Matrices

(i) In the given matrix, the number of rows is 3 and the number of columns is 4.

Therefore, the order of the matrix is 3 × 4.

(ii) Since, the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements in it.

(iii) a₁₃ = 19, a₂₁ = 35, a₃₃ = -5, a₂₄ = 12, a₂₃ = .

It is known that if a matrix is of the order m × n, then it has mn elements.

Therefore, to find all the possible orders of a matrix having 24 elements, we had to find all the ordered pairs of natural numbers whose product is 24.

The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and (6, 4).

Therefore, the possible orders of a matrix having 24 elements are;

1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, 6 × 4

(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.

Therefore, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.

It is known that if a matrix s of the order m × n, then it has mn elements.

Therefore, to find all the possible orders of a matrix having 18 elements, we had to find all the ordered pairs of natural numbers whose product is 18.

The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6) and (6, 3).

Therefore, the possible orders of a matrix having 24 elements are;

1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, 6 × 3

(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.

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

In general, a 2 × 2 matrix is given by A =

a_ij =

Therefore,

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Therefore, the required matrix is A =

In general, a 2 × 2 matrix is given by A =

a_ij =

Therefore,

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Therefore, the required matrix is A = .

In general, a 2 × 2 matrix is given by A =

a_ij =

Therefore,

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Therefore, the required matrix is A =

In general 3 × 4 matrix is given by A =

Therefore,

a₃₂ =

a₁₃ =

a₂₃ =

a₃₃ =

a₁₄ =

a₂₄ =

a₃₄ =

Therefore, required matrix is A =

In general 3 × 4 matrix is given by A =

a_ij = 2i-j, i = 1,2,3 and j = 1,2,3,4

Therefore,

a₁₁ = 2 × 1 - 1 = 2 - 1 = 1

a₂₁ = 2 × 2 - 1 = 4 - 1 = 3

a₃₁ = 2 × 3 - 1 = 6 - 1 = 5

a₁₂ = 2 × 1 - 2 = 2 - 2 = 0

a₂₂ = 2 × 2 - 2 = 4 - 2 = 2

a₃₂ = 2 × 3 - 2 = 6 - 2 = 4

a₁₃ = 2 × 1 - 3 = 2 - 3 = -1

a₂₃ = 2 × 2 - 3 = 4 - 3 = 1

a₃₃ = 2 × 3 - 3 = 6 - 3 = 3

a₁₄ = 2 × 1 - 4 = 2 - 4 = -2

a₂₄ = 2 × 2 - 4 = 4 - 4 = 0

a₃₄ = 2 × 3 - 4 = 6 - 4 = 2

Therefore, required matrix is A =

Since, the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding element, we have:

x = 1, y = 4 and z = 3

Since, the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding element, we have:

x + y = 6, xy = 8, 5 + z = 5

Now, 5 + z = 5
⇒ z = 0

Since, the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding element, we have:

x + y + z = 9...(1)

x + z = 5.....(2)

y + z = 7.......(3)

y + 5 = 9

⇒ y = 4

Then, putting the value of y in equation 3, we get,

4 + z = 7

⇒ z = 3

Therefore, x + z = 5

⇒ x = 2

Therefore, x =2, y = 4 and z = 3.

Since, the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding element, we have:

a –b = -1 …(1)

2a – b = 0 …(2)

2a + c = 5 …(3)

3c + d = 13 …(4)

From equation (2), we get:

b = 2a

Then, from eq. (1), we get,

a -2a = -1

⇒ a = 1

⇒ b = 2

Now, from eq. (3), we get:

2 × 1 + c = 5

⇒ c = 3

From (4), we get,

3 × 3 + d = 13

⇒ 9 + d = 13

⇒ d = 4

Therefore, a = 1, b = 2, c = 3 and d = 4.

We know that if a given matrix is said to be square matrix if the number of rows is equal to the number of columns.

Therefore, A = [a_ij]_{m × n} is a square matrix, if m = n.

Now,

Since, the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding element, we have:

3x + 7 = 0

5 = y – 2

⇒ y = 7

y + 1 = 8

⇒ y = 7

And 2 – 3x = 4

Thus, on comparing the corresponding elements of the two matrices, we get different values of x, which is not possible.

Therefore, it is not possible to find the values of x and y for which the given matrices are equal.

The given matrix of the order 3 × 3 has 9 elements with each entry 0 or 1.

Now, each of the 9 elements can be filled in two possible ways.

Hence, the required number of possible matrices is 2⁹ = 512.

i) A + B =

ii) A - B








iii) 3A - C







iv) AB





v) BA

A – B =

3A – C =

AB

BA =

.

.

.

Now, A+ B

B – C

A + (B –C)

.

+ B) – C =

.

Therefore, A + (B –C) = (A + B) – C

3A – 5B

.

Now, X + Y = …(1)

Adding (1) and (2), we get,

2X =

.

Now, X + Y =

2X + 3Y …(1)

3X + 2Y …(2)

Now, multiply equation (1) by 2 and equation (2) by 3, we get,

4X + 6Y …(3)

9X + 6Y …(4)

Subtracting equation (4) from (3), we get,

(4X + 6Y) – (9X + 6Y)

Now, 2X + 3Y

2X + Y

Now, on comparing elements of these two matrices, we get,

2 + y = 5

=> y = 3

And 2x + 2 = 8

=> x = 3

Therefore, x = 3 and y = 3.

On comparing the elements of these two matrices, we get,

2x + 3 = 9

⇒ 2x = 6

⇒ x = 3

2y = 12

⇒ y = 6

2z -3 = 15

⇒ 2z = 18

⇒ z = 9

2t +6 = 18

⇒ 2t = 12

⇒ t = 6

Therefore, x = 3, y =6, z = 9 and t = 6.

On comparing the corresponding elements of these two matrices, we get,

2x –y = 10 and 3x + y =5

Now, adding above two equations, we get

5x = 15

⇒ x = 3

Now, 3x + y = 5

⇒ y = 5 - 3x

⇒ y = 5 - 9 = -4

Therefore, x = 3 and y = -4.

On comparing the corresponding elements of these two matrices, we get,

3x = x + 4

⇒ 2x + 4

⇒ x =2

3y = 6 + x + y

⇒ 2y = 6 + x = 6 + 2 = 8

⇒ y = 4

3w = 2w + 3

⇒ w = 3

3z = -1 + z + w

⇒ 2z = -1 + w = -1 +3 = 2

⇒ z = 1

Therefore, x = 2, y = 4, z = 1 and w = 3.

Therefore, F(x)F(y) = F(x+y)

A² = A.A

Now, A² – 5A + 6I

A² = A.A

Now, A³ = A². A

Now, A³ – 6A² + 7A + 2I

Therefore, A³ – 6A² + 7A + 2I = 0

A² = A.A

Now, A² = kA – 2I

Comparing the corresponding elements, we get,

3k -2 = 1

⇒ 3k = 3

⇒ k = 1

Therefore, the value of k is 1.

We know that I

Now, LHS = I + A

=

And on RHS = (I – A)

As we know,

cos 2θ = 1 - 2 sin² θ

cos 2θ = 2 cos² θ - 1

and sin2θ = 2 sinθ cosθ

As tanθ = sinθ/cosθ

=

= LHS

Hence Proved.

(a) Let Rs x be invested in the first bond.

Then, the sum of money invested in the second bond will be Rs. (30000 – x).

It is given that the first bond pays 5% interest per year, and the second bond pays 7% interest per year.

Thus, in order to obtain an annual total interest of Rs. 1800, we get:

⇒ 5x + 210000 -7x = 180000

⇒ 210000 -2x = 180000

⇒ 2x = 210000 – 180000

⇒ 2x = 30000

⇒ x = 15000

Therefore, in order to obtain an annual total interest of Rs. 1800, the trust fund should invest Rs. 15000 in the first bond and the remaining Rs. 15000 in the second bond.

(a) Let Rs x be invested in the first bond.

Then, the sum of money invested in the second bond will be Rs (30000 – x).

It is given that the first bond pays 5% interest per year, and the second bond pays 7% interest per year.

Thus, in order to obtain an annual total interest of Rs 1800, we get:

⇒ 5x + 210000 -7x = 200000

⇒ 210000 -2x = 200000

⇒ 2x = 210000 – 200000

⇒ 2x = 10000

⇒ x = 5000

Therefore, in order to obtain an annual total interest of Rs 2000, the trust fund should invest Rs 5000 in the first bond and the remaining Rs 25000 in the second bond.

It is given that the bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books.

Number of chemistry book = 10 × 12
= 120

Number of physics book = 8 × 12

= 96

Number of economics book = 10 × 12

= 120

Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively.

The total amount of money = number of books × selling price

= [120 × 80 + 96 × 60 + 120 × 40]

= (9600 + 5760 + 4800)

= 20160

Therefore, the bookshop will receive Rs. 20160 from the sale of all these books.

Matrices P and Y are of the orders p × k and 3 × k respectively.

Therefore, matrix PY will be defined if k = 3.

Then, PY will be of the order p × k.

Matrices W and Y are of the orders n × 3 and 3 × k respectively.

As, the number of columns in W is equal to the number of rows in Y, Matrix WY is well defined and is of the order n × k.

Matrices PY and WY can be added only when their orders are the same.

Therefore, PY is of the order p × k and WY is of the order n × k.

Thus, we must have p = n.

Therefore, k = 3 and p = n are the restrictions on n, k and p so that

PY + WY will be defined.

Matrix X is of the order 2 × n.

Therefore, matrix 7X is also of the same order.

Matrix Z is of order 2 × p

⇒ 2 × n

⇒ matrix 5Z is also of the same order.

Now, both the matrices 7X and 5Z are of the order 2 × n.

Thus, matrix 7X – 5Z is well- defined and is of the order 2 × n.

We know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if

So, let

Therefore, transpose of the given matrix A is denoted by A’

Hence,

The transpose of the given matrix is .

We know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if

So, let

Therefore, transpose of the given matrix B is denoted by B’.

The transpose of the given matrix is .

We know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if

Therefore, the transpose of the given matrix C is denoted by C’.

The transpose of the given matrix is

(A+B)’ = A’+B’

Explanation: We will first calculate L.H.S i.e. (A+B)’ and then consecutively we will calculate R.H.S and verify that both are equal.

Therefore,

Now,

So,

From equation 1 & 2 we verify that

(A+B)’ = A’+B’. Hence verified.

We will first calculate L.H.S i.e. (A+B)’ and then consecutively we will calculate R.H.S and verify that both are equal.

From equation 1 & 2 we verify that

(A-B)’ = A’-B’. Hence verified.

(A+B)’ = A’+B’

Explanation: Calculate matrix A by taking transpose of A’ and also find the transpose of matrix B and the solve L.H.S and R.H.S separately and then compare the result.

So, A = transpose of A’

B’ = transpose of B

From equation 1 & 2 we verify that

(A+B)’ = A’+B’. Hence verified.

(A-B)’ = A’-B’

Explanation: Calculate matrix A by taking transpose of A’ and also find the transpose of matrix B and the solve L.H.S and R.H.S separately and then compare the result.

From equation 1 & 2 we verify that

(A - B)’ = A’ - B’. Hence verified.

Explanation: Whenever a constant term is multiplied with a matrix then it implies that every elements in each rows and columns are to be multiplied with that constant term. So in order to solve this question we have to multiplied every element of the matrix B with constant term that is 2.

Now, (A+2B)’ = transpose of A+2B

Explanation: The product of two matrices is defined or possible only if the number of columns of the former matrix is equal to number of rows of the latter matrix.

From equation 1 & 2 we verify that

(AB)’ = B’A’. Hence verified.

Explanation: The product of two matrices is defined or possible only if the number of columns of the former matrix is equal to number of rows of the latter matrix.

From equation 1 & 2 we verify that

(AB)’ = B’A’. Hence verified.

We know A’ can be calculated by taking the transpose of the given matrix A.

Now multiply A and A’. So,

And we know ‘I’ represents an identity matrix

From equation 1 & 2 we can say that

AA’ = I

AA’ = I. Hence verified.

.

And we know ‘I’ represents an identity matrix

From equation 1 & 2 we can say that

AA’ = I

AA’ = I. Hence verified.

A matrix is aid to be symmetric only if the transpose of matrix and the matrix itself are equal or same. This means that A = A’.

Now, we know that the transpose of matrix A is

So, from equation 1 & 2 we get

A = A’, hence we can say that Matrix A is a symmetric matrix.

Hence proved.

A matrix is said to be skew symmetric if the transpose of the matrix is equal to the negative of the matrix. This means that A’ = -A.

Now, we know that the transpose of matrix A is

Now if we carefully look at the equation 2 we can rewrite it as

So, A’ = (-1) × A (from equation 1)

⇒ A’ = -A, hence we can say that Matrix A is a skew symmetric matrix.

Hence proved

(A + A’) is a symmetric matrix.

On adding them we get,

Explanation: Now to show that the matrix obtained i.e. (A + A’) is symmetric we need to calculate its transpose and prove that the matrix (A + A’) and its transpose are equal. This means that (A + A’) = (A + A’)’.

. → eqⁿ 2

So, from equation 1 & 2 we get,

(A + A’) = (A + A’)’, hence we can say that (A + A’) is a symmetric matrix.

Ans. Hence proved.

(A – A’) is a skew symmetric matrix.

.

subtracting A’ from A, we get,

Explanation: Now to show that the matrix obtained i.e. (A + A’) is skew symmetric we need to calculate its transpose and prove that the matrix (A + A’) is equal to the negative of its transpose are equal. This means that (A + A’) = -(A + A’)’.

We can rewrite above equation as

Also, (A – A’)’ = (-1) × (A – A’) (from equation 1)

(A – A’)’ = -(A – A’), hence we can say that Matrix A is a skew symmetric matrix.

Hence proved.

(i)

On adding A and A’, we get,

Therefore, (A + A’) = 0

(ii)

On adding A and A’, we get,

We can rewrite the above equation as’

Therefore, (A - A’) = 2(A) (from equation 1)

As per Theorem 2 “Any square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.” So in order to prove this we will be using Theorem 1 which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.”

Now, on adding A and A’ we will get,

.

.

.

.

on subtracting A’ from A we will get,

Now, Add M and N, we get,

.

us, A is represented as the sum of a symmetric matrix M and a skew symmetric matrix N.

Ans. Hence proved

As per Theorem 2 “Any square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.” So in order to prove this we will be using Theorem 1 which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.”

Now, on adding A and A’ we will get,

.

Now, on subtracting A’ from A we will get,

⇒ N’ = -N

Now, Add M and N, we get,

.

Thus, A is represented as the sum of a symmetric matrix M and a skew symmetric matrix N.

As per Theorem 2 “Any square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.” So in order to prove this we will be using Theorem 1 which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.”

Now, on adding A and A’ we will get,

Now, on subtracting A’ from A we will get,

.

w, Add M and N, we get,

Thus, A is represented as the sum of a symmetric matrix M and a skew symmetric matrix N.

Explanation: As per Theorem 2 “Any square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.” So in order to prove this we will be using Theorem 1 which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.”

Now, on adding A and A’ we will get,

Now, on subtracting A’ from A we will get,

Now, Add M and N, we get,

So we see here,

Thus, A is represented as the sum of a symmetric matrix M and a skew symmetric matrix N.

Given A and B are symmetric matrix of same order

⇒ A = A’ → eqn1

⇒ B = B’ → eqn2

So, AB – BA = A’B’ – B’A’ (from 1 & 2)

⇒ AB –BA = (BA)’ – (AB)’ ()

⇒ AB – BA = (-1) ((AB)’ – (BA)’) (taking -1 common)

⇒ AB – BA = -(AB – BA)’ ()

Here we see that the relation between (AB – BA) and its transpose i.e. (AB – BA)’ is (AB –BA) = -(AB – BA)’, this implies that (AB – BA is a skew symmetric matrix.

Hence proved.

Given A =

Therefore, A’ =

Also given that A + A’ = I

(Putting the values in the above equation)

We know when two matrices are equal only when all their corresponding elements or entries are equal i.e. if A = B, then a_ij = b_ij for all i and j.

This implies,

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (1)(3) – (-1)(2) = 5

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – 2R₁

→

Apply row operation- R₂→ R₂/5

→

Apply row operation- R₁→ R₁ + R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(1) – (1)(1) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₁→ R₁ – R₂

→

Apply row operation- R₂→ R₂ – R₁

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (1)(7) – (2)(3) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – 2R₁

→

Apply row operation- R₁→ R₁ - 3R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(7) – (5)(3) = -1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – R₁

→

Apply row operation- R₁→ R₁/2

→

Apply row operation- R₁→ R₁ + 3R₂

→

Apply row operation- R₂→ -2R₂

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(4) – (1)(7) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – R₁

→

Apply row operation- R₂→ 2R₂

→

Apply row operation- R₁→ R₁ –R₂

→

Apply row operation- R₁→ R₁

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(3) – (1)(5) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – R₁

→

Apply row operation- R₁→ R₁/2

→

Apply row operation- R₁→ R₁ - 5R₂

→

Apply row operation- R₂→ 2R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (3)(2) – (5)(1) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ –R₁

→

Apply row operation- R₂→ 3R₂

→

Apply row operation- R₁→ R₁ - R₂

→

Apply row operation- R₁→ R₂

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (4)(4) – (5)(3) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation-

→

Apply row operation- R₁→ R₁/4

→

Apply row operation- R₁→ R₁ - 5R₂

→

Apply row operation- R₂→ 4R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix -

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (3)(7) – (2)(10) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [A : I]

→

Apply row operation- R₂→ R₂ – R₁

→

Apply row operation- R₁→ R₁/3

→

Apply row operation- R₁→ R₁ - 10R₂

→

Apply row operation- R₂→ 3R₂

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (3)(2) – (-1)(-4) = 2

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ + R₁

→

Apply row operation- R₁→ R₁/3

→

Apply row operation- R₁→ R₁ + R₂

→

Apply row operation- R₂→ R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(-2) – (-6)(1) = 2

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – R₁

→

Apply row operation- R₁→ R₁/2

→

Apply row operation- R₁→ R₁ + 3R₂

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (6)(1) – (-2)(-3) = 0

So the matrix is not invertible.

∴ the inverse of the given matrix does not exist.

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(2) – (-1)(-3) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ + R₁

→

Apply row operation- R₁→ R₁/2

→

Apply row operation- R₁→ R₁ + R₂

→

Apply row operation- R₂→ 2R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(2) – (1)(4) = 0

So the matrix is not invertible.

∴ the inverse of the given matrix does not exist.

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = [ (2) {2×2- 3×(-2)} – (-3) {2 × 2 - 3×3} + (3) {2× (-2) – 2×3}]

= [2 {4-(-6)} + 3 {4-9} + 3 { -4-6}]

= [2(10) + 3(-5) + 3(-10)]

= [20-15-30]

= -25

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₁→ R₁

→

Apply row operation- R₂→ R₂ -2R₁

→

Apply row operation- R₃→ R₃ - 3R₁

→.

Apply row operation- R₂→ R₂

Apply row operation- R₁→ R₁ + R₂

→

Apply row operation- R₃→ R₃ - R₂

→

Apply row operation- R₃→ R₃

→

Apply row operation- R₁→ R₁ R₃

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = [ (1) {0- 5 × (-5)} – (3) {(-3) × 0 – (-5) × 2} + (-2) {5 × (-3) – 2 × 0}]

= [1 {25} - 3 {0 + 10} - 2 {-15}]

= [1(25) - 3(10) – 2 (-15)]

= [25-30+30]

= 25

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [A : I]

→

Apply row operation- R₂→ R₂ + 3R₁

→

Apply row operation- R₃→ R₃ -2R₁

→

Apply row operation- R₂→ R₂

→

Apply row operation- R₁→ R₁ – 3R₂

→

Apply row operation- R₃→ R₃ + R₂

→

Apply row operation- R₁→ R₁ - R₃

→

Apply row operation- R₂→ R₂ + R₃

→

Apply row operation- R₃→ R₃

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here |A| = [(2) {1×3 – 1 × (0)} – (0) {3 × 5 – 0 × 0} + (-1) {5 × 1 – 1 × 0}]

= [2{3} - 0 {15} - 1 {5}]

= [6-0-5]

= 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [A : I]

→

Apply row operation- R₁→ R₁

→

Apply row operation- R₂→ R₂ -5R₁

→

Apply row operation- R₃→ R₃ – R₂

→

Apply row operation- R₁→ R₁ + R₃

→

Apply row operation- R₂→ R₂ –5R₃

→

Apply row operation- R₃ → 2R₃

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→

Here it is given that A & B are inverse of each other.

∴ A^-1 = B -(i)

Also B^-1 = A -(ii)

From definition of inverse matrix, we know that-

→ AA^-1 = I

∵ A^-1 = B {from eq(i)}

→ AB = I -(iii)

Similarly, BB^-1 = I

∵ B^-1 = A {from eq(ii)}

→ BA = I -(iv)

So AB = BA = I {from eq(iii) and eq(iv)}

Hence option D is the correct answer.

To prove: (aI + bA)ⁿ = aⁿI + na^n-1bA

Proof: Given A

We will be proving the above equation using mathematical induction.

Steps involved in mathematical inductionare-

1. Prove the equation for n=1

2. Assume the equation to be true for n=k, where kϵN

3. Finally prove the equation for n=k+1

I is the identity matrix of order 2,

i.e. I

Let P(n): (aI + bA)ⁿ = aⁿI + na^n-1bA, n ϵ N

For n=1,

L.H.S: (aI + bA)¹ = aI + bA

R.H.S: a¹I + 1a^1-1bA= aI + a⁰bA= aI + bA

So, L.H.S = R.H.S

∴ P(n) is true for n=1

Now assuming P(n) to be true for n=k, where k ϵ N

P(k) : (aI + bA)^k = a^kI + ka^k-1bA …… (1)

Now proving for n=k+1, i.e. P(k+1) is also true

L.H.S = (aI + bA)^k+1

= (aI + bA)^k . (aI + bA)¹

= (a^kI + ka^k-1bA). (aI + bA) ……from (1)

= aI(a^kI + ka^k-1bA) + bA(a^kI + ka^k-1bA)

= aI(a^kI) + aI(ka^k-1bA) + bA(a^kI) + bA(ka^k-1bA)

= (a.a^k) (I×I) + kb(a.a^k-1)(IA) + (ba^k)(AI) + (bb) ka^k-1(AA)

= a^k+1 I² + ka^1+k-1 bA + ba^kA + b²ka^k-1A² (IA = AI= A & I² = I)

= a^k+1 I + ka^k bA + ba^kA + b²ka^k-1A²

Calculating A²

A² = A.A

∴ A² = O (O is the null matrix)

Putting value of A² in L.H.S

L.H.S = a^k+1 I + ka^k bA + ba^kA + b²ka^k-1(O)

= a^k+1 I + ka^k bA + ba^kA + 0

= a^k+1 I + ka^k bA + ba^kA

= a^k+1 I + (k+1)a^k bA

Putting n=k+1 in R.H.S

R.H.S = a^k+1 I + (k+1)a^k bA

∴ L.H.S = R.H.S

All conditions are proved. Hence P(k+1) is true.

∴ By mathematical induction we have proved that P(n) is true for all n ϵ N.

Thus,(aI + bA)ⁿ = aⁿI + na^n-1bA, where I is the identity matrix of order 2 and nϵN.

We will be proving the above equation by putting different values of n (i.e. n = 1, 2, 3 ….n)

For n=1,

For n=2,

A² = A.A

For n=3,

A³ = A².A

r n=4,

A⁴ = A³.A

d so on for other values of n.

If we notice each result, then we will see that it is of same type that we are trying to prove.

So we can generalise the above results for all n ϵ N

Hence Proved

We will be proving the above equation by putting different values of n. Because n is a positive integer so it will take values which are greater than 0 i.e. n = 1, 2, 3 …n

For n=1,

L.H.S = Aⁿ = A¹ = A =

R.H.S =

∴ L.H.S = R.H.S

For n=2,

L.H.S = A²

= A.A

∴ L.H.S = R.H.S

For n = 3,

L.H.S = A³

= A².A

∴ L.H.S = R.H.S

For n = 4,

L.H.S = A⁴

= A³.A

=

==

R.H.S =

=

=

∴ L.H.S = R.H.S

And so on for other values of n.

If we notice each result then we will see that it is of same type that we are trying to prove.

So we can generalise the above results for all positive integer values of n.

Hence Proved

To prove: AB – BA is a skew symmetric matrix.

Symmetric matrix: A symmetric matrix is a square matrix that is equal to its transpose. In simple words, matrix A is symmetric if

A = A’

where A’ is the transpose of matrix A.

Skew Symmetric matrix: A skewsymmetric matrix is a square matrix that is equal to minus of its transpose. In simple words, matrix A is skew symmetric if

A = -A’

Given: A and B are symmetric matrices i.e.

A = A’ …(1)

B = B’ …(2)

Now calculating the transpose of AB – BA,

(AB – BA)’ = (AB)’ – (BA)’

(By property of transpose i.e. (A – B)’ = A’ – B’)

= B’A’ – A’B’

(By property of transpose i.e. (AB)’ = B’A’)

= BA – AB

= -(AB – BA)

Or we can say that: (AB – BA) = - (AB – BA)’

Clearly it satisfies the condition of skew symmetric matrix.

Hence AB–BA is a skew symmetric matrix.

Case 1: When A is a symmetric matrix i.e.

A = A’……. (1)

where A’ is the transpose of A

To prove: B’AB is also a symmetric matrix.

Calculating the transpose of B’AB

(B’AB)’= B’A’(B’)’ (By property of transpose i.e. (AB)’ = B’A’)

= B’A’B (By property of transpose i.e. (A’)’ = A)

= B’AB (from (1))

It satisfies the condition of symmetric matrix as matrix B’AB is equal to its transpose.

HenceB’ABis a symmetric matrix when A is symmetric.

Case 2: When A is a skew symmetric matrix i.e.

A = -A’……. (2)

where A’ is the transpose of A.

To prove: B’AB is also a skew symmetric matrix.

Calculating the transpose of B’AB

(B’AB)’= B’A’(B’)’ (By property of transpose i.e. (AB)’ = B’A’)

= B’A’B (By property of transpose i.e. (A’)’ = A)

= B’(-A)B (from (2))

= - (B’AB)

It satisfies the condition of skew symmetric matrix as matrix (B’AB) is equal to its transpose.

Hence(B’AB)is a skew symmetric matrix when A is skew symmetric.

∴ Both results are proved.

Given A

Transpose of a matrix: If A be an m×n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. It is denoted by A’ or A^T.

∴ Transpose of A = A’ =

Given equation

A’A =I

=

=

. =

As these matrices are equal to each other that means each element of matrix on L.H.S is equal each element of matrix on R.H.S.

∴ On comparing elements on both sides we get

4y² + z² = 1 …… (1)

2y² – z² = 0 …… (2)

x² + y² + z² = 1 …… (3)

– y² – z² = 0 …… (4)

From equation (4) we get,

x² = y² + z² …… (5)

Substituting this value in equation (3) we get,

2y² + 2z² = 1 …… (6)

Subtracting equation (2) and (6) we get,

3z² = 1

z² = 1/3

z =1/3

Substituting value of z in equation (2) we get,

2y² = 1/3

y² = 1/6

y = �1/6

Substituting values of y and z in equation (5) we get,

x² = 1/2

x = 1/2

Hence values of x, y, z are 1/2, 1/6, 1/3 respectively.

Multiplying matrices on the left hand side

L.H.S =

= [6.0 + 2.2 + 4.x]

= [4 + 4x]

R.H.S = O (Ois the null matrix)

= 0

∴ L.H.S = R.H.S, so we get,

[4+4x] = O

4 + 4x = 0

4x = -4

x = -1

Hence value of x is equal to -1.

To prove: A² – 5A + 7I = 0

Given: A =

L.H.S: A² – 5A + 7I

R.H.S = 0

I =

Calculating value of A²:

A² = A.A

=

=

=

=

Substituting value in L.H.S we get,

= A² – 5A + 7I

.

= 0 = R.H.S

L.H.S = R.H.S

Hence A² – 5A + 7I = 0 is proved.

Multiplying matrices on the left hand side

L.H.S =

=

=

=

=

=

=

R.H.S = O

=

∴ L.H.S = R.H.S we get,

x² – 48 =0

x² = 48

Hence the required value of

(a) Given the unit sale prices of x, y and z as Rs. 2.50, Rs. 1.50 and Rs. 1.00 respectively.

Unit sale prices can be represented in form of matrix as:

Calculating total revenue in market I:

Number of products in the form of matrix:

So the total revenue is given by:

=

=

=

=

∴ Total revenue in market is Rs 46000.

Calculating total revenue in market II:

Number of products in the form of matrix:

So the total revenue is given by:

=

=

=

=.

∴ Total revenue in market is Rs. 53000.

(b) Given the unit cost prices of x, y and z as Rs. 2.00, Rs. 1.00 and 50 paisa respectively.

Calculating gross profit in market I:

Unit cost prices can be represented in form of matrix as:

So the total cost of products in market I is given by:

=

=

=

Since the total revenue in market I is Rs. 46000, the gross profit in this market is given by:

(Rs. 46000 – Rs. 31000)

= Rs. 15000.

Calculating gross profit in market II:

The total cost of products in market II is given by:

=

=

=

=

Since the total revenue in market II is Rs. 53000, the gross profit in this market is given by:

(Rs. 53000 – Rs. 36000)

= Rs. 17000.

Given X

From above equation it can be observed that matrix on R.H.S is a 2×3 matrix and that on the L.H.S is also a 2×3 matrix. Therefore, X must be a 2×2 matrix.

Let X

So the equation is given by:

Now equating the corresponding elements of both the matrices we get,

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

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

Now, a + 4b = -7 ⇒ a = -4b -7

∴ 2a + 5b = -8 ⇒ 2.(-4b -7) + 5b = -8

⇒ -8b -14 + 5b = -8

⇒ -3b = 6

⇒ b = -2

∴ a = -4b -7 ⇒ a = -4.(-2) -7

⇒ a = 1

Now, c + 4d = 2 ⇒ c = -4d + 2

∴ 2c + 5d = 4 ⇒ 2.(-4d + 2) + 5d = 4

⇒ -8d + 4 + 5d = 4

⇒ -3d = 0

⇒ d = 0

∴ c = -4d + 2 ⇒ c = -4.0 + 2

⇒ c = 2

Thus, a = 1, b = -2, c = 2, d = 0.

Hence X becomes

To prove: ABⁿ = BⁿA

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

We have to prove it using mathematical induction.

Steps involved in mathematical induction are-

1. Prove the equation for n=1

2. Assume the equation to be true for n=k, where kϵN

3. Finally prove the equation for n=k+1

Let P(n): ABⁿ = BⁿA

For n=1,

L.H.S: ABⁿ = AB¹ = AB

R.H.S: BⁿA = B¹A = BA = AB

So, L.H.S = R.H.S

∴ P(n) is true for n=1.

Now assuming P(n) to be true for n=k, where k ϵ N

P(k): AB^k = B^kA …… (1)

Now proving for n=k+1, i.e. P(k+1) is also true

L.H.S = ABⁿ

= AB^k+1

= (AB^k).B

= (B^kA).B …… from (1)

= B^k(A.B)

= B^k(BA) (∵AB = BA)

= B^k+1A

R.H.S = BⁿA

= B^k+1A

∴ L.H.S = R.H.S

All conditions are proved. Hence P(k+1) is true.

∴ By mathematical induction we have proved that ABⁿ = BⁿA.

Now, to prove: (AB)ⁿ = AⁿBⁿ for all n ϵ N

For n=1,

L.H.S = (AB)ⁿ = (AB)¹ = AB

R.H.S = AⁿBⁿ = A¹B¹ = AB

∴ L.H.S = R.H.S

∴ It is true for n=1

Assuming it to be true for n=k then,

(AB)^k = A^kB^k ……(2)

Now proving for n=k+1,

L.H.S = (AB)ⁿ

= (AB)^k+1

= (AB)^k(AB)¹

= (A^kB^k)AB

= A^k(B^k.A)B

= A^k(A.B^k)B (ABⁿ = BⁿA)

= (A^kA)(B^kB)

= A^k+1B^k+1

R.H.S = AⁿBⁿ

= A^k+1B^k+1

∴ L.H.S = R.H.S

All conditions are proved. Hence P(k+1) is true.

∴ By mathematical induction we have proved that (AB)ⁿ = AⁿBⁿ for all nϵN

Hence proved.

Given A =

Calculating A²:

A² = A.A

=

=

=

=

And given that A² = I

Then

Comparing corresponding elements we get,

Given A is both symmetric and skew symmetric matrix then, A = A’ and also A = -A’

⇒ A’ = -A’

⇒ 2A’ = 0

⇒ A’ = O

Clearly it is observed that transpose of A is a null matrix or zero matrix then matrix A must also be a zero matrix.

Hence A is a zero matrix.

Given that A² = A

Calculating value of (I + A)³ – 7 A:

⇒ I³ + A³ + 3I²A + 3IA² – 7A

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

⇒ I + A.A + 3A + 3A – 7A (A² = A)

⇒ I + A² + 3A + 3A – 7A

⇒ I + A - A

⇒ I

Hence(I + A)³ – 7 A = I.