= (x2 – x + 1) × (x + 1) – (x + 1) × (x – 1)
= x (x2 – x + 1) + 1 (x2 – x + 1) – (x2 – 1)
[∵ (a – b)(a + b) = (a2 – b2)]
= x3 – x2 + x + x2 – x + 1 – x2 + 1
= x3 – x2 + 2
Using the properties of determinants in evaluate:
Let
Expanding |A| along C1, we get
= (a + x) [(a + y)(a + z) – yz] – y [(x)(a + z) – xz] + z [xy – x (a + y)]
= (a + x) [a2 + az + ya + yz – yz] – y [ax + xz – xz] + z [xy – xa + xy]
= (a + x) [a2 + az + ya] – y [ax] + z [– xa]
= a(a2 + az + ya) + x(a2 + az + ya) – yax – zxa
= a3 + a2z + ya2 + xa2 + xaz + xya – yax – zxa
= a3 + a2z + ya2 + xa2
= a2 (a + z + y + x)
Using the properties of determinants in evaluate:
Let
Expanding |A| along C1, we get
= 0 – xy2 [(x2y)(0) – (x2z)(yz2)] + xz2 [(x2y)(zy2) – 0]
= – xy2 [– (x2yz3)] + xz2 [(x2zy3)]
= x3y3z3 + x3z3y3
= 2x3y3z3
Using the properties of determinants in evaluate:
Let
By applying C1→ C1 + C2 + C3, we have
Taking (x + y + z) common from Column C1, we get
By applying R2→ R2 – R1, we get
By applying R3→ R3 – R1, we get
Applying C2→ C2 – C3, we get
Now, expanding along C1, we get
= (x + y + z) [1×{(3y)(2z + x) – (-3z)(x – y)}]
= (x + y + z) [6yz + 3yx + (3z)(x – y)]
= (x + y + z) [6yz + 3yx + 3zx – 3zy]
= (x + y + z) [3yz + 3zx + 3yx]
= 3(x + y + z)(yz + zx + yx)
Using the properties of determinants in evaluate:
By applying C1→ C1 + C2 + C3, we get
Taking (3x + 4) common from first column, we get
By applying R2→ R2 – R1, we get
By applying R3→ R3 – R1, we get
Now, expanding along first column, we get
= (3x + 4) [1×{(16) – 0}]
= (3x + 4)(16)
= 16(3x + 4)
Using the properties of determinants in evaluate:
By applying R1→ R1 + R2 + R3, we get
Taking (a + b +c) common from the first row, we get
By applying C2→ C2 – C1, we get
By applying C3→ C3 – C1, we get
Now, expanding along first row, we get
= (a + b+ c)[1×{-(a + b + c)×{-(a + b + c)} – 0}]
= (a + b + c)[(a + b + c)2]
= (a + b + c)3
Using the properties of determinants in prove that:
Taking LHS,
Firstly, multiply and divide R1, R2, R3 by x, y, z respectively, we get
[rearrange the terms]
Taking xyz common from the first and second column, we get
Applying C3→ C3 + C1, we get
Taking common (xy + yz + xz) common from C3, we get
If any two columns (or rows) of a determinant are identical (all corresponding elements are same), then the value of determinant is zero.
Here, C2 and C3 are identical.
Hence,
∴ LHS = RHS
Hence Proved
Using the properties of determinants in prove that:
Taking LHS,
By applying R1→ R1 + R2 + R3, we get
Taking 2 common from the first row, we get
Applying R1→ R1 – R2, we get
Applying R3→ R3 - R1, we get
Applying R2→ R2 – R1, we get
Taking y, z, x common from R1, R2 and R3 respectively, we get
Expanding along C1, we get
= 2xyz [(1){(1) – 0} – (1){0 – 1} + 0}]
= 2xyz [1 + 1]
= 4xyz
= RHS
Hence,
∴ LHS = RHS
Hence Proved
Using the properties of determinants in prove that:
Taking LHS,
Applying R1→ R1 – R2, we get
[∵(a2 – b2) = (a – b)(a + b)]
Taking (a – 1) common from the first row, we get
Applying R2→ R2 – R3, we get
Taking (a – 1) common from the second row, we get
Now, expanding along C3, we get
= (a – 1)2 [1{(a + 1) – 2}]
= (a – 1)2 [a + 1 – 2]
= (a – 1)3
= RHS
Hence, LHS = RHS
Hence Proved
If A + B + C = 0, then prove that
Given: A + B + C = 0
To Prove:
Taking LHS,
Expanding along the first row, we get
= [1{1 – cos2A} – cos C {cos C – cos B cos A} + cos B {cos C cos A – cos B}]
= {1 – cos2A} – {cos2C – cos A cos B cos C} + {cos A cos B cos C – cos2B}
= {sin2A} – cos2C + cos A cos B cos C + cos A cos B cos C – cos2B
[∵ cos2x + sin2x = 1]
= sin2A – cos2C – cos2B + 2cos A cos B cos C
= -(cos2B – sin2A)– cos2C + 2cos A cos B cos C
= -[cos(B + A)cos(B – A)] + cos C [2 cos A cos B – cos C]
[∵ cos2B – sin2A = cos(B + A)cos(B – A)]
= -[cos(B + A)cos(B – A)] + cos C [cos(A + B) + cos(A – B) – cos C]…(i)
[∵ 2cos A cos B = cos(A + B) + cos(A – B)]
It is given that A + B + C = 0
⇒ A + B = - C
Putting the value of A + B in eq (i), we get
= -[cos(- C) cos(B – A)] + cos C [cos(-C) + cos (A – B) – cos C]
= -cos C cos(B – A) + cos C[cos C + cos(A – B) – cos C]
[∵ cos(-C) = cos C]
Now, cos(A – B) = cos A cos B + sin A sin B
= -cos C{cos B cos A + sin B sin A} + cos C [cos A cos B + sin A sin B]
= 0 = RHS
Hence Proved
If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are (x1, y1), (x2, y2), (x3, y3), then
The coordinates of the vertices of an equilateral triangle are (x1, y1), (x2, y2), (x3, y3).
So, the area of triangle with given vertices is given by
Given that the length of the sides of an equilateral triangle = a
Also, area of equilateral triangle =
Now, squaring both the sides, we get
Hence Proved
Find the value of θ satisfying
We have,
Expanding along R1, we get
⇒ (1){-6 – {(-7) cos2θ}} – 1{8 – 7cos2θ} + sin3θ {28 – 21} = 0
⇒ – 6 + 7cos2θ – 8 + 7cos2θ + 7sin3θ = 0
⇒ 14cos2θ + 7sin3θ – 14 = 0
⇒ 2cos2θ + sin3θ – 2 = 0
Now, we know that
cos 2θ = 1 – 2sin2θ
sin 3θ = 3sinθ – 4sin3θ
⇒ 2(1 – 2sin2θ) + (3sinθ – 4sin3θ) – 2 = 0
⇒ 2 – 4sin2θ + 3sinθ – 4sin3θ – 2 = 0
⇒ -2 + 4sin2θ - 3sinθ + 4sin3θ + 2 = 0
⇒ sinθ (4sinθ – 3 + 4sin2θ) = 0
⇒ sinθ (4sin2θ – 6sinθ + 2sinθ – 3) = 0
⇒ sinθ [2sinθ(2sinθ – 3) + 1(2sinθ – 3)] = 0
⇒ sinθ (2sinθ + 1)(2sinθ – 3) = 0
⇒ sinθ = 0 or 2sinθ + 1 = 0 or 2sinθ – 3 = 0
⇒ θ = nπ or 2sinθ = -1 or 2sinθ = 3
⇒
⇒ θ = nπ ; m, n ∈ Z
If then find values of x.
We have,
By applying C1→ C1 + C2 + C3, we get
Taking (12 + x) common from the first column, we get
By applying C2→ C2 + C3, we get
Applying R2→ R2 – R3, we get
Applying R3→ R3 – R1, we get
Expanding along first column, we get
⇒ (12 + x)[(1){0 – (2x)(-2x)}] = 0
⇒ (12 + x)(4x2) = 0
⇒ 12 + x = 0 or 4x2 = 0
⇒ x = -12 or x = 0
Hence, the value of x = -12 and 0
If a1, a2, a3, ..., ar are in G.P., then prove that the determinant is independent of r.
Given: a1, a2…, ar are in G.P
We know that, ar+1 = AR(r+1)-1 = ARr …(i)
[∵an = arn-1, where a = first term and r = common ratio]
where A = First term of given G.P
and R = common ratio of G.P
…[from(i)]
Taking ARr, ARr+6 and ARr+10 common from R1, R2 and R3 respectively, we get
If any two columns (or rows) of a determinant are identical (all corresponding elements are same), then the value of determinant is zero.
Here, R1 and R2 are identical.
Hence Proved
Show that the points (a + 5, a – 4), (a – 2, a + 3) and (a, a) do not lie on a straight line for any value of a.
Given points are (a + 5, a – 4), (a – 2, a + 3) and (a, a)
To Prove: these points don’t lie on straight line for any value of a
So, we have to show that these points form a triangle.
Area of triangle:-
Applying R2→ R2 – R1, we get
Applying R3→ R3 – R1, we get
Now, expanding along third column, we get
Hence, given points form a triangle i.e. points do not lie on a straight line.
Hence Proved
Show that the Δ ABC is an isosceles triangle if the determinant
We have,
Applying C2→ C2 – C1, we get
Taking common cos B – cos A from second column, we get
Applying C3→ C3 – C1, we get
⇒
Taking common cos C – cos A from column third, we get
Now, expanding along first row, we get
⇒ (cos B – cos A)(cos C – cos A)[(1){cos C + cos A + 1 – (cos B + cos A + 1)}] = 0
⇒ (cos B – cos A)(cos C – cos A)[cos C + cos A + 1 – cos B – cos A – 1] = 0
⇒ (cos B – cos A)(cos C – cos A)(cos C – cos B) = 0
⇒ cos B – cos A = 0 or cos C – cos A = 0 or cos C – cos B = 0
⇒ cos B = cos A or cos C = cos A or cos C = cos B
⇒ B = A or C = A or C = B
Hence, ΔABC is an isosceles triangle.
Find A–1 if and show that
We have,
We have to find A-1 and
Firstly, we find |A|
Expanding |A| along C1, we get
= 0 – 1 {0 – 1} + 1 {1 – 0}
= 1 + 1
= 2
Now, we have to find adj A and for that we have to find co-factors:
Now, we have to show that
= A-1
Hence Proved
If find A–1. Using A–1, solve the system of linear equations x – 2y = 10, 2x – y – z = 8, –2y + z = 7.
We have,
We have to find A-1 and
Firstly, we find |A|
Expanding |A| along C1, we get
= (-1 + 2) + 2 (0) + 0
= 1
Now, we have to find adj A and for that we have to find co-factors:
Now, the system of linear equation is
x – 2y = 10
2x – y – z = 8
-2y + z = 7
We know that, AX = B
Here,
and we can see that this matrix is the transpose of the given matrix. So, transpose of A-1 is
⇒ X = A-1B
∴ x = 0, y = -5 and z = -3
Using matrix method, solve the system of equations 3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2
Given system of equations is:
3x + 2y – 2z = 3,
x + 2y + 3z = 6,
2x – y + z = 2
We know that,
AX = B
i.e.
∴ X = A-1 B
So, firstly, we have to find the A-1 and
Firstly, we find |A|
Expanding |A| along C1, we get
= 3(2 + 3) – 1(2 – 2) + 2(6 + 4)
= 3(5) + 2(10)
= 15 + 20
= 35
Now, we have to find adj A and for that we have to find co-factors:
Now, X = A-1B
∴ x = 1, y = 1 and z = 1
Given find BA and use this to solve the system of equations y + 2z = 7, x – y = 3, 2x + 3y + 4z = 17.
We have,
BA = 6I …(i)
Now, given system of equations is:
y + 2z = 7,
x – y = 3,
2x + 3y + 4z = 17
So,
Applying R1→ R2, we get
Applying R2→ R3, we get
Now, …(ii)
So, BA = 6I [from eq(i)]
and
Now, putting the value of B-1, in eq. (ii), we get
⇒
∴ x = 4 , y = 2 and z = -1
If a + b + c ≠ 0 and then prove that a = b = c.
Let
Applying C1→ C1 + C2 + C3, we get
Taking (a + b + c) common from the first column, we get
Now, Expanding along C1, we get
= (a + b + c)[(1)(bc – a2) – (1)(b2 – ac) + (1)(ba – c2)]
= (a + b + c)[bc – a2 – b2 + ac + ab – c2]
= (a + b + c)[-(a2 + b2 + c2 – ab – bc – ac)]
[∵ (a – b)2 = a2 + b2 – 2ab]
Given that Δ = 0
⇒ (a + b + c)[(a – b)2 + (b – c)2 + (c – a)2] = 0
Either (a + b + c) = 0 or (a – b)2 + (b – c)2 + (c – a)2 = 0
but it is given that (a + b + c) ≠ 0
∴(a – b)2 + (b – c)2 + (c – a)2 = 0
⇒ a – b = b – c = c – a = 0
⇒ a = b = c
Hence Proved
Prove that is divisible by a + b + c and find the quotient.
We have,
Applying R1→ R1 – R2, we get
Taking (a + b+ c) common from first row, we get
Applying R2→ R2 – R3, we get
Taking (a + b+ c) common from second row, we get
Applying C1→ C1 + C2 + C3, we get
Now, expanding along C1, we get
= (a + b + c)2[ab + bc + ca – (a2 + b2 + c2){(c – b)(b – a) – (a – c)2}]
= (a + b + c)2[ab + bc + ca – (a2 + b2 + c2){(cb – ac – b2 + ab – (a + c2 – 2ac)}]
= (a + b + c)2[ab + bc + ca – (a2 + b2 + c2){(cb – ac – b2 + ab – a - c2 + 2ac)}]
= (a + b + c)2[ab + bc + ca – (a2 + b2 + c2){ac + bc + ab – (a2 + b2 + c2)}]
=(a + b + c)2[ab + bc + ca – (a2 + b2 + c2)]2
=(a + b + c)(a + b + c)[ab + bc + ca – (a2 + b2 + c2)
Hence, given determinant is divisible by (a + b + c) and Quotient is (a + b + c)[ab + bc + ca – (a2 + b2 + c2)
If x + y + z = 0, prove that
Given: x + y + z = 0
To Prove:
Taking LHS,
Expanding along the first row, we get
= xa{(za)(ya) – (xc)(xb)} – (yb){(yc)(ya) – (zb)(xb)} + (zc){(yc)(xc) – (zb)(za)}
= xa{a2yz – x2bc} – yb{y2ac – b2xz} + zc{c2xy – z2ab}
= a3xyz – x3abc – y3abc + b3xyz + c3xyz – z3abc
= xyz(a3 + b3 + c3) – abc(x3 + y3 + z3)
It is given that x + y + z = 0
⇒ x3 + y3 + z3 = 3xyz
= xyz(a3 + b3 + c3) – abc (3xyz)
= xyz(a3 + b3 + c3 – 3abc)
Hence Proved
If then value of x is
A. 3
B. ± 3
C. ± 6
D. 6
We have,
⇒ (2x)(x) – (5)(8) = (6)(3) – (7)(-2)
⇒ 2x2 – 40 = 18 – (-14)
⇒ 2x2 – 40 = 18 + 14
⇒ x2 – 20 = 9 + 7
⇒ x2 – 20 = 16
⇒ x2 = 16 + 20
⇒ x2 = 36
⇒ x = √36
⇒ x = ±6
Hence, the correct option is (c)
The value of determinant
A. a3 + b3 + c3
B. 3 bc
C. a3 + b3 + c3 – 3abc
D. none of these
We have,
Applying C2→ C2 + C3, we get
Taking (a + b + c) common from second column, we get
Applying C1 → C1 – C3, we get
Expanding along first row, we get
= (a + b + c)[(-b){c – b} – (1){-c2 – (-ab)} + a{-c – (-a)}]
= (a + b + c)(-bc + b2 + c2 – ab – ac + a2)
= a(-bc + b2 + c2 – ab – ac + a2) + b(-bc + b2 + c2 – ab – ac + a2) + c(-bc + b2 + c2 – ab – ac + a2)
= -abc + ab2 + ac2 – a2b – a2c + a3 – b2c + b3 + bc2 – ab2 – abc + a2b – bc2 + b2c + c3 – abc – ac2 + a2c
= a3 + b3 + c3 – 3abc
Hence, the correct option is (c)
The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be
A. 9
B. 3
C. – 9
D. 6
We know that, the area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is given by
[given]
Now, expanding along second column, we get
⇒ -(k) {-3 – 3} = 18
⇒ -k (-6) = 18
⇒ 6k = 18
⇒ k = 3
Hence, the correct option is (b)
The determinant equals
A. abc (b–c) (c – a) (a – b)
B. (b–c) (c – a) (a – b)
C. (a + b + c) (b – c) (c – a) (a – b)
D. None of these
We have,
Taking (b – a) common from C1 and C3, we get
Applying C1→ C1 – C3, we get
If any two columns (or rows) of a determinant are identical (all corresponding elements are same), then the value of determinant is zero.
Here, C1 and C2 are identical.
Hence, the correct option is (d).
The number of distinct real roots of in the interval is
A. 0
B. –1
C. 1
D. None of these
We have,
Applying C1→ C1 + C2 + C3, we get
Taking (2cos X + sin X) common from the first column, we get
Applying R2→ R2 – R1, we get
Applying R3→ R3 – R1, we get
Expanding |A| along C1, we get
⇒ (2cos X + sin X) [(1){(sin X – cos X)(sin X – cos X)}]
⇒ (2cos X + sin X)(sin X – cos X)2 = 0
⇒ 2cos X = -sin X or (sin X – cos X)2 = 0
⇒ tan X = -2 or tan X = 1
but tan X = -2 is not possible as for
So, tan X = 1
Hence, only one real distinct root exist.
Hence, the correct option is (c)
If A, B and C are angles of a triangle, then the determinant is equal to
A. 0
B. –1
C. 1
D. None of these
We have,
Expanding along C1, we get
= [(-1){1 – cos2A} – cos C{-cos C – cos Acos B} + cos B{cos A cos C + cos B}]
= -1 + cos2A + cos2C + cos A cos B cos C + cos A cos B cos C + cos2B
= -1 + cos2A + cos2B + cos2C + 2cos A cos B cos C
Here, we use formula
1 + cos2A = 2cos2A
Taking L.C.M, we get
Now, we use the formula:
cos(A + B) cos(A – B) = 2cos Acos B
so, cos2A + cos2B = 2 cos(A + B) cos(A – B)
…(i)
Since, A, B and C are the angles of a triangle and we know that the sum of the angles of a triangle = 180° or π
⇒ A + B + C = π
⇒ A + B = π – C
Putting the value of (A+B) in eq. (i), we get
[∵ cos(π – x) = -cos X]
= -cos C{cos(A – B) – cos C} + 2cos Acos Bcos C
= -cos C[cos(A – B) – cos{π – (A + B)}] + 2cos Acos Bcos C
= -cos C[cos(A – B) + cos(A + B)] + + 2cos Acos Bcos C
= -cos C[2cos Acos B] + 2cos Acos Bcos C
= 0
Let then is equal to
A. 0
B. –1
C. 2
D. 3
We have,
Dividing R2 and R3 by ‘t’
Expanding along the first row, we get
= (1)(1 – 2) + (1)(2 – 1)
= - 1 + 1
= 0
Hence, the correct option is (A)
The maximum value of is (θ is a real number)
A.
B.
C.
D.
Given
Performing operations C1→ C2 – C3 and C2→ C2 – C3
= 0 – 0 + 1 (sin θ. cos θ)
Multiply and divide by 2,
= 1/2 (2sin θ cos θ)
We know that 2 sin θ cos θ = sin 2θ
= 1/2 (sin 2θ)
Since the maximum value of sin 2θ is 1, θ = 45°.
∴ Δ = 1/2 (sin 2(45°))
= 1/2 sin 90°
= 1/2 (1)
∴ Δ = 1/2
If ,
A. f (a) = 0
B. f (b) = 0
C. f (0) = 0
D. f (1) = 0
Given
Option (A):
= 0 – 0 + (a – b) [2a (a + c) – 0 (a + b)]
= (a – b) [2a2 + 2ac – 0]
= (a – b) (2a2 + 2ac) ≠ 0
Option (B):
= 0 – (b - a) [(b + a) (0) – (b – c) (2b)] + 0
= - (b – a) [0 - 2b2 + 2bc]
= (a – b) (2b2 – 2bc) ≠ 0
Option (C):
= 0 + a [a (0) – (-bc)] – b [ac – b (0)]
= a [bc] – b [ac]
= abc – abc = 0
Option (D):
= 0 – (1 - a) [(1 + a) (0) – (1 – c) (1 + b)] + (1 – b) [ (1 + a) (1 + c) – 0 (1 + b)]
= - (1 – a) [- (1 – c) (1 + b)] + (1 – b) [(1 + a) (1 + c)]
= (1 – a) (1 – c) (1 + b) + (1 – b) (1 + a) (1 + c) ≠ 0
Hence, option (C) satisfies.
If , then A-1 exists if
A. λ = 2
B. λ ≠ 2
C. λ ≠ -2
D. None of these
Given
⇒ |A| = 2 (6 – 5) - λ (0 – 5) + (-3) (0 – 2)
= 2 + 5λ + 6
= 5λ + 8
We know that A-1 exists if A is non – singular matrix i.e. |A| ≠ 0.
∴ 5λ + 8 ≠ 0
⇒ 5λ ≠ -8
So, A-1 exists if and only if .
If A and B are invertible matrices, then which of the following is not correct?
A. adj A = |A|. A-1
B. det (A)-1 = [det (A)]-1
C. (AB)-1 = B-1 A-1
D. (A + B)-1 = B-1 + A-1
Given A and B are invertible matrices.
Consider (AB) B-1 A-1
⇒ (AB) B-1 A-1 = A(BB-1) A-1
= AIA-1 = (AI) A-1
= AA-1 = I
⇒ (AB)-1 = B-1 A-1 … option (C)
Also AA-1 = I
⇒ |AA-1| = |I|
⇒ |A| |A-1| = 1
∴ det (A)-1 = [det (A)]-1 … (B)
We know that
⇒ adj A = |A|. A-1 … option (A)
But
∴ (A + B)-1 ≠ B-1 + A-1
Hence, option (D) is not correct.
If x, y, z are all different from zero and , then value of x-1 + y-1 + z-1 is
A. x y z
B. x-1 y-1 z-1
C. –x –y –z
D. -1
Given
Applying C1→ C �1 – C3 and C2→ C2 – C3,
⇒
Expanding along R1,
⇒ x [y (1 + z) + z] – 0 + 1 (yz) = 0
⇒ xy + xyz + xz + yz = 0
Dividing by xyz on both sides,
Hence, option (D) satisfies.
The value of the determinant is
A. 9x2 (x + y)
B. 9y2 (x + y)
C. 3y2 (x + y)
D. 7x2 (x + y)
Given matrix
= x [x2 – (x + y) (x + 2y)] – (x + y) [(x + 2y) (x) – (x + y)2] + (x + 2y) [(x + 2y)2 – x (x + y)]
= x [x2 – x2 – 3xy – 2y2] – (x + y) [x2 + 2xy – x2 – 2xy – y2] + (x + 2y) [x2 + 4xy + 4y2 – x2 – xy]
= x [-3xy – 2y2] – (x + y) [-y2] + (x + 2y) [3xy + 4y2]
= -3x2y – 2xy2 + xy2 + y3 +3x2y + 4xy2 + 6xy2 + 8y3
= 9y3 + 9xy2
= 9y2 (x + y) … option (B)
Hence, option B satisfies.
There are two values of a which makes determinant, ,then sum of these number is
A. 4
B. 5
C. -4
D. 9
Given
= 1 [2a2 – (-4)] + 2 [4a – 0] + 5 [8 – 0] = 86
= 1 [2a2 + 4] + 2 [4a] + 5 [8] = 86
= 2a2 + 4 + 8a + 40 = 86
= 2a2 + 8a + 44 = 86
= 2a2 + 8a = 42
= 2 (a2 + 4a) = 42
= (a2 + 4a) = 21
⇒ a2 + 4a – 21 = 0
⇒ (a + 7) (a – 3) = 0
∴ a = -7 or 3
Sum of these numbers is -7 + 3 = -4. [Option (C)]
Fill in the blanks
If A is a matrix of order 3 × 3, then |3A| = ___.
If A is a matrix of order 3 × 3, then |3A| = 27 |A|.
Explanation:
We know that if A = [aij]3×3, then |k.A| = k3|A|
∴ |3A| = 33 |A| = 27 |A|
Fill in the blanks
If A is invertible matrix of order 3 × 3, then |A-1|= ____.
If A is invertible matrix of order 3 × 3, then |A-1| = |A|-1
Explanation:
Given A is invertible matrix of order 3 × 3.
We know that AA-1 = I.
⇒|A||A-1| = 1
|A|-1
Fill in the blanks
If x, y, z ∈ R, then the value of determinant is equal to ___.
If x, y, z ∈ R, then the value of determinant is equal to 0.
Explanation:
Given
Performing the operation C1→ C1 – C2,
We know that (a + b)2 – (a – b)2 = 4ab.
C1 and C3 are proportional to each other.
Fill in the blanks
If cos 2θ = 0, then
If cos 2θ = 0, then
Explanation:
Given cos 2θ = 0
⇒ cos 2θ = cos π/2
⇒ 2θ = π/2
∴ θ = π/4
Then
Fill in the blanks
If A is a matrix of order 3 × 3, then (A2)-1 = ____.
If A is a matrix of order 3 × 3, then (A2)-1 = (A-1)2.
Explanation:
We know that if A is a matrix of order 3 × 3, then (A2)-1 = (A-1)2.
Fill in the blanks
If A is a matrix of order 3 × 3, then number of minors in determinant of A are ___.
If A is a matrix of order 3 × 3, then number of minors in determinant of A are 9.
Explanation:
In a 3 × 3 matrix, there are 9 elements.
So, there are 9 minors of these elements.
Fill in the blanks
The sum of the products of elements of any row with the co-factors of corresponding elements is equal to ___.
The sum of the products of elements of any row with the co-factors of corresponding elements is equal to Δ (Determinant).
Explanation:
If ,
then |A| = a11C11 + a12C12 + a13C13
= Sum of products of elements of R1 with their corresponding cofactors
= Δ (Determinant)
Fill in the blanks
If x = -9 is a root of , then other two roots are ___.
If x = -9 is a root of , then other two roots are 2 and 7.
Explanation:
Given x = -9 is a root of .
⇒ x [x2 – 12] – 3 [2x – 14] + 7 [12 – 7x] = 0
⇒ x3 – 12x – 6x + 42 + 84 – 49x = 0
⇒ x3 – 67x + 126 =0
Here, 126 × 1 = 9 × 2 × 7
For x = 2,
⇒ 23 – 67 (2) + 126 = 134 – 134 = 0
∴ x = 2 is one root.
For x = 7,
⇒ 73 – 67 (7) + 126 = 469 – 469 = 0
∴ x = 7 is also one root.
Fill in the blanks
Explanation:
Given
Performing the operation C1→ C1 – C3
Taking (z – x) common from C1,
= (z – x) [1 [0 – (y – z) (z – y)] - (xyz) [0 - (y - z)] + (x – z) [(z – y) – 0]]
= (z – x) (z – y) (-y + z – xyz + x – z)
= (z – x) (z – y) (x – y – xyz)
= (z – x) (y – z) (y – x + xyz)
Fill in the blanks
If , then A = ____.
If , then A = 0.
Explanation:
Given
Here R1 and R3 are identical.
∴ A = 0
State True or False for the statements
(A3)-1 = (A-1)3, where A is a square matrix and |A| ≠ 0.
True
We know that (An)-1 = (A-1)n, where n∈N.
∴ (A3)-1 = (A-1)3
State True or False for the statements
(aA)-1 = (1/a) A-1, where a is any real number and A is a square matrix.
False
We know that if A is a non-singular square matrix, then for any scalar a, aA is invertible such that
i.e. , where a is any non-zero scalar.
In the above statement a is any real number So, we can conclude that above statement is false.
State True or False for the statements
|A-1| ≠ |A|-1, where A is non-singular matrix.
False
Given A is non-singular matrix.
We know that AA-1 = I.
⇒|A||A-1| = 1
∴ |A-1| = 1/|A| = |A|-1
State True or False for the statements
If A and B are matrices of order 3 and |A| = 5, |B| = 3, then |3AB| = 27 × 5 × 3 = 405.
True
We know that |AB| = |A|. |B| and if A = [aij]3×3, then |k.A| = k3|A|.
∴ |3A| = 27 |AB|
= 27 |A| |B|
= 27 (5) (3)
= 405
Hence proved.
The above statement is true.
State True or False for the statements
If the value of a third order determinant is 12, then the value of the determinant formed by replacing each element by its co-factor will be 144.
True
Let A be the determinant.
∴ |A| = 12
We know that if A is a square matrix of order n, then |adj A| = |A|n-1
For n = 3, |adj A| = |A|3-1
= |A|2
= 122 = 144
Hence the above statement is true.
State True or False for the statements
, where a, b, c are in A.P.
True
Since a, b, c are in AP, 2b = a + c.
Performing the operation R1→ R1 + R3
Since 2b = a + c,
Here R1 and R2 are proportional to each other,
⇒ 0 = 0
Hence the above statement is true.
State True or False for the statements
|adj. A| = |A|2, where A is a square matrix of order two.
False
We know that if A is a square matrix of order n, then |adj A| = |A|n-1
Here n =2,
⇒ |adj A| = |A|n-1 = |A|
Hence the statement is false.
State True or False for the statements
The determinant is equal to zero.
True
Here in first determinant C1 and C3 are identical.
Taking cos A and cos B common from C2 and C3.
Here C2 and C3 are identical.
= 0
Hence the statement is true.
State True or False for the statements
If the determinant splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.
True
Given
Splitting first row,
Splitting second row,
Similarly, we can split these 4 determinants in 8 determinants by splitting each one in two determinants further.
So given statement is true.
State True or False for the statements
Let ,then
True
Given
We have to prove that
Performing the operation,
Taking 2 common from C1 and performing the operation C1 → C1 – C2 and C2→ C2 – C3
Here in the second determinant C1 and C2 ae identical.
Here in the second determinant C1 and C3 are identical.
= 2 (16) = 32
Hence the above statement is true.
State True or False for the statements
The maximum value of is 1/2.
True
Given
Performing operations R2→ R2 – R1 and R3→ R3 – R1
= cos θ. sin θ
Multiply and divide by 2,
= 1/2 (2sin θ cos θ)
We know that 2 sin θ cos θ = sin 2θ
= 1/2 (sin 2θ)
Since the maximum value of sin 2θ is 1, θ = 45°.
∴ Δ = 1/2 (sin 2(45°))
= 1/2 sin 90°
= 1/2 (1)
∴ Δ = 1/2
Hence the above statement is true.