Matrices

Given,

$$\begin{bmatrix*}[r] x + 2 & 7 \\ y + 3 & a - 2 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & b - 3 \\ 4 & 3 \end{bmatrix*}$$.

∴ x + 2 = 4

⇒ x = 4 - 2 = 2.

∴ y + 3 = 4

⇒ y = 4 - 3 = 1.

∴ b - 3 = 7

⇒ b = 7 + 3 = 10.

∴ a - 2 = 3

⇒ a = 3 + 2 = 5.

Hence, Option 1 is the correct option.

Given,

A = $$\begin{bmatrix*}[r] 5 & -5 \\ 3 & -3 \end{bmatrix*} \text{ and } B = \begin{bmatrix*}[r] -5 & 5 \\ -3 & 3 \end{bmatrix*}$$

$$A -B = \begin{bmatrix*}[r] 5 & -5 \\ 3 & -3 \end{bmatrix*} - \begin{bmatrix*}[r] -5 & 5 \\ -3 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 - (-5) & -5 - 5 \\ 3 - (-3) & -3 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 & -10 \\ 6 & -6 \end{bmatrix*}$$

Hence, Option 2 is the correct option.

Given,

A = $$\begin{bmatrix*}[r] 5 & 5 \\ 4 & 0 \end{bmatrix*}, B = \begin{bmatrix*}[r] 3 & 2 \\ 1 & 4 \end{bmatrix*} \text{ and } C = \begin{bmatrix*}[r] -2 & 3 \\ 2 & 1 \end{bmatrix*}$$

$$(A + B - C) = \begin{bmatrix*}[r] 5 & 5 \\ 4 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 2 \\ 1 & 4 \end{bmatrix*} - \begin{bmatrix*}[r] -2 & 3 \\ 2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 + 3 - (-2) & 5 + 2 - 3 \\ 4 + 1 - 2 & 0 + 4 - 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 8 + 2 & 7 - 3 \\ 5 - 2 & 4 - 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 & 4 \\ 3 & 3 \end{bmatrix*}.$$

Hence, Option 3 is the correct option.

Given,

⇒ A + P = B

⇒ P = B - A

Substituting values we get :

$$\Rightarrow P = \begin{bmatrix*}[r] -2 & 5 \\ 1 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 7 & 5 \\ -3 & 3 \end{bmatrix*} \\[1em] \Rightarrow P = \begin{bmatrix*}[r] -2 - 7 & 5 - 5 \\ 1 - (-3) & 0 - 3 \end{bmatrix*} \\[1em] \Rightarrow P = \begin{bmatrix*}[r] -9 & 0 \\ 4 & -3 \end{bmatrix*}.$$

Hence, Option 4 is the correct option.

Additive inverse of a matrix M is given by -M.

So, additive inverse of (A + B) = -(A + B).

Substituting values we get :

$$\Rightarrow -(A + B) = -\Big(\begin{bmatrix*}[r] 4 & 2 \\ 7 & -2 \end{bmatrix*} + \begin{bmatrix*}[r] -2 & 1 \\ 3 & -4 \end{bmatrix*}\Big) \\[1em] = -\Big(\begin{bmatrix*}[r] 4 + (-2) & 2 + 1 \\ 7 + 3 & -2 + (-4) \end{bmatrix*}\Big) \\[1em] = -\Big(\begin{bmatrix*}[r] 2 & 3 \\ 10 & -6 \end{bmatrix*}\Big) \\[1em] = \begin{bmatrix*}[r] -2 & -3 \\ -10 & 6 \end{bmatrix*}$$

Hence, Option 1 is the correct option.

(i) False

Reason — For addition the order of both the matrices must be same.

Here, A and B have different orders.

Hence, the statement is false because the orders of both matrices are different.

(ii) True

Reason — For subtraction the order of both the matrices must be same.

Here, A and B have same orders.

Hence, the statement is true.

(iii) False

Reason — On transposing, the no. of rows and columns gets interchanged.

Hence, transpose of a 2 × 1 matrix will be of order 1 × 2.

Hence, the statement is false.

(iv) True

Reason — The transpose of a square matrix is also a square matrix.

Hence, the statement is true.

(v) False

Reason — A column matrix has only one column and can have many rows.

Hence, the statement is false.

Given, $$\begin{bmatrix} -4 & a + 5 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} b + 4 & 2 \\ 3 & c - 1 \end{bmatrix}$$

By definition of equality of matrices we get,

-4 = b + 4 ⇒ b = -4 - 4 = -8,

a + 5 = 2 ⇒ a = 2 - 5 = -3,

2 = c - 1 ⇒ c = 2 + 1 = 3.

Hence, a = -3, b = -8 and c = 3.

Given, $$\begin{bmatrix} a & a - b \\ b + c & 0 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix}$$

By definition of equality of matrices we get,

a = 3,

a - b = -1
⇒ 3 - b = -1 [∵ a = 3]
⇒ b = 3 + 1 = 4

b + c = 2
⇒ 4 + c = 2
⇒ c = -2.

Hence, a = 3, b = 4 and c = -2.

(i) Given,

$$\phantom{\Rightarrow} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -1 & -2 \\ 1 & -7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 + (-1) & 2 + (-2) \\ 3 + 1 & 4 + (-7) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 \\ 4 & -3 \end{bmatrix}.$$

Hence, resultant matrix = $$\begin{bmatrix} 0 & 0 \\ 4 & -3 \end{bmatrix}$$.

(ii) Given,

$$\phantom{\Rightarrow} \begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix} - \begin{bmatrix} 0 & 2 & 3 \\ 6 & -1 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 - 0 & 3 - 2 & 4 - 3 \\ 5 - 6 & 6 - (-1) & 7 - 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 & 1 & 1 \\ -1 & 7 & 7 \end{bmatrix}.$$

Hence, resultant matrix = $$\begin{bmatrix} 2 & 1 & 1 \\ -1 & 7 & 7 \end{bmatrix}$$.

(iii) Given,

$$\begin{bmatrix} 0 & 1 & 2 \\ 4 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}$$

The above calculation is not possible because for addition the order of both matrices must be equal.

Given,

$$\Rightarrow \begin{bmatrix} 5 & 2 \\ -1 & y - 1 \end{bmatrix} - \begin{bmatrix} 1 & x - 1 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 - 1 & 2 - (x - 1) \\ -1 - 2 & y - 1 - (-3) \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 3 - x \\ -3 & y + 2 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix} \\[1em]$$

By definition of equality of matrices we get,

⇒ 3 - x = 7
⇒ x = 3 - 7 = -4.

⇒ y + 2 = 2 ⇒ y = 0.

Hence, x = -4 and y = 0.

Given,

$$\Rightarrow \begin{bmatrix} -8 & x \end{bmatrix} + \begin{bmatrix} y & -2 \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -8 + y & x + (-2) \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix}$$

By definition of equality of matrices we get,

-8 + y = -3
⇒ y = -3 + 8 = 5.

x - 2 = 2
⇒ x = 2 + 2 = 4.

Hence, x = 4 and y = 5.

M = $$\begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}$$

M^t = $$\begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix}$$

(i)

$$M + M^t = \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} + \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 5 + 5 & -3 + (-2) \\ -2 + (-3) & 4 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 10 & -5 \\ -5 & 8 \end{bmatrix}.$$

Hence, $$M + M^t = \begin{bmatrix} 10 & -5 \\ -5 & 8 \end{bmatrix}.$$

(ii)

$$M^t - M = \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} - \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 5 - 5 & -2 - (-3) \\ -3 - (-2) & 4 - 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.$$

Hence, M^t - M = $$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.$$

(i) Given,

⇒ X + B = C - A

⇒ X = C - A - B

Substituting values of A, B and C in above equation we get,

$$\Rightarrow X = \begin{bmatrix} -1 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -3 \end{bmatrix} - \begin{bmatrix} 0 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 - 2 - 0 & 4 - (-3) - 2 \end{bmatrix} \\[1em] = \begin{bmatrix} -3 & 5 \end{bmatrix}.$$

Hence, X = $$\begin{bmatrix} -3 & 5 \end{bmatrix}.$$

(ii) Given,

⇒ A - X = B + C

⇒ X = A - (B + C)

Substituting values of A, B and C in above equation we get,

$$\Rightarrow X = \begin{bmatrix} 2 & -3 \end{bmatrix} - \Big(\begin{bmatrix} 0 & 2 \end{bmatrix} + \begin{bmatrix} -1 & 4 \end{bmatrix}\Big) \\[1em] = \begin{bmatrix} 2 & -3 \end{bmatrix} - \Big(\begin{bmatrix} 0 + (-1) & 2 + 4 \end{bmatrix}\Big) \\[1em] = \begin{bmatrix} 2 & -3 \end{bmatrix} - \begin{bmatrix} -1 & 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 - (-1) & -3 - 6 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 & -9 \end{bmatrix}.$$

Hence, X = $$\begin{bmatrix} 3 & -9 \end{bmatrix}$$.

(i) Given,

⇒ A + X = B

⇒ X = B - A

$$\Rightarrow X = \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix} - \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 - (-1) & -3 - 0 \\ -2 - 2 & 0 - (-4) \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & -3 \\ -4 & 4 \end{bmatrix}.$$

Hence, X = $$\begin{bmatrix} 4 & -3 \\ -4 & 4 \end{bmatrix}.$$

(ii) Given,

⇒ A - X = B

⇒ X = A - B

$$\Rightarrow X = \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} - \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 - 3 & 0 - (-3) \\ 2 - (-2) & -4 - 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -4 & 3 \\ 4 & -4 \end{bmatrix}.$$

Hence, X = $$\begin{bmatrix} -4 & 3 \\ 4 & -4 \end{bmatrix}.$$

(iii) Given,

⇒ X - B = A

⇒ X = A + B

$$\Rightarrow X = \begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 + 3 & 0 + (-3) \\ 2 + (-2) & -4 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 & -3 \\ 0 & -4 \end{bmatrix}.$$

Hence, X = $$\begin{bmatrix} 2 & -3 \\ 0 & -4 \end{bmatrix}.$$

Given,

$$\Rightarrow 4\begin{bmatrix*}[r] 5 & x \end{bmatrix*} - 5\begin{bmatrix*}[r] y & -2 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 20 & 4x \end{bmatrix*} - \begin{bmatrix*}[r] 5y & -10 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 20 - 5y & 4x - (-10) \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 20 - 5y & 4x + 10 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 22 \end{bmatrix*}$$

∴ 20 - 5y = 10 and 4x + 10 = 22

⇒ 5y = 20 - 10 and 4x = 22 - 10

⇒ 5y = 10 and 4x = 12

⇒ y = 2 and x = 3.

Hence, Option 2 is the correct option.

Given,

A - B = $$\begin{bmatrix*}[r] 6 & 4 \\ -3 & 0 \end{bmatrix*}$$

Substituting value of A in above equation we get :

$$\Rightarrow \begin{bmatrix*}[r] -3 & -7 \\ 0 & -8 \end{bmatrix*} - B = \begin{bmatrix*}[r] 6 & 4 \\ -3 & 0 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] -3 & -7 \\ 0 & -8 \end{bmatrix*} - \begin{bmatrix*}[r] 6 & 4 \\ -3 & 0 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] -3 - 6 & -7 - 4 \\ 0 - (-3) & -8 - 0 \end{bmatrix*} \\[1em] \Rightarrow B = \begin{bmatrix*}[r] -9 & -11 \\ 3 & -8 \end{bmatrix*}$$

Hence, Option 4 is the correct option.

As, I is a unit matrix of order 2.

∴ I = $$\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$$

Given,

$$\Rightarrow M + 4I = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} \\[1em] \Rightarrow M + 4\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} \\[1em] \Rightarrow M + \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 8 & -3 \\ 4 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 8 - 4 & -3 - 0 \\ 4 - 0 & 2 - 4 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 4 & -3 \\ 4 & -2 \end{bmatrix*}.$$

Hence, Option 4 is the correct option.

Given,

$$\Rightarrow 2\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 3 \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 2x \\ 0 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 9 \\ 3y & 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 3 & 2x + 9 \\ 0 + 3y & 2 + 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 2x + 9 \\ 3y & 8 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em]$$

From above equation we get :

⇒ z = 9, 3y = 15 and 2x + 9 = -7

⇒ z = 9, y = $\frac{15}{3}$ and 2x = -7 - 9

⇒ z = 9, y = 5 and 2x = -16

⇒ z = 9, y = 5 and x = $-\frac{16}{2}$

⇒ z = 9, y = 5 and x = -8.

Hence, Option 2 is the correct option.

Substituting values of A and B in A - 2B, we get :

$$\Rightarrow A - 2B = \begin{bmatrix*}[r] 4 & 7 \\ 3 & -2 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 2 \\ -1 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 7 \\ 3 & -2 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 4 \\ -2 & 8 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 - 2 & 7 - 4 \\ 3 - (-2) & -2 - 8 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 3 \\ 3 + 2 & -10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 3 \\ 5 & -10 \end{bmatrix*}.$$

Hence, Option 3 is the correct option.

(i) Given,

$$\Rightarrow 3\begin{bmatrix*}[r] 4 & x \end{bmatrix*} + 2\begin{bmatrix*}[r] y & -3 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 12 & 3x \end{bmatrix*} + \begin{bmatrix*}[r] 2y & -6 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 12 + 2y & 3x + (-6) \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 12 + 2y & 3x - 6 \end{bmatrix*} = \begin{bmatrix*}[r] 10 & 0 \end{bmatrix*}$$

By equality of matrices we get,

12 + 2y = 10
⇒ 2y = 10 - 12
⇒ 2y = -2
⇒ y = -1.

3x - 6 = 0
⇒ 3x = 6
⇒ x = 2

Hence, x = 2 and y = -1.

(ii) Given,

$$\Rightarrow x\begin{bmatrix*}[r] -1 \\ 2 \end{bmatrix*} - 4\begin{bmatrix*}[r] -2 \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x \\ 2x \end{bmatrix*} - \begin{bmatrix*}[r] -8 \\ 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x -(-8) \\ 2x - 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x + 8 \\ 2x - 4y \end{bmatrix*} = \begin{bmatrix*}[r] 7 \\ -8 \end{bmatrix*}$$

By definition of equality of matrices we get,

-x + 8 = 7 ........(i)

2x - 4y = -8 ......(ii)

Solving eq. (i) we get,

⇒ x = 8 - 7 = 1.

Substituting x = 1 in eq. (ii) we get,

⇒ 2x - 4y = -8
⇒ 2(1) - 4y = -8
⇒ 2 - 4y = -8
⇒ 4y = 2 + 8
⇒ 4y = 10
⇒ y = $\frac{10}{4} = \frac{5}{2} = 2.5$.

Hence, x = 1 and y = 2.5

(i) Given,

2A - 3B + C

Substituting values of A, B and C in above equation we get,

$$\Rightarrow 2A - 3B + C = 2\begin{bmatrix*}[r] 2 & 1 \\ 3 & 0 \end{bmatrix*} - 3\begin{bmatrix*}[r] 1 & 1 \\ 5 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] -3 & -1 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 2 \\ 6 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 15 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] -3 & -1 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 - 3 + (-3) & 2 - 3 + (-1) \\ 6 - 15 + 0 & 0 - 6 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & -2 \\ -9 & -6 \end{bmatrix*}.$$

Hence 2A - 3B + C = $$\begin{bmatrix*}[r] -2 & -2 \\ -9 & -6 \end{bmatrix*}.$$

(ii) Given,

A + 2C - B

Substituting values of A, B and C in above equation we get,

$$\Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 3 & 0 \end{bmatrix*} + 2\begin{bmatrix*}[r] -3 & -1 \\ 0 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 1 \\ 5 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 1 \\ 3 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] -6 & -2 \\ 0 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 1 & 1 \\ 5 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + (-6) - 1 & 1 + (-2) - 1 \\ 3 + 0 - 5 & 0 + 0 - 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -5 & -2 \\ -2 & -2 \end{bmatrix*}.$$

Hence, A + 2C - B = $$\begin{bmatrix*}[r] -5 & -2 \\ -2 & -2 \end{bmatrix*}.$$

Given,

$$\Rightarrow \begin{bmatrix*}[r] 4 & -2 \\ 4 & 0 \end{bmatrix*} + 3A = \begin{bmatrix*}[r] -2 & -2 \\ 1 & -3 \end{bmatrix*} \\[1em] \Rightarrow 3A = \begin{bmatrix*}[r] -2 & -2 \\ 1 & -3 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & -2 \\ 4 & 0 \end{bmatrix*} \\[1em] \Rightarrow 3A = \begin{bmatrix*}[r] -2 - 4 & -2 - (-2) \\ 1 - 4 & -3 - 0 \end{bmatrix*} \\[1em] \Rightarrow 3A = \begin{bmatrix*}[r] -6 & 0 \\ -3 & -3 \end{bmatrix*} \\[1em] \Rightarrow A = \frac{1}{3}\begin{bmatrix*}[r] -6 & 0 \\ -3 & -3 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] -2 & 0 \\ -1 & -1 \end{bmatrix*}.$$

Hence, A = $$\begin{bmatrix*}[r] -2 & 0 \\ -1 & -1 \end{bmatrix*}$$.

(i)

$$2A + B = 2\begin{bmatrix*}[r] 1 & 4 \\ 2 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & -1 \\ -3 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 8 \\ 4 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] -4 & -1 \\ -3 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + (-4) & 8 + (-1) \\ 4 + (-3) & 6 + (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & 7 \\ 1 & 4 \end{bmatrix*}.$$

Hence, 2A + B = $$\begin{bmatrix*}[r] -2 & 7 \\ 1 & 4 \end{bmatrix*}$$.

(ii) Given,

$$\Rightarrow C + B = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} - B \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] -4 & -1 \\ -3 & -2 \end{bmatrix*} \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 0 - (-4) & 0 - (-1) \\ 0 - (-3) & 0 - (-2) \end{bmatrix*} \\[1em] \Rightarrow C = \begin{bmatrix*}[r] 4 & 1 \\ 3 & 2 \end{bmatrix*}.$$

Hence, C = $$\begin{bmatrix*}[r] 4 & 1 \\ 3 & 2 \end{bmatrix*}$$.

Given,

$$\Rightarrow 2\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 3 \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 2x \\ 0 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 9 \\ 3y & 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 3 & 2x + 9 \\ 0 + 3y & 2 + 6 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 2x + 9 \\ 3y & 8 \end{bmatrix*} = \begin{bmatrix*}[r] z & -7 \\ 15 & 8 \end{bmatrix*}$$

By definition of equality of matrices we get,

z = 9,

2x + 9 = -7
⇒ 2x = -7 - 9
⇒ 2x = -16
⇒ x = -8,

3y = 15
⇒ y = 5.

Hence, x = -8, y = 5 and z = 9.

$$A = \begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} \text{ and } A^t = \begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*}$$.

(i) Substituting value of A and A^t in 2A + 3A^t we get,

$$\Rightarrow 2A + 3A^t = 2\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} + 3\begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 & 12 \\ 0 & -18 \end{bmatrix*} + \begin{bmatrix*}[r] -9 & 0 \\ 18 & -27 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + (-9) & 12 + 0 \\ 0 + 18 & -18 + (-27) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -15 & 12 \\ 18 & -45 \end{bmatrix*}.$$

Hence, 2A + 3A^t = $$\begin{bmatrix*}[r] -15 & 12 \\ 18 & -45 \end{bmatrix*}.$$

(ii) Substituting value of A and A^t in 2A^t - 3A we get,

$$\Rightarrow 2A^t - 3A = 2\begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} - 3\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 & 0 \\ 12 & -18 \end{bmatrix*} - \begin{bmatrix*}[r] -9 & 18 \\ 0 & -27 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 - (-9) & 0 - 18 \\ 12 - 0 & -18 - (-27) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 3 & -18 \\ 12 & 9 \end{bmatrix*}.$$

Hence, $$2A^t - 3A = \begin{bmatrix*}[r] 3 & -18 \\ 12 & 9 \end{bmatrix*}.$$

(iii) Substituting value of A and A^t in $\frac{1}{2}A - \frac{1}{3}A^t$ we get,

$$\Rightarrow \frac{1}{2}A - \frac{1}{3}A^t = \frac{1}{2}\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} - \frac{1}{3}\begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\frac{3}{2} & 3 \\ 0 & -\frac{9}{2} \end{bmatrix*} - \begin{bmatrix*}[r] -1 & 0 \\ 2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\frac{3}{2} - (-1) & 3 - 0 \\ 0 - 2 & -\frac{9}{2} - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\frac{3}{2} + 1 & 3 \\ -2 & -\frac{9}{2} + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -\frac{1}{2} & 3 \\ -2 & -\frac{3}{2} \end{bmatrix*}$$

Hence, $$\frac{1}{2}A - \frac{1}{3}A^t = \begin{bmatrix*}[r] -\frac{1}{2} & 3 \\ -2 & -\frac{3}{2} \end{bmatrix*}.$$

(iv) Substituting value of A and A^t in $A^t - \frac{1}{3}A$ we get,

$$\Rightarrow A^t - \frac{1}{3}A = \begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} - \frac{1}{3}\begin{bmatrix*}[r] -3 & 6 \\ 0 & -9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 & 0 \\ 6 & -9 \end{bmatrix*} - \begin{bmatrix*}[r] -1 & 2 \\ 0 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 - (-1) & 0 - 2 \\ 6 - 0 & -9 - (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & -2 \\ 6 & -6 \end{bmatrix*}.$$

Hence, $$A^t - \frac{1}{3}A = \begin{bmatrix*}[r] -2 & -2 \\ 6 & -6 \end{bmatrix*}.$$

(i) Given,

$$\Rightarrow X + 2A = B \\[1em] \Rightarrow X = B - 2A \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 1 \\ -2 & 0 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 2 \\ -4 & 0 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 2 - 2 & -1 - 2 \\ 1 - (-4) & 1 - 0 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] 0 & -3 \\ 5 & 1 \end{bmatrix*}$$

Hence, X = $$\begin{bmatrix*}[r] 0 & -3 \\ 5 & 1 \end{bmatrix*}.$$

(ii) Given,

$$\Rightarrow 3X + B + 2A = 0 \\[1em] \Rightarrow 3X = -(B + 2A) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} + 2\begin{bmatrix*}[r] 1 & 1 \\ -2 & 0 \end{bmatrix*}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & 2 \\ -4 & 0 \end{bmatrix*}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 2 + 2 & -1 + 2 \\ 1 + (-4) & 1 + 0 \end{bmatrix*}\Big) \\[1em] \Rightarrow 3X = -\Big(\begin{bmatrix*}[r] 4 & 1 \\ -3 & 1 \end{bmatrix*}\Big) \\[1em] \Rightarrow X = -\frac{1}{3}\Big(\begin{bmatrix*}[r] 4 & 1 \\ -3 & 1 \end{bmatrix*}\Big) \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -\frac{4}{3} & -\frac{1}{3} \\ 1 & -\frac{1}{3} \end{bmatrix*}$$

Hence, X = $$\begin{bmatrix*}[r] -\frac{4}{3} & -\frac{1}{3} \\ 1 & -\frac{1}{3} \end{bmatrix*}$$.

(iii) Given,

$$\Rightarrow 3A - 2X = X - 2B \\[1em] \Rightarrow X + 2X = 3A + 2B \\[1em] \Rightarrow 3X = 3A + 2B \\[1em] \Rightarrow 3X = 3\begin{bmatrix*}[r] 1 & 1 \\ -2 & 0 \end{bmatrix*} + 2\begin{bmatrix*}[r] 2 & -1 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 3 & 3 \\ -6 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & -2 \\ 2 & 2 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 3 + 4 & 3 + (-2) \\ -6 + 2 & 0 + 2 \end{bmatrix*} \\[1em] \Rightarrow 3X = \begin{bmatrix*}[r] 7 & 1 \\ -4 & 2 \end{bmatrix*} \\[1em] \Rightarrow X = \frac{1}{3}\begin{bmatrix*}[r] 7 & 1 \\ -4 & 2 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] \frac{7}{3} & \frac{1}{3} \\ -\frac{4}{3} & \frac{2}{3} \end{bmatrix*}.$$

Hence, X = $$\begin{bmatrix*}[r] \frac{7}{3} & \frac{1}{3} \\ -\frac{4}{3} & \frac{2}{3} \end{bmatrix*}.$$

I = $$\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$$

Given,

$$\Rightarrow 5M + 3I = 4\begin{bmatrix*}[r] 2 & -5 \\ 0 & -3 \end{bmatrix*} \\[1em] \Rightarrow 5M + 3\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -20 \\ 0 & -12 \end{bmatrix*} \\[1em] \Rightarrow 5M + \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 8 & -20 \\ 0 & -12 \end{bmatrix*} \\[1em] \Rightarrow 5M = \begin{bmatrix*}[r] 8 & -20 \\ 0 & -12 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] \\[1em] \Rightarrow 5M = \begin{bmatrix*}[r] 8 - 3 & -20 - 0 \\ 0 - 0 & -12 - 3 \end{bmatrix*} \\[1em] \Rightarrow 5M = \begin{bmatrix*}[r] 5 & -20 \\ 0 & -15 \end{bmatrix*} \\[1em] \Rightarrow M = \frac{1}{5}\begin{bmatrix*}[r] 5 & -20 \\ 0 & -15 \end{bmatrix*} \\[1em] \Rightarrow M = \begin{bmatrix*}[r] 1 & -4 \\ 0 & -3 \end{bmatrix*}.$$

Hence, M = $$\begin{bmatrix*}[r] 1 & -4 \\ 0 & -3 \end{bmatrix*}.$$

Given,

AB = R

A_{m × 3}B_{3 × 2} = R_{5 × n}

We know that,

For matrix multiplication :

No. of columns in A must be equal to the number of rows in B.

Resultant matrix order = No. of rows in A × No. of columns in B.

∴ m = 5 and n = 2.

Hence, Option 2 is the correct option.

Substituting value of A and B in A = B², we get :

$$\Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 12 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 \times 2 + 12 \times 0 & 2 \times 12 + 12 \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times 12 + 1 \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 + 0 & 24 + 12 \\ 0 + 0 & 0 + 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*}$$

∴ x = 36.

Hence, Option 4 is the correct option.

Let resultant matrix of CA be R.

C_{3 × 3} × A_{3 × 3} = R_{m × n}

We know that,

For matrix multiplication :

No. of columns in C must be equal to the number of rows in A.

Resultant matrix order = No. of rows in C × No. of columns in A.

∴ m = 3 and n = 3.

∴ Order of matrix R will be 3 × 3.

Let resultant matrix of B² be S.

B_{3 × 3} × B_{3 × 3} = S_{g × h}

Resultant matrix order = No. of rows in B × No. of columns in B.

∴ g = 3 and h = 3.

∴ Order of matrix S will be 3 × 3.

⇒ CA + B²

⇒ R + S

Addition of matrix possible between matrix of same order and same is the order of resultant matrix.

Hence, resultant matrix's order = 3 × 3.

Hence, Option 2 is the correct option.

BA is not possible because no. of columns in B (1) is not equal to the no. of rows in A (2).

Hence, Option 3 is the correct option.

Substituting values of A, B and C in A² + 2B - 3C, we get :

$$\Rightarrow A^2 + 2B - 3C = \begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 1 & 1 \end{bmatrix*} + 2\begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} - 3\begin{bmatrix*}[r] 1 & 1 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 1 & 1 \times 0 + 0 \times 1 \\ 1 \times 1 + 1 \times 1 & 1 \times 0 + 1 \times 1 \end{bmatrix*} + \begin{bmatrix*}[r] 0 & 2 \\ 2 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 0 & 0 + 0 \\ 1 + 1 & 0 + 1 \end{bmatrix*} + \begin{bmatrix*}[r] 0 & 2 \\ 2 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 0 & 2 \\ 2 & 0 \end{bmatrix*} - \begin{bmatrix*}[r] 3 & 3 \\ 0 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 0 - 3 & 0 + 2 - 3 \\ 2 + 2 - 0 & 1 + 0 - 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 & -1 \\ 4 & 1 \end{bmatrix*}$$

Hence, Option 1 is the correct option.

$$\begin{bmatrix*}[r] 3 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 0 \end{bmatrix*}$$

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

Hence, matrix multiplication is possible.

$$\Rightarrow \begin{bmatrix*}[r] 3 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 3 \times 2 + 2 \times 0 \\ \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 \\ \end{bmatrix*}.$$

Hence, $$\begin{bmatrix*}[r] 3 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 0 \end{bmatrix*} = \begin{bmatrix*}[r] 6 \end{bmatrix*}.$$

$$\begin{bmatrix*}[r] 1 & -2 \\ \end{bmatrix*}\begin{bmatrix*}[r] -2 & 3 \\ -1 & 4 \end{bmatrix*}$$

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

$$\Rightarrow \begin{bmatrix*}[r] 1 & -2 \\ \end{bmatrix*}\begin{bmatrix*}[r] -2 & 3 \\ -1 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 1 \times (-2) + (-2) \times (-1) & 1 \times 3 + (-2) \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -2 + 2 & 3 + (-8) \\ \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & -5 \\ \end{bmatrix*}.$$

Hence, $$\begin{bmatrix*}[r] 1 & -2 \\ \end{bmatrix*}\begin{bmatrix*}[r] -2 & 3 \\ -1 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -5 \\ \end{bmatrix*}.$$

$$\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 3 \end{bmatrix*}$$

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

$$\Rightarrow \begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 6 \times (-1) + 4 \times 3 \\ 3 \times (-1) + (-1) \times 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + 12 \\ -3 + (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 \\ -6 \end{bmatrix*}.$$

Hence, $$\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 6 \\ -6 \end{bmatrix*}.$$

$$\begin{bmatrix*}[r] 6 & 4 \\ 3 & -1 \end{bmatrix*}\begin{bmatrix*}[r] -1 & 3 \\ \end{bmatrix*}$$

For matrix multiplication, the no. of columns in first matrix should be equal to no. of rows in the second matrix.

The above matrix multiplication is not possible as the no. of columns in first matrix is not equal to no. of rows in the second matrix.

(i) Substituting value in AB,

$$\Rightarrow AB = \begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -1 \\ 3 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 \times 1 + 2 \times 3 & 0 \times (-1) + 2 \times 2 \\ 5 \times 1 + (-2) \times 3 & 5 \times (-1) + (-2) \times 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 + 6 & 0 + 4 \\ 5 - 6 & -5 - 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 & 4 \\ -1 & -9 \end{bmatrix*}.$$

Hence, AB = $$\begin{bmatrix*}[r] 6 & 4 \\ -1 & -9 \end{bmatrix*}.$$

(ii) Substituting value in BA,

$$\Rightarrow BA = \begin{bmatrix*}[r] 1 & -1 \\ 3 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 0 + (-1) \times 5 & 1 \times 2 + (-1) \times (-2) \\ 3 \times 0 + 2 \times 5 & 3 \times 2 + 2 \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 - 5 & 2 + 2 \\ 0 + 10 & 6 - 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -5 & 4 \\ 10 & 2 \end{bmatrix*}.$$

Hence, BA = $$\begin{bmatrix*}[r] -5 & 4 \\ 10 & 2 \end{bmatrix*}.$$

(iii) Substituting value in AI,

$$\Rightarrow AI = \begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 \times 1 + 2 \times 0 & 0 \times 0 + 2 \times 1 \\ 5 \times 1 + (-2) \times 0 & 5 \times 0 + (-2) \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 2 \\ 5 & -2 \end{bmatrix*} = A$$

Hence, AI = matrix A.

Given,

$$\Rightarrow A^2 = B \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 3 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times 3 + x \times 0 & 3 \times x + x \times 1 \\ 0 \times 3 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 + 0 & 3x + x \\ 0 + 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 4x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 9 & 16 \\ 0 & -y \end{bmatrix*}$$

By definition of equality of matrices we get,

4x = 16
⇒ x = 4.

1 = -y
⇒ y = -1.

Hence, x = 4 and y = -1.

Given,

$$\Rightarrow \begin{bmatrix*}[r] x & 0 \\ -3 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 1 \\ 0 & y \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 2 \\ -3 & -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 1 + 0 \times 0 & x \times 1 + 0 \times y \\ -3 \times 1 + 1 \times 0 & -3 \times 1 + 1 \times y \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 2 \\ -3 & -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x + 0 & x + 0 \\ -3 + 0 & -3 + y \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 2 \\ -3 & -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & x \\ -3 & -3 + y \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 2 \\ -3 & -2 \end{bmatrix*}$$

By definition of equality of matrices we get,

x = 2

-3 + y = -2
⇒ y = -2 + 3
⇒ y = 1.

Hence, x = 2 and y = 1.

(i) Substituting value of A, B and C in (AB)C we get,

$$\Rightarrow \Big(\begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 2 \\ 4 & 3 \end{bmatrix*}\Big)\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \Big(\begin{bmatrix*}[r] 1 \times 1 + 3 \times 4 & 1 \times 2 + 3 \times 3 \\ 2 \times 1 + 4 \times 4 & 2 \times 2 + 4 \times 3 \end{bmatrix*}\Big)\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \Big(\begin{bmatrix*}[r] 1 + 12 & 2 + 9 \\ 2 + 16 & 4 + 12 \end{bmatrix*}\Big)\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 13 & 11 \\ 18 & 16 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 13 \times 4 + 11 \times 1 & 13 \times 3 + 11 \times 2 \\ 18 \times 4 + 16 \times 1 & 18 \times 3 + 16 \times 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 52 + 11 & 39 + 22 \\ 72 + 16 & 54 + 32 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}.$$

Hence, (AB)C = $$\begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}.$$

(ii) Substituting value of A, B and C in A(BC) we get,

$$\Rightarrow \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\Big(\begin{bmatrix*}[r] 1 & 2 \\ 4 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 3 \\ 1 & 2 \end{bmatrix*}\Big) \\[1em] = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\Big(\begin{bmatrix*}[r] 1 \times 4 + 2\times 1 & 1 \times 3 + 2 \times 2 \\ 4 \times 4 + 3 \times 1 & 4 \times 3 + 3 \times 2 \end{bmatrix*}\Big) \\[1em] = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 4 + 2 & 3 + 4 \\ 16 + 3 & 12 + 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 6 & 7 \\ 19 & 18 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 6 + 3 \times 19 & 1 \times 7 + 3 \times 18 \\ 2 \times 6 + 4 \times 19 & 2 \times 7 + 4 \times 18 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 6 + 57 & 7 + 54 \\ 12 + 76 & 14 + 72 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}$$

Hence, A(BC) = $$\begin{bmatrix*}[r] 63 & 61 \\ 88 & 86 \end{bmatrix*}$$.

Hence, A(BC) = (AB)C.

Substituting values of A, B and C in A² + AC - 5B we get,

$$\Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & 1 \\ 0 & -2 \end{bmatrix*}\begin{bmatrix*}[r] -3 & 2 \\ -1 & 4 \end{bmatrix*} - 5\begin{bmatrix*}[r] 4 & 1 \\ -3 & -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 2 + 1 \times 0 & 2 \times 1 + 1 \times (-2) \\ 0 \times 2 + (-2) \times 0 & 0 \times 1 + (-2) \times (-2) \end{bmatrix*} + \begin{bmatrix*}[r] 2 \times (-3) + 1 \times (-1) & 2 \times 2 + 1 \times 4 \\ 0 \times (-3) + (-2) \times (-1) & 0 \times 2 + (-2) \times 4 \end{bmatrix*} - \begin{bmatrix*}[r] 20 & 5 \\ -15 & -10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 0 & 2 - 2 \\ 0 + 0 & 0 + 4 \end{bmatrix*} + \begin{bmatrix*}[r] -6 - 1 & 4 + 4 \\ 0 + 2 & 0 - 8 \end{bmatrix*} - \begin{bmatrix*}[r] 20 & 5 \\ -15 & -10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 0 \\ 0 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] -7 & 8 \\ 2 & -8 \end{bmatrix*} - \begin{bmatrix*}[r] 20 & 5 \\ -15 & -10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + (-7) - 20 & 0 + 8 - 5 \\ 0 + 2 - (-15) & 4 + (-8) - (-10) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -23 & 3 \\ 17 & 6 \end{bmatrix*}$$

Hence, A² + AC - 5B = $$\begin{bmatrix*}[r] -23 & 3 \\ 17 & 6 \end{bmatrix*}$$.

I = $$\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$$

M = $$\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*}$$

$$\text{L.H.S. }= M^2 = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 2 + 1 \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 4 & 2 + 2 \\ 2 + 2 & 4 + 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 & 4 \\ 4 & 5 \end{bmatrix*} \\[1em] \text{R.H.S.} = 2M + 3I \\[1em] = 2\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} + 3\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 4 \\ 4 & 2 \end{bmatrix*} + \begin{bmatrix*}[r] 3 & 0 \\ 0 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 + 3 & 4 + 0 \\ 4 + 0 & 2 + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 & 4 \\ 4 & 5 \end{bmatrix*}.$$

Since, L.H.S. = R.H.S.

Hence, proved that M² = 2M + 3I.

Given,

$$\Rightarrow BA = M^2 \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -b \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] a & 0 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & -1 \\ 1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -1 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 \times a + (-b) \times 0 & 0 \times 0 + (-b) \times 2 \\ 1 \times a + 0 \times 0 & 1 \times 0 + 0 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 \times 1 + (-1) \times 1 & 1 \times (-1) + (-1) \times 1 \\ 1 \times 1 + 1 \times 1 & 1 \times (-1) + 1 \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & 0 - 2b \\ a + 0 & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 1 - 1 & -1 - 1 \\ 1 + 1 & -1 + 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -2b \\ a & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -2 \\ 2 & 0 \end{bmatrix*}$$

By definition of equality of matrices we get,

-2b = -2
⇒ b = 1.

a = 2.

Hence, a = 2 and b = 1.

Given,

$\Rightarrow B^2 = B + \frac{1}{2}A$

Substituting value of B in above equation we get,

$$\Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \frac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 1 \times 0 & 2 \times 1 + 1 \times 1\ 0 \times 2 + 1 \times 0 & 0 \times 1 + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \frac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 0 & 2 + 1 \\ 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \frac{1}{2}A \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} + \frac{1}{2}A \\[1em] \Rightarrow \frac{1}{2}A = \begin{bmatrix*}[r] 4 & 3 \\ 0 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 1 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \frac{1}{2}A = \begin{bmatrix*}[r] 4 - 2 & 3 - 1 \\ 0 - 0 & 1 - 1 \end{bmatrix*} \\[1em] \Rightarrow \frac{1}{2}A = \begin{bmatrix*}[r] 2 & 2 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow A = 2\begin{bmatrix*}[r] 2 & 2 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow A = \begin{bmatrix*}[r] 4 & 4 \\ 0 & 0 \end{bmatrix*}.$$

Hence, A = $$\begin{bmatrix*}[r] 4 & 4 \\ 0 & 0 \end{bmatrix*}.$$

Given, A² = I

$$\therefore \begin{bmatrix*}[r] -1 & 1 \\ a & b \end{bmatrix*}\begin{bmatrix*}[r] -1 & 1 \\ a & b \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -1 \times -1 + 1 \times a & -1 \times 1 + 1 \times b \\ a \times (-1) + b \times a & a \times 1 + b \times b \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 + a & -1 + b \\ -a + ab & a + b^2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$$

By definition of equality of matrices we get,

1 + a = 1
⇒ a = 0.

-1 + b = 0
⇒ b = 1.

Hence, a = 0 and b = 1

Given,

$$\Rightarrow \begin{bmatrix*}[r] 2 & 5 \\ 5 & 2 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -7 \\ 14 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times x + 5 \times y \\ 5 \times x + 2 \times y \end{bmatrix*} = \begin{bmatrix*}[r] -7 \\ 14 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 5y \\ 5x + 2y \end{bmatrix*} = \begin{bmatrix*}[r] -7 \\ 14 \end{bmatrix*}$$

By definition of equality of matrices we get,

2x + 5y = -7

⇒ 2x = -(7 + 5y)

⇒ x = $\frac{-(7 + 5y)}{2}$ ......(i)

5x + 2y = 14

Substituting value of x from (i) in above equation we get,

$$\begin{aligned}\Rightarrow 5 \times \frac{-(7 + 5y)}{2} + 2y = 14 \\[1em] \Rightarrow \frac{-35 - 25y}{2} + 2y = 14 \\[1em] \Rightarrow \frac{-35 - 25y + 4y}{2} = 14 \\[1em] \Rightarrow -35 - 21y = 28 \\[1em] \Rightarrow -21y = 28 + 35 \\[1em] \Rightarrow -21y = 63 \\[1em] \Rightarrow y = -3.\end{aligned}$$

Substituting y = -3 in (i) we get,

$x = \frac{-(7 + 5(-3))}{2} = \frac{-(7 - 15)}{2} = \frac{8}{2} = 4$.

Hence, x = 4 and y = -3.

Given,

$$\Rightarrow \begin{bmatrix*}[r] x + y & x - 4 \end{bmatrix*}\begin{bmatrix*}[r] -1 & -2 \\ 2 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] (x + y) \times -1 + (x - 4) \times 2 & (x + y) \times -2 + (x - 4) \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x - y + 2x - 8 & -2x - 2y + 2x - 8 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x - y - 8 & - 2y - 8 \end{bmatrix*} = \begin{bmatrix*}[r] -7 & -11 \end{bmatrix*}$$

By definition of equality of matrices we get,

-2y - 8 = -11

⇒ -2y = -3

⇒ y = $\frac{3}{2}$.

x - y - 8 = -7

⇒ x - y = 1

⇒ x = 1 + y

⇒ x = $1 + \frac{3}{2} = \frac{5}{2}$.

Hence, $x = \frac{5}{2}, y = \frac{3}{2}.$

Given,

$$\Rightarrow \begin{bmatrix*}[r] -2 & 0 \\ 3 & 1 \end{bmatrix*}\begin{bmatrix*}[r] -1 \\ 2x \end{bmatrix*} + 3\begin{bmatrix*}[r] -2 \\ 1 \end{bmatrix*} = 2\begin{bmatrix*}[r] y \\ 3 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -2 \times -1 + 0 \times 2x \\ 3 \times (-1) + 1 \times 2x \end{bmatrix*} + \begin{bmatrix*}[r] -6 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \\ -3 + 2x \end{bmatrix*} + \begin{bmatrix*}[r] -6 \\ 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] 2 + (-6) \\ -3 + 2x + 3 \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*} \\[1em] \begin{bmatrix*}[r] -4 \\ 2x \end{bmatrix*} = \begin{bmatrix*}[r] 2y \\ 6 \end{bmatrix*}$$

By definition of equality of matrices we get,

2y = -4
⇒ y = -2.

2x = 6
⇒ x = 3.

Hence, x = 3 and y = -2.

(i) Let order of matrix M be a × b.

$$M_{a \times b} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*}_{2 \times 2} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*}_{1 \times 2}$$

Since, the product of matrices is possible, only when the number of columns in the first matrix is equal to the number of rows in the second.

∴ b = 2

Also, the no. of rows of product (resulting) matrix is equal to no. of rows of first matrix.

∴ a = 1

Order of matrix M = a × b = 1 × 2.

Let M = $$\begin{bmatrix*}[r] x & y \end{bmatrix*}$$.

$$\Rightarrow \begin{bmatrix*}[r] x & y \end{bmatrix*} \times \begin{bmatrix*}[r] 1 & 1 \\ 0 & 2 \end{bmatrix*} =\begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 1 + y \times 0 & x \times 1 + y \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x & x + 2y \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \end{bmatrix*}$$

By definition of equality of matrices we get,

x = 1

x + 2y = 2
⇒ 1 + 2y = 2
⇒ 2y = 1
⇒ y = $\frac{1}{2}$.

∴ M = $$\begin{bmatrix*}[r] x & y \end{bmatrix*} = \begin{bmatrix*}[r] 1 & \frac{1}{2} \end{bmatrix*}$$.

Hence, M = $$\begin{bmatrix*}[r] 1 & \frac{1}{2} \end{bmatrix*}.$$

(ii) Let order of matrix M be a × b.

i.e. $$\begin{bmatrix*}[r] 1 & 4 \\ 2 & 1 \end{bmatrix*}_{2 \times 2} \times M_{a \times b} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*}_{2 \times 1}$$

Since product of matrix is possible, only when the number of columns in the first matrix is equal to no. of rows in second.

∴ a = 2.

Also the no. of columns of product (resulting matrix) is equal to no. of columns of second matrix.

∴ b = 1.

Hence, order of matrix = 2 × 1.

Let M = $$\begin{bmatrix*}[r] x \\ y \end{bmatrix*}$$

$$\Rightarrow \begin{bmatrix*}[r] 1 & 4 \\ 2 & 1 \end{bmatrix*} \times \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 \times x + 4 \times y \\ 2 \times x + 1 \times y \end{bmatrix*} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x + 4y \\ 2x + y \end{bmatrix*} = \begin{bmatrix*}[r] 13 \\ 5 \end{bmatrix*}$$

By definition of equality of matrices we get,

x + 4y = 13
⇒ x = 13 - 4y ......(i)

2x + y = 5
⇒ 2(13 - 4y) + y = 5
⇒ 26 - 8y + y = 5
⇒ -7y = -21
⇒ y = 3.

⇒ x = 13 - 4y = 13 - 4(3) = 13 - 12 = 1.

$$\therefore M = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 1 \\ 3 \end{bmatrix*}.$$

Hence, $$M = \begin{bmatrix*}[r] 1 \\ 3 \end{bmatrix*}.$$

Given,

⇒ A² = B

$$\Rightarrow \begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 2 & x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + x \times 0 & 2 \times x + x \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 + 0 & 2x + x \\ 0 & 0 + 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 3x \\ 0 & 1 \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 36 \\ 0 & 1 \end{bmatrix*}$$

By definition of equality of matrices we get,

3x = 36
⇒ x = 12.

Hence, x = 12.

Since, AB = BA = B and B is not a zero matrix.

It means that A is a unit or identity matrix.

Given,

$$\Rightarrow AB = A + B \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 & 0 \\ 0 & 4 \end{bmatrix*}\begin{bmatrix*}[r] a & b \\ 0 & c \end{bmatrix*} = \begin{bmatrix*}[r] 3 & 0 \\ 0 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] a & b \\ 0 & c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times a + 0 \times 0 & 3 \times b + 0 \times c \\ 0 \times a + 4 \times 0 & 0 \times b + 4 \times c \end{bmatrix*} = \begin{bmatrix*}[r] 3 + a & b \\ 0 & 4 + c \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3a & 3b \\ 0 & 4c \end{bmatrix*} = \begin{bmatrix*}[r] 3 + a & b \\ 0 & 4 + c \end{bmatrix*}$$

By definition of equality of matrices we get,

3a = 3 + a
⇒ 3a - a = 3
⇒ 2a = 3
⇒ a = $\frac{3}{2}$.

3b = b
⇒ b = 0.

4c = 4 + c
⇒ 4c - c = 4
⇒ 3c = 4
⇒ c = $\frac{4}{3}$

Hence, a = $\frac{3}{2}, b = 0 \text{ and c} = \frac{4}{3}$.

Given,

$$\Rightarrow AX = B \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 1 & 3 \end{bmatrix*}X = \begin{bmatrix*}[r] 3 \\ -11 \end{bmatrix*}$$

X will be a matrix of order 2 × 1. So, let X = $$\begin{bmatrix*}[r] a \\ b \end{bmatrix*}$$.

$$\Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 1 & 3 \end{bmatrix*}\begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times a + 1 \times b \\ 1 \times a + 3 \times b \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ -11 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2a + b \\ a + 3b \end{bmatrix*} = \begin{bmatrix*}[r] 3 \\ -11 \end{bmatrix*}$$

By definition of equality of matrices we get,

2a + b = 3
⇒ b = 3 - 2a .......(i)

a + 3b = -11

Substituting value of b from (i) in above equation we get,

⇒ a + 3(3 - 2a) = -11
⇒ a + 9 - 6a = -11
⇒ -5a = -11 - 9
⇒ -5a = -20
⇒ a = 4.

b = 3 - 2a = 3 - 2(4) = 3 - 8 = -5.

Hence, X = $$\begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ -5 \end{bmatrix*}.$$

$$M^2 = \begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 \times 4 + 1 \times (-1) & 4 \times 1 + 1 \times 2 \\ -1 \times 4 + 2 \times (-1) & -1 \times 1 + 2 \times 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 16 - 1 & 4 + 2 \\ -4 + (-2) & -1 + 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 15 & 6 \\ -6 & 3 \end{bmatrix*}.$$

Substituting value of M² in L.H.S. of 6M - M² = 9I,

$$\phantom{=} 6\begin{bmatrix*}[r] 4 & 1 \\ -1 & 2 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & 6 \\ -6 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 & 6 \\ -6 & 12 \end{bmatrix*} - \begin{bmatrix*}[r] 15 & 6 \\ -6 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 24 - 15 & 6 - 6 \\ -6 - (-6) & 12 - 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 0 \\ 0 & 9 \end{bmatrix*} \\[1em] = 9\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = 9\text{I}$$

Since, L.H.S. = R.H.S.

Hence, proved that 6M - M² = 9I.

Given,

PQ = null matrix.

$$\therefore \begin{bmatrix*}[r] 2 & 6 \\ 3 & 9 \end{bmatrix*}\begin{bmatrix*}[r] 3 & x \\ y & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 3 + 6 \times y & 2 \times x + 6 \times 2 \\ 3 \times 3 + 9 \times y & 3 \times x + 9 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6 + 6y & 2x + 12 \\ 9 + 9y & 3x + 18 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}.$$

By definition of equality of matrices we get,

6 + 6y = 0
⇒ 6y = -6
⇒ y = -1.

2x + 12 = 0
⇒ 2x = -12
⇒ x = -6.

Hence, x = -6 and y = -1.

Given,

$$\Rightarrow \begin{bmatrix*}[r] 2cos60° & -2sin30° \\ -tan 45° & cos0° \end{bmatrix*}\begin{bmatrix*}[r] cot 45° & cosec 30° \\ sec60° & sin 90° \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times \frac{1}{2} & -2 \times \frac{1}{2} \\ -1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 2 \\ 2 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 \times 1 + (-1) \times 2 & 1 \times 2 + (-1) \times 1 \\ -1 \times 1 + 1 \times 2 & -1 \times 2 + 1 \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 - 2 & 2 - 1 \\ -1 + 2 & -2 + 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -1 & 1 \\ 1 & -1 \end{bmatrix*}.$$

Hence, $$\begin{bmatrix*}[r] 2cos60° & -2sin30° \\ -tan 45° & cos0° \end{bmatrix*}\begin{bmatrix*}[r] cot 45° & cosec 30° \\ sec60° & sin 90° \end{bmatrix*} = \begin{bmatrix*}[r] -1 & 1 \\ 1 & -1 \end{bmatrix*}.$$

(i) A + B = B + A

The above statement is true because addition of matrices is commutative.

(ii) A - B = B - A

The above statement is false because subtraction of matrices is not commutative.

(iii) (B.C).A = B.(C.A)

The above statement is true because multiplication of matrices is associative.

(iv) (A + B).C = A.C + B.C

The above statement is true because multiplication of matrices is distributive over addition.

(v) A.(B - C) = A.B - A.C

The above statement is true because multiplication of matrices is distributive over subtraction.

(vi) (A - B).C = A.C - B.C

The above statement is true because multiplication of matrices is distributive over subtraction.

(vii) A² - B² = (A + B)(A - B)

The above statement is false because laws of algebra for factorization and expansion are not applicable to matrices.

(viii) (A - B)² = A² - 2A.B + B²

The above statement is false because laws of algebra for factorization and expansion are not applicable to matrices.

Since, addition and subtraction of matrices requires each matrix to be of same order.

∴ A + B and A - B are not possible.

AB is possible because no. of columns in A (2) is equal to no. of rows in B (2).

BA is not possible because no. of columns in B (1) is not equal to no. of rows in A (2).

Hence, Option 3 is the correct option.

We know that,

For matrix multiplication :

No. of columns in 1st matrix must be equal to the number of rows in 2nd matrix.

Resultant matrix order = No. of rows in 1st matrix × No. of columns in 2nd matrix.

Let order of matrix M be a × b.

⇒ M_{a × b} × $$\begin{bmatrix*}[r] 3 & 2 \\ -1 & 0 \end{bmatrix*}_{2 \times 2} = \begin{bmatrix*}[r] 3 & -1 \end{bmatrix*}_{1 \times 2}$$

b = 2 and a = 1.

Order of matrix M = 1 × 2.

Hence, Option 3 is the correct option.

Given,

$$\begin{bmatrix*}[r] 2x - y \\ x + y \end{bmatrix*} = \begin{bmatrix*}[r] 9 \\ 9 \end{bmatrix*}$$

⇒ 2x - y = 9 .......(1)

⇒ x + y = 9 ........(2)

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

⇒ 2x - y + x + y = 9 + 9

⇒ 3x = 18

⇒ x = $\frac{18}{3}$ = 6.

Substituting value of x in equation (2), we get :

⇒ 6 + y = 9

⇒ y = 9 - 6 = 3.

Hence, Option 4 is the correct option.

Substituting values of A and B in A + B, we get :

$$\Rightarrow A + B = \begin{bmatrix*}[r] x - y & x + y \\ y - x & y + x \end{bmatrix*} + \begin{bmatrix*}[r] x + y & y - x \\ x - y & y + x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x - y + x + y & x + y + y - x \\ y - x + x - y & y + x + y + x \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2x & 2y \\ 0 & 2(x + y) \end{bmatrix*}.$$

Hence, Option 3 is the correct option.

We know that,

Resultant matrix order = No. of rows in 1st matrix × No. of columns in 2nd matrix.

Resultant matrix (P) on multiplication of AB has order = 3 × 3.

Resultant matrix (Q) on multiplication of BA has order = 5 × 5.

Both P and Q are square matrix of order 3 and 5 respectively.

Hence, Option 4 is the correct option.

It is not necessarily true in case of matrices that if A X B = 0

Then, either A = 0 or B = 0.

So, assertion (A) is true.

According to reason, $$A = \begin{bmatrix*}[r] 2 & 2 \\ 5 & 5 \end{bmatrix*}$$ and $$B = \begin{bmatrix*}[r] -4 & 3 \\ 4 & -3 \end{bmatrix*}$$

$$\Rightarrow AB = \begin{bmatrix*}[r] 2 & 2 \\ 5 & 5 \end{bmatrix*}.\begin{bmatrix*}[r] -4 & 3 \\ 4 & -3 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 2 \times (-4) + 2 \times 4 & 2 \times 3 + 2 \times (-3)\ 5 \times (-4) + 5 \times 4 & 5 \times 3 + 5 \times (-3) \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] -8 + 8 & 6 - 6\ -20 + 20 & 15 - 15 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 0 & 0\ 0 & 0 \end{bmatrix*}\\[1em]$$

So, reason (R) is true.

Thus, Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

∴ Order of matrix A = 1 x 2

∴ Order of matrix B = 2 x 1

Since product of matrix is possible, only when the number of columns in the first matrix is equal to number of rows in second.

Since,

Number of columns in B = 1

Number of rows in A = 1

∴ Product BA is possible.

We know that,

The no. of rows in the resulting matrix is equal to the no. of rows in first matrix and no. of columns equals to the no. of columns in second matrix.

∴ Order of matrix BA = 2 x 2

So, assertion is true. But reason says the product BA is possible only if number of rows in matrix B is same as number of columns in matrix A, which is false.

Hence, option 1 is the correct option.

We know that,

Matrix multiplication does not satisfy the cancellation law.

∴ Statement 2 is false.

Thus, if A x B = A x C, we cannot conclude that B = C.

∴ Statement 1 is false.

Hence, option 2 is the correct option.

Given,

Matrix A = $$\begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*}$$ and matrix B = $$\begin{bmatrix*}[r] 5 & 5 \\ 5 & 5 \end{bmatrix*}$$

Solving,

$$\Rightarrow AB = \begin{bmatrix*}[r] 2 & -2 \\ -2 & 2 \end{bmatrix*}. \begin{bmatrix*}[r] 5 & 5 \\ 5 & 5 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 2 \times 5 + (-2) \times 5 & 2 \times 5 + (-2) \times 5 \\ (-2) \times 5 + 2 \times 5 & (-2) \times 5 + 2 \times 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 10 - 10 & 10 - 10 \\ -10 + 10 & -10 + 10 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}$$

So, AB = 0 and also from above example it is proved that AB can be equal to zero even if A ≠ 0 and B ≠ 0.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Given,

$$\Rightarrow \begin{bmatrix*}[r] 3 & -2 \\ -1 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 2x \\ 1 \end{bmatrix*} + 2\begin{bmatrix*}[r] -4 \\ 5 \end{bmatrix*} = 4\begin{bmatrix*}[r] 2 \\ y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3 \times 2x + (-2) \times 1 \\ -1 \times 2x + 4 \times 1 \end{bmatrix*} + \begin{bmatrix*}[r] -8 \\ 10 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6x - 2 \\ -2x + 4 \end{bmatrix*} + \begin{bmatrix*}[r] -8 \\ 10 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6x - 2 + (-8) \\ -2x + 4 + 10 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 6x - 10 \\ -2x + 14 \end{bmatrix*} = \begin{bmatrix*}[r] 8 \\ 4y \end{bmatrix*}$$

By definition of equality of matrices we get,

6x - 10 = 8
⇒ 6x = 18
⇒ x = 3

-2x + 14 = 4y
⇒ -2(3) + 14 = 4y
⇒ -6 + 14 = 4y
⇒ 8 = 4y
⇒ y = 2.

Hence, x = 3 and y = 2.

Given,

$$\Rightarrow \begin{bmatrix*}[r] 3x \times 1 + 8 \times 3 & 3x \times 4 + 8 \times 7 \end{bmatrix*} - \begin{bmatrix*}[r] 6 & -21 \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3x + 24 & 12x + 56 \end{bmatrix*} - \begin{bmatrix*}[r] 6 & -21 \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3x + 24 - 6 & 12x + 56 - (-21) \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 3x + 18 & 12x + 77 \end{bmatrix*} = \begin{bmatrix*}[r] 15 & 10y \end{bmatrix*}$$

By definition of equality of matrices we get,

3x + 18 = 15
⇒ 3x = -3
⇒ x = -1.

12x + 77 = 10y
⇒ 12(-1) + 77 = 10y
⇒ 65 = 10y
⇒ y = 6.5

Hence, x = -1 and y = 6.5

Given,

$$\Rightarrow \begin{bmatrix*}[r] x & y \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 25 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times x + y \times y \end{bmatrix*} = \begin{bmatrix*}[r] 25 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x^2 + y^2 \end{bmatrix*} = \begin{bmatrix*}[r] 25 \end{bmatrix*}$$

By definition of equality of matrices we get,

x² + y² = 25
⇒ x² = 25 - y² ......(i)

Given,

$$\Rightarrow \begin{bmatrix*}[r] -x & y \end{bmatrix*}\begin{bmatrix*}[r] 2x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -x \times 2x + y \times y \end{bmatrix*} = \begin{bmatrix*}[r] -2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -2x^2 + y^2 \end{bmatrix*} = \begin{bmatrix*}[r] -2 \end{bmatrix*}$$

By definition of equality of matrices we get,

-2x² + y² = -2 ......(ii)

Substituting value of x² from (i) in (ii) we get,

⇒ -2(25 - y²) + y² = -2
⇒ -50 + 2y² + y² = -2
⇒ 3y² = -2 + 50
⇒ 3y² = 48
⇒ y² = 16
⇒ y = ± 4.

⇒ x² = 25 - y²
⇒ x² = 25 - 16
⇒ x² = 9
⇒ x = ± 3.

(i) Since, x, y ∈ W

∴ x = 3, y = 4.

Hence, x = 3 and y = 4.

(ii) Since, x, y ∈ Z

∴ x = ±3, y = ±4.

Hence, x = ±3 and y = ±4.

Given,

$$\Rightarrow \begin{bmatrix*}[r] cos 45° & sin 30° \\ \sqrt{2}cos 0° & sin 0° \end{bmatrix*}\begin{bmatrix*}[r] sin 45° & cos 90° \\ sin 90° & cot 45° \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \frac{1}{\sqrt{2}} & \frac{1}{2} \\ \sqrt{2}(1) & 0 \end{bmatrix*}\begin{bmatrix*}[r] \frac{1}{\sqrt{2}} & 0 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \frac{1}{\sqrt{2}} & \frac{1}{2} \\ \sqrt{2} & 0 \end{bmatrix*}\begin{bmatrix*}[r] \frac{1}{\sqrt{2}} & 0 \\ 1 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \times 1 & \frac{1}{\sqrt{2}} \times 0 + \frac{1}{2} \times 1 \\ \sqrt{2} \times \frac{1}{\sqrt{2}} + 0 \times 1 & \sqrt{2} \times 0 + 0 \times 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] \frac{1}{2} + \frac{1}{2} & 0 + \frac{1}{2} \\ 1 + 0 & 0 + 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & \frac{1}{2} \\ 1 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 1 & 0.5 \\ 1 & 0 \end{bmatrix*}$$

Hence, $$\begin{bmatrix*}[r] cos 45° & sin 30° \\ \sqrt{2}cos 0° & sin 0° \end{bmatrix*}\begin{bmatrix*}[r] sin 45° & cos 90° \\ sin 90° & cot 45° \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 0.5 \\ 1 & 0 \end{bmatrix*}.$$

Let order of matrix M be a × b.

Given,

$$\Rightarrow 3A \times M = 2B .......(i) \\[1em] \Rightarrow 3\begin{bmatrix*}[r] 0 & -1 \\ 4 & -3 \end{bmatrix*}_{2 \times 2} \times M_{a \times b} = 2\begin{bmatrix*}[r] -5 \\ 6 \end{bmatrix*}_{2 \times 1}$$

Since product of matrix is possible, only when the number of columns in the first matrix is equal to no. of rows in second.

∴ a = 2.

Also the no. of columns of product (resulting matrix) is equal to no. of columns of second matrix.

∴ b = 1.

Hence, order of matrix M = 2 × 1.

Let M = $$\begin{bmatrix*}[r] a \\ b \end{bmatrix*}$$

Substituting value of A, M and B in (i) we get,

$$\Rightarrow 3\begin{bmatrix*}[r] 0 & -1 \\ 4 & -3 \end{bmatrix*} \times \begin{bmatrix*}[r] a \\ b \end{bmatrix*} = 2\begin{bmatrix*}[r] -5 \\ 6 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -3 \\ 12 & -9 \end{bmatrix*} \times \begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] -10 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 \times a + (-3) \times b \\ 12 \times a + (-9) \times b \end{bmatrix*} = \begin{bmatrix*}[r] -10 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -3b \\ 12a - 9b \end{bmatrix*} = \begin{bmatrix*}[r] -10 \\ 12 \end{bmatrix*}$$

By definition of equality of matrices we get,

-3b = -10
⇒ b = $\frac{10}{3}$ ....(i)

12a - 9b = 12

Substituting value of b from (i) in above equation we get,

$$\begin{aligned}\Rightarrow 12a - 9 \times \frac{10}{3} = 12 \\[1em] \Rightarrow 12a - 30 = 12 \\[1em] \Rightarrow 12a = 42 \\[1em] \Rightarrow a = \frac{7}{2}.\end{aligned}$$

$$\therefore \begin{bmatrix*}[r] a \\ b \end{bmatrix*} = \begin{bmatrix*}[r] \frac{7}{2} \\ \frac{10}{3} \end{bmatrix*}.$$

Hence, M = $$\begin{bmatrix*}[r] \frac{7}{2} \\ \frac{10}{3} \end{bmatrix*}.$$

$$\Rightarrow \begin{bmatrix*}[r] x & 3x \\ y & 4y \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ 1 \end{bmatrix*} = \begin{bmatrix*}[r] 5 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x \times 2 + 3x \times 1 \\ y \times 2 + 4y \times 1 \end{bmatrix*}= \begin{bmatrix*}[r] 5 \\ 12 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 5x \\ 6y \end{bmatrix*}= \begin{bmatrix*}[r] 5 \\ 12 \end{bmatrix*} \\[1em]$$

By definition of equality of matrices we get,

5x = 5
⇒ x = 1.

6y = 12
⇒ y = 2.

Hence, x = 1 and y = 2.

Given,

$$X = \begin{bmatrix*}[r] -3 & 4 \\ 2 & -3 \end{bmatrix*}\begin{bmatrix*}[r] 2 \\ -2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 \times 2 + 4 \times (-2) \\ 2 \times 2 + (-3) \times (-2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -6 + (-8) \\ 4 + 6 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -14 \\ 10 \end{bmatrix*}.$$

Given,

$$\Rightarrow 2X - 3Y = \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 2\begin{bmatrix*}[r] -14 \\ 10 \end{bmatrix*} - 3Y = \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] -28 \\ 20 \end{bmatrix*} - 3Y = \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 3Y = \begin{bmatrix*}[r] -28 \\ 20 \end{bmatrix*} - \begin{bmatrix*}[r] 10 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 3Y = \begin{bmatrix*}[r] -28 - 10 \\ 20 - (-8) \end{bmatrix*} \\[1em] \Rightarrow 3Y = \begin{bmatrix*}[r] -38 \\ 28 \end{bmatrix*} \\[1em] \Rightarrow Y = \frac{1}{3}\begin{bmatrix*}[r] -38 \\ 28 \end{bmatrix*}.$$

Hence, X = $$\begin{bmatrix*}[r] -14 \\ 10 \end{bmatrix*} \text{ and Y} = \frac{1}{3}\begin{bmatrix*}[r] -38 \\ 28 \end{bmatrix*}$$.

$$A = \begin{bmatrix*}[r] 2 & 5 \\ 1 & 3 \end{bmatrix*}, A^t = \begin{bmatrix*}[r] 2 & 1 \\ 5 & 3 \end{bmatrix*}, I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$$.

$$\Rightarrow A^t.B + BI = \begin{bmatrix*}[r] 2 & 1 \\ 5 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 4 & -2 \\ -1 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & -2 \\ -1 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 4 + 1 \times (-1) & 2 \times (-2) + 1 \times 3 \\ 5 \times 4 + 3 \times (-1) & 5 \times (-2) + 3 \times 3 \end{bmatrix*} + \begin{bmatrix*}[r] 4 \times 1 + (-2) \times 0 & 4 \times 0 + (-2) \times 1 \\ -1 \times 1 + 3 \times 0 & -1 \times 0 + 3 \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 8 - 1 & -4 + 3 \\ 20 - 3 & -10 + 9 \end{bmatrix*} + \begin{bmatrix*}[r] 4 + 0 & 0 - 2 \\ -1 + 0 & 0 + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 7 & -1 \\ 17 & -1 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & -2 \\ -1 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 7 + 4 & -1 + (-2) \\ 17 + (-1) & -1 + 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 11 & -3 \\ 16 & 2 \end{bmatrix*}.$$

Hence, $$A^t.B + BI = \begin{bmatrix*}[r] 11 & -3 \\ 16 & 2 \end{bmatrix*}.$$

$$A^2 = \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 0 \times 2 & 1 \times 0 + 0 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 0 + 1 \times 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 0 & 0 + 0 \\ 2 + 2 & 0 + 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} \\[1em] B^2 = \begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 2 + 3 \times (-1) & 2 \times 3 + 3 \times 0 \\ -1 \times 2 + 0 \times (-1) & -1 \times 3 + 0 \times 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 - 3 & 6 + 0 \\ -2 + 0 & -3 + 0 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*}.$$

Substituting value of A² and B² in A² + AB + B² we get,

$$A^2 + AB + B^2 = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 0 \\ 2 & 1 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 3 \\ -1 & 0 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 1 \times 2 + 0 \times (-1) & 1 \times 3 + 0 \times 0 \\ 2 \times 2 + 1 \times (-1) & 2 \times 3 + 1 \times 0 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 2 + 0 & 3 + 0 \\ 4 - 1 & 6 + 0 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 4 & 1 \end{bmatrix*} + \begin{bmatrix*}[r] 2 & 3 \\ 3 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & 6 \\ -2 & -3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 2 + 1 & 0 + 3 + 6 \\ 4 + 3 + (-2) & 1 + 6 + (-3) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 9 \\ 5 & 4 \end{bmatrix*}.$$

Hence, A² + AB + B² = $$\begin{bmatrix*}[r] 4 & 9 \\ 5 & 4 \end{bmatrix*}.$$

Given,

$$\Rightarrow 3A - 2C = 6B \\[1em] \Rightarrow 3\begin{bmatrix*}[r] 3 & a \\ -4 & 8 \end{bmatrix*} - 2\begin{bmatrix*}[r] -1 & 4 \\ 3 & b \end{bmatrix*} = 6\begin{bmatrix*}[r] c & 4 \\ -3 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 & 3a \\ -12 & 24 \end{bmatrix*} - \begin{bmatrix*}[r] -2 & 8 \\ 6 & 2b \end{bmatrix*} = \begin{bmatrix*}[r] 6c & 24 \\ -18 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 9 - (-2) & 3a - 8 \\ -12 - 6 & 24 - 2b \end{bmatrix*} = \begin{bmatrix*}[r] 6c & 24 \\ -18 & 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 11 & 3a - 8 \\ -18 & 24 - 2b \end{bmatrix*} = \begin{bmatrix*}[r] 6c & 24 \\ -18 & 0 \end{bmatrix*}$$

By definition of equality of matrices we get,

6c = 11
⇒ c = $\frac{11}{6} = 1\frac{5}{6}.$

3a - 8 = 24
⇒ 3a = 32
⇒ a = $\frac{32}{3} = 10\frac{2}{3}$

24 - 2b = 0
⇒ 2b = 24
⇒ b = 12.

Hence, a = $10\frac{2}{3}, b = 12, c = 1\frac{5}{6}.$

Given,

$$\Rightarrow BA = C^2 \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -q \\ 1 & 0 \end{bmatrix*}\begin{bmatrix*}[r] p & 0 \\ 0 & 2 \end{bmatrix*} = \begin{bmatrix*}[r] 2 & -2 \\ 2 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 2 & -2 \\ 2 & 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 \times p + (-q) \times 0 & 0 \times 0 + (-q) \times 2 \\ 1 \times p + 0 \times 0 & 1 \times 0 + 0 \times 2 \end{bmatrix*} = \begin{bmatrix*}[r] 2 \times 2 + (-2) \times 2 & 2 \times (-2) + (-2) \times 2 \\ 2 \times 2 + 2 \times 2 & 2 \times (-2) + 2 \times 2 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 0 & -2q \\ p & 0 \end{bmatrix*} = \begin{bmatrix*}[r] 0 & -8 \\ 8 & 0 \end{bmatrix*}$$

By definition of equality of matrices we get,

-2q = -8
⇒ q = 4.

p = 8.

Hence, p = 8 and q = 4.

$$\Rightarrow \begin{bmatrix*}[r] 4 sin 30° & 2cos 60° \\ sin 90° & 2 cos 0° \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 4 \times \frac{1}{2} & 2 \times \frac{1}{2} \\ 1 & 2 \times 1 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 & 1 \\ 1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 2 \times 4 + 1 \times 5 & 2 \times 5 + 1 \times 4 \\ 1 \times 4 + 2 \times 5 & 1 \times 5 + 2 \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 8 + 5 & 10 + 4 \\ 4 + 10 & 5 + 8 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 13 & 14 \\ 14 & 13 \end{bmatrix*}.$$

Hence, $$\begin{bmatrix*}[r] 4 sin 30° & 2cos 60° \\ sin 90° & 2 cos 0° \end{bmatrix*}\begin{bmatrix*}[r] 4 & 5 \\ 5 & 4 \end{bmatrix*} = \begin{bmatrix*}[r] 13 & 14 \\ 14 & 13 \end{bmatrix*}.$$

Given,

$$\Rightarrow A^2 = 9A + mI \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 0 \\ -1 & 7 \end{bmatrix*}\begin{bmatrix*}[r] 2 & 0 \\ -1 & 7 \end{bmatrix*} = 9\begin{bmatrix*}[r] 2 & 0 \\ -1 & 7 \end{bmatrix*} + m\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 \times 2 + 0 \times (-1) & 2 \times 0 + 0 \times 7 \\ -1 \times 2 + 7 \times (-1) & -1 \times 0 + 7 \times 7 \end{bmatrix*} = \begin{bmatrix*}[r] 18 & 0 \\ -9 & 63 \end{bmatrix*} + \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 & 0 \\ -9 & 49 \end{bmatrix*} = \begin{bmatrix*}[r] 18 & 0 \\ -9 & 63 \end{bmatrix*} + \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} = \begin{bmatrix*}[r] 4 & 0 \\ -9 & 49 \end{bmatrix*} - \begin{bmatrix*}[r] 18 & 0 \\ -9 & 63 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} = \begin{bmatrix*}[r] 4 - 18 & 0 - 0 \\ -9 - (-9) & 49 - 63 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] m & 0 \\ 0 & m \end{bmatrix*} = \begin{bmatrix*}[r] -14 & 0 \\ 0 & -14 \end{bmatrix*}$$

By definition of equality of matrices we get,

⇒ m = -14.

Hence, m = -14.

(i) Let order of matrix X be a × b.

i.e. $$\begin{bmatrix*}[r] 4 sin 30° & cos 0° \\ cos 0° & 4 sin 30° \end{bmatrix*}_{2 \times 2} \times X_{a \times b} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*}_{2 \times 1}$$

Since product of matrix is possible, only when the number of columns in the first matrix is equal to no. of rows in second.

∴ a = 2.

Also the no. of columns of product (resulting matrix) is equal to no. of columns of second matrix.

∴ b = 1.

Hence, order of matrix X = 2 × 1.

(ii) Let matrix X = $$\begin{bmatrix*}[r] x \\ y \end{bmatrix*}$$

Given,

$$\Rightarrow AX = B \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 sin 30° & cos 0° \\ cos 0° & 4 sin 30° \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times \frac{1}{2} & 1 \\ 1 & 4 \times \frac{1}{2} \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2 & 1 \\ 1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + y \\ x + 2y \end{bmatrix*} = \begin{bmatrix*}[r] 4 \\ 5 \end{bmatrix*}$$

By definition of equality of matrices we get,

2x + y = 4
⇒ y = 4 - 2x .......(i)

x + 2y = 5

Substituting value of y from (i) in above equation we get,

⇒ x + 2(4 - 2x) = 5
⇒ x + 8 - 4x = 5
⇒ -3x = 5 - 8
⇒ -3x = -3
⇒ x = 1.

⇒ y = 4 - 2x = 4 - 2(1) = 2.

$$\therefore X = \begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 1 \\ 2 \end{bmatrix*}$$

Hence, X = $$\begin{bmatrix*}[r] 1 \\ 2 \end{bmatrix*}.$$

Quantity matrix :

Q = $$\begin{bmatrix*}[r] 15 & 3 & 8 \\ 10 & 4 & 12 \\ 5 & 5 & 9 \end{bmatrix*}$$

Price matrix :

P = $$\begin{bmatrix*}[r] 170 \\ 150 \\ 250 \end{bmatrix*}$$

The total amount :

$$\Rightarrow QP = \begin{bmatrix*}[r] 15 & 3 & 8 \\ 10 & 4 & 12 \\ 5 & 5 & 9 \end{bmatrix*} \times \begin{bmatrix*}[r] 170 \\ 150 \\ 250 \end{bmatrix*} \\[1em] \Rightarrow QP = \begin{bmatrix*}[r] 15 \times 170 + 3 \times 150 + 8 \times 250 \\ 10 \times 170 + 4 \times 150 + 12 \times 250 \\ 5 \times 170 + 5 \times 150 + 9 \times 250 \end{bmatrix*} \\[1em] \Rightarrow QP = \begin{bmatrix*}[r] 2550 + 450 + 2000 \\ 1700 + 600 + 3000 \\ 850 + 750 + 2250 \end{bmatrix*} \\[1em] \Rightarrow QP = \begin{bmatrix*}[r] 5000 \\ 5300 \\ 3850 \end{bmatrix*}.$$

Hence, the total amounts paid by Neeta, Salma and Vivian are ₹ 5,000, ₹ 5,300 and ₹ 3,850.