skew-symmetric matrix

C. diagonal matrix

D. scalar matrix

Answer:

A’ = A & B’ = B

(ABA)’ = A’ (AB)’

= A’B’A’

= ABA

Symmetric Matrix

Option (A) is the answer.

Question 21.

If and A = A^T, then
A. x = 0, y = 5

B. x + y = 5

C. x = y

D. none of these

Answer:

A = A^T

x = y

Option (C) is the answer.

Question 22.

If A is 3 × 4 matrix and B is a matrix such that A^TB and BA^T are both defined. Then, B is of the type
A. 3 × 4

B. 3 × 3

C. 4 × 4

D. 4 × 3

Answer:

Order of A = 3 × 4

Order of A’ = 4 × 3

As A^TB and BA^T are both defined, so the number of columns in B should be equal to the number of rows in A’ for BA’ and also the number of columns in A’ should be equal to the number of rows in A’ for BA’.

So the order of matrix B = 3 × 4.

Option (A) is the answer.

Question 23.

If A = [a_ij] is a square matrix of even order such that a_ij = i² – j², then
A. A is a skew-symmetric matrix and |A| = 0

B. A is symmetric matrix and |A| is a square

C. A is symmetric matrix and |A| = 0

D. none of these

Answer:

a_ij = i² – j²

a11 = 12 – 12 = 0

a12 = 12 – 22 = -3

a21 = 22 – 12 = 3

a22 = 22 – 22 = 0

So, A^T = -A

|A| = 0(0) – (-3)(3) = 0 + 9 = 9 ≠ 0

So, none of these.

Option (D) is the answer.

Question 24.

If , then A^T + A = I₂, if
A.

B.

C.

D. none of these

Answer:

{n Є Z}

Option (C) is the answer.

Question 25.

If is expressed as the sum of a symmetric and skew-symmetric matrix, then the symmetric matrix is
A.

B.

C.

D.

Answer:

As, sum is expressed as

{Newly formed Symmetric Matrix}

Option (A) is the answer.

Question 26.

Out of the following matrices, choose that matrix which is a scalar matrix:
A.

B.

C.

D.

Answer:

Scalar Matrix is a matrix whose all off-diagonal elements are zero and all on-diagonal elements are equal.

Option (A) is the answer.

Question 27.

The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is
A. 27

B. 18

C. 81

D. 512

Answer:

Let

Elements = 9 Order = 3 × 3

Every item in this matrix can be filled in two ways either by 0 or by 1.

Possible Matrices = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 512

Option (D) is the answer.

Question 28.

Which of the given values of x and y make the following pairs of matrices equal?

and,

A.

B.

C.

D. Not possible to find

Answer:

As the given matrices are equal,

3x + 7 = 0

y – 2 = 5

y = 7

y + 1 = 8

y = 7

2 – 3x = 4

3x = -2

These values of x are not equal to each other, so it is not possible to find.

Option (D) is the answer.

Question 29.

If and , then the values of k, a, b, are respectively
A. –6, –12, –18

B. –6, 4, 9

C. –6, –4, –9

D. –6, 12, 18

Answer:

Comparing the equations,

-4k = 24

k = -6

3k = 2b

3(-6) = 2b

2b = -18

b= -9

3a = 2k

3a = 2(-6)

3a = -12

a = -4

Values are

k = -6, a = -4 & b = -9

Option (C) is the answer.

Question 30.

If and , then B equals
A. I cos θ + J sin θ

B. I sin θ + J cos θ

C. I cos θ – J sin θ

D. – I cos θ + J sin θ

Answer:

So,

Option (A) is the answer.

Question 31.

The trace of the matrix is
A. 17

B. 25

C. 3

D. 12

Answer:

As the trace of a matrix is the sum of on – diagonal elements,

So, 1 + 7 + 9 = 17

Trace = 17

Option (A) is the answer.

Question 32.

If A = [a_ij] is scalar matrix of order n × n such that a_ij = k for all i, then trace of A is equal to
A. nk

B. n + k

C.

D. none of these

Answer:

Trace of A, i.e., tr (A) =

= k + k + k + k + k + ................... (n times)

= k(n)

= nk

Option (A) is the answer.

Question 33.

The matrix is a
A. square matrix

B. diagonal matrix

C. unit matrix

D. none of these

Answer:

None of these

Option (D) is the answer.

Question 34.

The number of possible matrices of order 3 × 3 with each entry 2 or 0 is
A. 9

B. 27

C. 81

D. none of these

Answer:

Let

Elements = 9 Order = 3 × 3

Every item in this matrix can be filled in two ways either by 0 or by 2.

Possible Matrices = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 512

Option (D) is the answer.

Question 35.

If , then the value of x + y is
A. x = 3, y = 1

B. x = 2, y = 3

C. x = 2, y = 4

D. x = 3, y = 3

Answer:

Comparing the equations,

2x + y = 7 & 4x = x + 6

3x = 6, x = 2

2(2) + y = 7

y = 7 – 4 = 3

x = 2 & y = 3

Option (B) is the answer.

Question 36.

If A is a square matrix such that A² = I, then (A – I)³ + (A + I)³ – 7A is equal to
A. A

B. I – A

C. I + A

D. 3A

Answer:

Expansion of the given expression,

A³ – I³ + 3AI² – 3A²I + A³ + I³ + 3AI² + 3A²I – 7A

2A³ – 7A + 6AI²

2A²A – 7A + 6A

2AI – A {A² = I}

2A – A

Option (A) is the answer.

Question 37.

If A and B are two matrices of order 3 × m and 3 × n respectively and m = n, then the order of 5A – 2B is
A. m × 3

B. 3 × 3

C. m × n

D. 3 × n

Answer:

m = n

If A and B are two matrices of order 3 × m and 3 × n respectively and m = n

Then, A & B have same orders as 3 × n each,

So the order of (5A – 2B) should be same as 3 × n.

Option (D) is the answer.

Question 38.

If A is a matrix of order m × n and B is a matrix such that AB^T and B^TA are both defined, then the order of matrix B is
A. m × n

B. n × n

C. n × m

D. m × n

Answer:

Let &

As AB’, is a defined matrix, {Given}

So n = q

BA’ is also a defined matrix, {Given}

So, p = m

Hence order of B is m × n.

Option (D) is the answer.

Question 39.

If A and B are matrices of the same order, then AB^T – B^TA is a
A. skew-symmetric matrix

B. null matrix

C. unit matrix

D. symmetric matrix

Answer:

A & B are matrices of same order,

Let K = (AB^T- BA^T)

= (AB^T)^T - (BA^T)^T

= (B^T)^T(A)^T- (A^T)^TB^T

= BA^T - AB^T

= -(AB^T - BA^T)

= -K

Hence, (AB^T – B^TA) is a skew - symmetric matrix .

Option (A) is the answer.

Question 40.

If matrix , where , then A² is equal to
A. I

B. A

C. O

D. – I

Answer:

As per the given conditions,

= I, which is an Identity Matrix.

Option (A) is the answer.

Question 41.

If , then A – B is equal to
A. I

B. 0

C. 2I

D.

Answer:

Option (D) is the answer.

Question 42.

If A and B are square matrices of the same order, then (A + B)(A – B) is equal to
A. A² – B²

B. A² – BA – AB – B²

C. A² – B² + BA – AB

D. A² – BA + B² + AB

Answer:

(A + B)(A – B) = A (A – B) + B (A – B)

= A.A – A.B + B.A – B.B

=A² – A.B + B.A – B.B

= A² – AB + BA – BB

Matrix multiplication does not have a commutative property i.e.., A.B ≠ B.A

Hence, Option (C) is the answer.

Question 43.

If and , then
A. only AB is defined

B. only BA is defined

C. AB and BA both are defined

D. AB and BA both are not defined

Answer:

As the matrices, A & B, both are defined, having orders as 2 × 3 & 3 × 2,

So multiplication of matrices is defined. (AB ≠ BA)

Matrix multiplication is defined only if

Hence, AB and BA both are defined.

Option (C) is the answer.

Question 44.

The matrix is a
A. diagonal matrix

B. symmetric matrix

C. skew-symmetric matrix

D. scalar matrix

Answer:

Skew-Symmetric Matrix

Option (C) is the answer.

Question 45.

The matrix is
A. identify matrix

B. symmetric matrix

C. skew-symmetric matrix

D. diagonal matrix

Answer:

As the elements off – diagonal are not zero unlike the on – diagonal elements, so it is a diagonal matrix.

Option (D) is the answer.

Algebra Of Matrices

Class 12th Mathematics RD Sharma Volume 1 Solution

Exercise 5.1

Exercise 5.2

Exercise 5.3

Exercise 5.4

Exercise 5.5

Very Short Answer

Mcq

Class 12^th Mathematics RD Sharma Volume 1 Solution