Matrices

(i) The matrix has 2 rows and 3 columns.

Hence, the order of matrix is (2 × 3).

(ii) The matrix has 3 rows and 2 columns.

Hence, the order of matrix is (3 × 2).

(iii) The matrix has 3 rows and 1 column.

Hence, the order of matrix is (3 × 1).

(iv) The matrix has 1 row and 2 columns.

Hence, the order of matrix is (1 × 2).

(v) The matrix has 2 rows and 2 columns.

Hence, the order of matrix is (2 × 2).

(vi) The matrix has 1 row and 1 column.

Hence, the order of matrix is (1 × 1).

(i) Since, there is only once column and three rows in the matrix.

Hence, it is column matrix of order 3 x 1.

(ii) Since, no. of rows = no. of columns in the matrix.

Hence, it is a square matrix of order 2.

(iii) Since, there is only one row and 3 columns in the matrix.

Hence, it is a row matrix of order 1 x 3.

(iv) Since, there are 2 rows and 2 columns in the matrix.

Also, only the main diagonal element is 1 and the rest elements are zero.

Hence, it is a unit matrix of order 2.

(v) Since, there are 3 rows and 3 columns in the matrix.

Also, all the elements apart from main diagonal is zero.

Hence , it is a diagonal matrix of order 3.

(vi) Since, there are 2 rows and 3 columns in the matrix and all elements are zero.

Hence , it is a zero matrix of order 2 x 3.

Given,

a_ij = (3i − j).

∴ a₁₁ = [3(1) - 1] = 2, a₁₂ = [3(1) - 2] = 1, a₁₃ = [3(1) - 3] = 0.

a₂₁ = [3(2) - 1] = 5, a₂₂ = [3(2) - 2]= 4, a₂₃ = [3(2) - 3] = 3.

Hence, required matrix = $$\begin{bmatrix} 2 & 1 & 0 \\ 5 & 4 & 3 \end{bmatrix}$$.

Given,

a_ij = (i × j)

⇒ a₁₁ = 1 × 1 = 1, a₁₂ = 1 × 2 = 2.

⇒ a₂₁ = 2 × 1 = 2, a₂₂ = 2 × 2 = 4.

⇒ a₃₁ = 3 × 1 = 3, a₃₂ = 3 × 2 = 6.

Hence, required matrix = $$\begin{bmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{bmatrix}$$.

If a matrix has 4 elements the possible orders are, 1 x 4, 2 x 2, 4 x 1.

Hence, possible orders are 1 x 4, 2 x 2, 4 x 1.

If a matrix has 6 elements the possible orders are, 1 × 6, 6 × 1, 2 × 3, 3 × 2.

Hence, possible orders are 1 × 6, 6 × 1, 2 × 3, 3 × 2.

Given,

$$\begin{bmatrix} a + 3 & 4 \\ -2 & b - 6 \end{bmatrix} =\begin{bmatrix} -1 & c - 3 \\ d + 1 & 2 \end{bmatrix}$$

∴ a + 3 = -1

⇒ a = -1 - 3 = -4.

∴ b - 6 = 2

⇒ b = 6 + 2 = 8

∴ c - 3 = 4

⇒ c = 4 + 3 = 7

∴ d + 1 = -2

⇒ d = -2 - 1 = -3

Hence , a = -4, b = 8, c = 7, d = -3.

Given,

$$\begin{bmatrix} 5x + 3y \\ 2x - y \end{bmatrix} =\begin{bmatrix} 12 \\ 7 \end{bmatrix}$$

∴ 5x + 3y = 12....(1)

∴ 2x - y = 7

y = 2x - 7....(2)

Substituting value of y from equation (2) in equation (1), we get :

⇒ 5x + 3(2x - 7) = 12

⇒ 5x + 6x - 21 = 12

⇒ 11x = 12 + 21

⇒ 11x = 33

⇒ x = $\frac{33}{11}$ = 3.

Substitute value of x in equation (2):

⇒ y = 2x - 7

⇒ y = 2(3) - 7

⇒ y = 6 - 7

⇒ y = -1.

Hence , x = 3 and y = -1.

Given,

$$\begin{bmatrix} x + y & a - b \\ a + b & 2x - 3y \end{bmatrix} =\begin{bmatrix} 5 & 3 \\ -1 & -5 \end{bmatrix}$$

∴ x + y = 5

⇒ y = 5 - x ........(1)

∴ 2x - 3y = -5 ........(2)

Substituting value of y from equation(1) in (2), we get :

⇒ 2x - 3(5 - x) = -5

⇒ 2x - 15 + 3x = -5

⇒ 5x - 15 = -5

⇒ 5x = -5 + 15

⇒ 5x = 10

⇒ x = $\frac{10}{5}$ = 2.

Substitute value of x in equation 1 :

⇒ y = 5 - x

⇒ y = 5 - 2

⇒ y = 3.

∴ a - b = 3

⇒ a = 3 + b .........(3)

∴ a + b = -1 .........(4)

Substituting value of a from equation (3) in (4), we get :

⇒ 3 + b + b = -1

⇒ 2b = -1 - 3

⇒ 2b = -4

⇒ b = $\frac{-4}{2}$ = -2.

Substitute value of b in equation (3), we get :

⇒ a = -2 + 3

⇒ a = 1.

Hence, x = 2, y = 3, a = 1, b = -2.

The matrix otained by interchanging the rows and column of a matrix is called the transpose of the matrix.

(i) Given,

$$\Rightarrow A = \begin{bmatrix} 2 & 3 \\ 5 & -4 \end{bmatrix} \\[1em] \Rightarrow A^T = \begin{bmatrix} 2 & 5 \\ 3 & -4 \end{bmatrix}.$$

Hence, the transpose of matrix A $$= \begin{bmatrix} 2 & 5 \\ 3 & -4 \end{bmatrix}$$.

(ii) Given,

$$\Rightarrow B = \begin{bmatrix} 5 & 7 & -3 \end{bmatrix} \\[1em] \Rightarrow B^T = \begin{bmatrix} 5 \\ 7 \\ -3 \end{bmatrix}.$$

Hence, the transpose of matrix B = $$\begin{bmatrix} 5 \\ 7 \\ -3 \end{bmatrix}$$

(iii) Given,

$$\Rightarrow C = \begin{bmatrix} -2 \\ 6 \end{bmatrix} \\[1em] \Rightarrow C^T = \begin{bmatrix} -2 & 6 \\ \end{bmatrix}$$

Hence, the transpose of matrix C $$= \begin{bmatrix} -2 & 6 \\ \end{bmatrix}$$.

(i) 5A

$$\Rightarrow 5 \times \begin{bmatrix} 2 & -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 10 & -15 & 20 \end{bmatrix}$$

Hence, 5A = $$\begin{bmatrix} 10 & -15 & 20 \end{bmatrix}$$.

(ii) (-4)A

$$\Rightarrow -4 \times \begin{bmatrix} 2 & -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -8 & 12 & -16 \end{bmatrix}$$

Hence, (-4)A = $$\begin{bmatrix} -8 & 12 & -16 \end{bmatrix}$$.

(iii) −A

$$\Rightarrow -1 \times \begin{bmatrix} 2 & -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -2 & 3 & -4 \end{bmatrix}$$

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

(i) 3M

$$\Rightarrow 3 \times \begin{bmatrix} 6 & -4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 18 & -12 \end{bmatrix}$$

Hence, 3M = $$\begin{bmatrix} 18 & -12 \end{bmatrix}$$.

(ii) $\frac{1}{2}M$

$$\Rightarrow \frac{1}{2} \times \begin{bmatrix} 6 & -4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 & -2 \end{bmatrix}$$

Hence, $$\frac{1}{2}M = \begin{bmatrix} 3 & -2 \end{bmatrix}$$.

(iii) −2M

$$\Rightarrow -2 \times \begin{bmatrix} 6 & -4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -12 & 8 \end{bmatrix}$$

Hence, −2M = $$\begin{bmatrix} -12 & 8 \end{bmatrix}$$.

(i) 2C

$$\Rightarrow 2 \times\begin{bmatrix} 3 & -6 \\ 0 & 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 & -12 \\ 0 & 18 \end{bmatrix}$$

Hence, 2C = $$\begin{bmatrix} 6 & -12 \\ 0 & 18 \end{bmatrix}$$.

(ii) $\frac{1}{3}C$

$$\Rightarrow \frac{1}{3} \times\begin{bmatrix} 3 & -6 \\ 0 & 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}$$

Hence, $$\frac{1}{3}C = \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}$$.

(iii) −C

$$\Rightarrow -1 \times\begin{bmatrix} 3 & -6 \\ 0 & 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix}.$$

Hence, -C = $$\begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix}$$.

Given,

$$\Rightarrow 6M = \begin{bmatrix} 0 & 6 \\ -12 & 24 \end{bmatrix} \\[1em] \Rightarrow M = \frac{1}{6} \times \begin{bmatrix} 0 & 6 \\ -12 & 24 \end{bmatrix} \\[1em] \Rightarrow M = \begin{bmatrix} 0 & 1 \\ -2 & 4 \end{bmatrix}.$$

Hence, M = $$\begin{bmatrix} 0 & 1 \\ -2 & 4 \end{bmatrix}$$.

(i) Given,

$$A = \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix}$$

$$B = \begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix}$$

Solving,

$$\Rightarrow A + B = \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix} + \begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix} \\[1em] \Rightarrow A + B = \begin{bmatrix} 2 + 1 & 5 + (-3) \\ -3 + 2 & 7 + 5 \end{bmatrix}\\[1em] \Rightarrow A + B = \begin{bmatrix} 3 & 2 \\ -1 & 12 \end{bmatrix}.$$

Hence, A + B = $$\begin{bmatrix} 3 & 2 \\ -1 & 12 \end{bmatrix}$$.

(ii) Given,

$$A = \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix}$$

$$B = \begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix}$$

Solving,

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

Hence, A - B = $$\begin{bmatrix} 1 & 8 \\ -5 & 2 \end{bmatrix}$$.

(iii) Given,

$$A = \begin{bmatrix} 2 & 5 \\ -3 & 7 \end{bmatrix}$$

$$B = \begin{bmatrix} 1 & -3 \\ 2 & 5 \end{bmatrix}$$

Solving,

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

Hence, B - A = $$\begin{bmatrix} -1 & -8 \\ 5 & -2 \end{bmatrix}$$.

(i) 2M + 5N

Given,

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

$$N = \begin{bmatrix} -1 & 6 \\ 3 & 2 \end{bmatrix}$$

Solving,

$$\Rightarrow 2M + 5N = 2 \times \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} + 5 \times \begin{bmatrix} -1 & 6 \\ 3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & -6 \\ 8 & 10 \end{bmatrix} + \begin{bmatrix} -5 & 30 \\ 15 & 10 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 - 5 & -6 + 30 \\ 8 + 15 & 10 + 10 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -1 & 24 \\ 23 & 20 \end{bmatrix}.$$

Hence, 2M + 5N = $$\begin{bmatrix} -1 & 24 \\ 23 & 20 \end{bmatrix}$$.

(ii) 4N − 3M

Given,

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

$$N = \begin{bmatrix} -1 & 6 \\ 3 & 2 \end{bmatrix}$$

Solving,

$$\Rightarrow 4N − 3M = 4 \times \begin{bmatrix} -1 & 6 \\ 3 & 2 \end{bmatrix} - 3 \times \begin{bmatrix} 2 & -3 \\ 4 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -4 & 24 \\ 12 & 8 \end{bmatrix} - \begin{bmatrix} 6 & -9 \\ 12 & 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -4 -6 & 24 - (-9) \\ 12 - 12 & 8 - 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -10 & 33 \\ 0 & -7 \end{bmatrix}.$$

Hence, 4N − 3M = $$\begin{bmatrix} -10 & 33 \\ 0 & -7 \end{bmatrix}$$

(i) 2A + 3B

Given,

$$A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$$

$$B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$$

$$\Rightarrow 2A + 3B = 2 \times \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} + 3 \times \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & 8 \\ 6 & 4 \end{bmatrix} + \begin{bmatrix} 3 & 9 \\ -6 & 15 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 + 3 & 8 + 9 \\ 6 - 6 & 4 + 15 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 & 17 \\ 0 & 19 \end{bmatrix}.$$

Hence, 2A + 3B = $$\begin{bmatrix} 7 & 17 \\ 0 & 19 \end{bmatrix}$$.

(ii) 3B − 2C

Given,

$$B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$$

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

Solving,

$$\Rightarrow 3B − 2C = 3 \times \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} - 2 \times \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 & 9 \\ -6 & 15 \end{bmatrix} - \begin{bmatrix} -4 & 10 \\ 6 & 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 - (-4) & 9 - 10 \\ -6 - 6 & 15 - 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & -1 \\ -12 & 7 \end{bmatrix}$$

Hence, 3B − 2C = $$\begin{bmatrix} 7 & -1 \\ -12 & 7 \end{bmatrix}$$.

(iii) 3A − 2B + 4C

Given,

$$A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$$

$$B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$$

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

Solving,

$$\Rightarrow 3A − 2B + 4C = 3 \times \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - 2 \times \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} + 4 \times \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} - \begin{bmatrix} 2 & 6 \\ -4 & 10 \end{bmatrix} + \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 - 2 & 12 - 6 \\ 9 - (-4) & 6 - 10 \end{bmatrix} + \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 6 \\ 13 & -4 \end{bmatrix} + \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 - 8 & 6 + 20 \\ 13 + 12 & -4 + 16 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -4 & 26 \\ 25 & 12 \end{bmatrix}.$$

Hence, 3A − 2B + 4C = $$\begin{bmatrix} -4 & 26 \\ 25 & 12 \end{bmatrix}$$.

Given,

$$A = \begin{bmatrix} 0 & 1 \\ 5 & -1 \end{bmatrix}$$

$$B = \begin{bmatrix} 1 & -3 \\ 0 & -2 \end{bmatrix}$$

$$C = \begin{bmatrix} 2 & -5 \\ 4 & 0 \end{bmatrix}$$

$$\Rightarrow 3A + 4B − 5C = 3 \times \begin{bmatrix} 0 & 1 \\ 5 & -1 \end{bmatrix} + 4 \times \begin{bmatrix} 1 & -3 \\ 0 & -2 \end{bmatrix} - 5 \times \begin{bmatrix} 2 & -5 \\ 4 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 3 \\ 15 & -3 \end{bmatrix} + \begin{bmatrix} 4 & -12 \\ 0 & -8 \end{bmatrix} - \begin{bmatrix} 10 & -25 \\ 20 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 + 4 & 3 - 12 \\ 15 + 0 & -3 - 8 \end{bmatrix} - \begin{bmatrix} 10 & -25 \\ 20 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & -9 \\ 15 & -11 \end{bmatrix} - \begin{bmatrix} 10 & -25 \\ 20 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 - 10 & -9 - (-25) \\ 15 - 20 & -11 - 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -6 & 16 \\ -5 & -11 \end{bmatrix}.$$

Hence, 3A + 4B − 5C = $$\begin{bmatrix} -6 & 16 \\ -5 & -11 \end{bmatrix}$$.

Given,

$$\Rightarrow X + \begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -5 & -2 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} 3 & 1 \\ -5 & -2 \end{bmatrix} - \begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 - 4& 1 - 6 \\ -5 - (-3) & -2 - 7 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 & -5 \\ -2 & -9 \end{bmatrix}.$$

Hence, X = $$\begin{bmatrix} -1 & -5 \\ -2 & -9 \end{bmatrix}$$.

Given,

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

$$B = \begin{bmatrix} 5 & 2 \\ -7 & 3 \end{bmatrix}$$

Since,

⇒ A + B - C = 0

⇒ C = A + B

$$\Rightarrow C = \begin{bmatrix} -2 & 3 \\ 4 & 5 \end{bmatrix} + \begin{bmatrix} 5 & 2 \\ -7 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix} -2 + 5 & 3 + 2 \\ 4 - 7 & 5 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 & 5 \\ -3 & 8 \end{bmatrix}$$

Hence, C = $$\begin{bmatrix} 3 & 5 \\ -3 & 8 \end{bmatrix}$$

Given,

$$A = \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix}, B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}, C = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$$

⇒ A + X = 2B + C

⇒ X = 2B + C - A

$$\Rightarrow X = 2 \times \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -6 & 4 \\ 8 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -6 + 1 & 4 + 0 \\ 8 + 0 & 0 + 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -5 & 4 \\ 8 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -5 - 2 & 4 - (-1) \\ (8 - 2) & 2 - 0 \end{bmatrix} \\[1em] = \begin{bmatrix} -7 & 5 \\ 6 & 2 \end{bmatrix}.$$

Hence, X = $$\begin{bmatrix} -7 & 5 \\ 6 & 2 \end{bmatrix}.$$

Solving,

$$\Rightarrow \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} + 2M = 3\begin{bmatrix} 3 & 2 \\ 0 & -3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} + 2M = \begin{bmatrix} 9 & 6 \\ 0 & -9 \end{bmatrix} \\[1em] \Rightarrow 2M = \begin{bmatrix} 9 & 6 \\ 0 & -9 \end{bmatrix} - \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} \\[1em] \Rightarrow M = \frac{1}{2} \begin{bmatrix} 9 - 1 & 6 - 4 \\ 0 - (-2) & -9 - 3 \end{bmatrix} \\[1em] \Rightarrow M = \frac{1}{2} \begin{bmatrix} 8 & 2 \\ 2 & -12 \end{bmatrix} \\[1em] \Rightarrow M = \begin{bmatrix} 4 & 1 \\ 1 & -6 \end{bmatrix}.$$

Hence, M = $$\begin{bmatrix} 4 & 1 \\ 1 & -6 \end{bmatrix}$$.

Given,

$$A = \begin{bmatrix} 6 & 2 \\ 5 & -4 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}$$.

⇒ 2A + 3B − 5X = 0

⇒ 2A + 3B = 5X

⇒ X = $\frac{1}{5}$ (2A + 3B)

Substituting values of A and B, we get :

$$\Rightarrow X = \frac{1}{5} \Big(2 \times \begin{bmatrix} 6 & 2 \\ 5 & -4 \end{bmatrix} + 3 \times \begin{bmatrix} 1 & 2 \\ -5 & 1 \end{bmatrix}\Big) \\[1em] = \frac{1}{5} \Big(\begin{bmatrix} 12 & 4 \\ 10 & -8 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ -15 & 3 \end{bmatrix}\Big) \\[1em] = \frac{1}{5} \Big(\begin{bmatrix} 12 + 3 & 4 + 6 \\ 10 + (-15) & -8 + 3 \end{bmatrix}\Big) \\[1em] = \frac{1}{5}\begin{bmatrix} 15 & 10 \\ -5 & -5 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 & 2 \\ -1 & -1 \end{bmatrix}.$$

Hence, X = $$\begin{bmatrix} 3 & 2 \\ -1 & -1 \end{bmatrix}$$.

Given,

$$Y = \begin{bmatrix} 1 & 2 \\ -1 & 5 \end{bmatrix} \\[1em] \Rightarrow 2X + Y = \begin{bmatrix} 5 & 0 \\ -3 & 3 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix} 5 & 0 \\ -3 & 3 \end{bmatrix} - Y \\[1em] \Rightarrow X = \frac{1}{2} \Big(\begin{bmatrix} 5 & 0 \\ -3 & 3 \end{bmatrix} - Y\Big)$$

Substituting value of Y, we get :

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

Hence, X = $$\begin{bmatrix} 2 & -1 \\ -1 & -1 \end{bmatrix}.$$

Given,

$$A + B = \begin{bmatrix} 5 & 4 \\ 7 & 3 \end{bmatrix}....(1)$$

$$A - B = \begin{bmatrix} 11 & 2 \\ -1 & 7 \end{bmatrix}....(2)$$

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

$$\Rightarrow (A + B) + (A - B) = \begin{bmatrix} 5 & 4 \\ 7 & 3 \end{bmatrix} + \begin{bmatrix} 11 & 2 \\ -1 & 7 \end{bmatrix} \\[1em] \Rightarrow A + B + A - B = \begin{bmatrix} 5 + 11 & 4 + 2 \\ 7 + (-1) & 3 + 7 \end{bmatrix} \\[1em] \Rightarrow 2A = \begin{bmatrix} 16 & 6 \\ 6 & 10 \end{bmatrix} \\[1em] \Rightarrow A = \frac{1}{2} \begin{bmatrix} 16 & 6 \\ 6 & 10 \end{bmatrix} \\[1em] \Rightarrow A = \begin{bmatrix} 8 & 3 \\ 3 & 5 \end{bmatrix}.$$

Substituting value of A in equation (1), we get :

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

Hence, A = $$\begin{bmatrix} 8 & 3 \\ 3 & 5 \end{bmatrix}$$ and B = $$\begin{bmatrix} -3 & 1 \\ 4 & -2 \end{bmatrix}$$.

Solving,

$$\Rightarrow 6 \begin{bmatrix} 4 & -5 \\ 3 & 2 \end{bmatrix} - 3\begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 24 & -30 \\ 18 & 12 \end{bmatrix} - \begin{bmatrix} 6 & -9 \\ -3 & 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 24 - 6 & -30 - (-9) \\ 18 - (-3) & 12 - 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 18 & -21 \\ 21 & 0 \end{bmatrix}.$$

Hence, resultant matrix = $$\begin{bmatrix} 18 & -21 \\ 21 & 0 \end{bmatrix}.$$

Solving,

$$\Rightarrow \sin A \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \sin^2 A & -\cos A \sin A \\ \sin A \cos A & \sin^2 A \end{bmatrix} + \begin{bmatrix} \cos^2 A & \sin A \cos A \\ -\sin A \cos A & \cos^2 A \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \sin^2 A + \cos^2 A & -\cos A \sin A + \sin A \cos A \\ \sin A \cos A - \sin A \cos A & \sin^2 A + \cos^2 A \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$

Hence, resultant matrix = $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$.

Solving L.H.S.,

$$\Rightarrow \cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \cos^2 \theta & \sin \theta \cos \theta \\ -\sin \theta \cos \theta & \cos^2 \theta \end{bmatrix} + \begin{bmatrix} \sin^2 \theta & -\cos \theta \sin \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & \sin \theta \cos \theta - \cos \theta \sin \theta \\ -\sin \theta \cos \theta + \sin \theta \cos \theta & \cos^2 \theta + \sin^2 \theta \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$

Hence, resultant matrix = $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$

Solving,

$$\Rightarrow 3 \begin{bmatrix} 2 & x \\ 1 & 0 \end{bmatrix} + 2 \begin{bmatrix} 4 & 3 \\ y & 2 \end{bmatrix} =\begin{bmatrix} z & -3 \\ 15 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 & 3x \\ 3 & 0 \end{bmatrix} + \begin{bmatrix} 8 & 6 \\ 2y & 4 \end{bmatrix} =\begin{bmatrix} z & -3 \\ 15 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 + 8 & 3x + 6 \\ 3 + 2y & 0 + 4 \end{bmatrix} =\begin{bmatrix} z & -3 \\ 15 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 14 & 3x + 6 \\ 3 + 2y & 4 \end{bmatrix} =\begin{bmatrix} z & -3 \\ 15 & 4 \end{bmatrix}.$$

∴ z = 14

∴ 3x + 6 = -3

⇒ 3x = -3 - 6

⇒ 3x = -9

⇒ x = $\frac{-9}{3}$

⇒ x = -3

∴ 3 + 2y = 15

⇒ 2y = 15 - 3

⇒ 2y = 12

⇒ y = $\frac{12}{2}$

⇒ y = 6.

Hence, x = -3, y = 6 and z = 14.

Given,

$$3\begin{bmatrix} 5 & 6 \\ 2 & x \end{bmatrix} - \begin{bmatrix} 8 & y \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 6 & 7 \end{bmatrix}$$

Solving,

$$\Rightarrow 3\begin{bmatrix} 5 & 6 \\ 2 & x \end{bmatrix} - \begin{bmatrix} 8 & y \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 6 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 15 & 18 \\ 6 & 3x \end{bmatrix} - \begin{bmatrix} 8 & y \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 6 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 15 - 8 & 18 - y \\ 6 - 0 & 3x - 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 6 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & 18 - y \\ 6 & 3x - 8 \end{bmatrix} = \begin{bmatrix} 7 & 8 \\ 6 & 7 \end{bmatrix}.$$

∴ 18 - y = 8

⇒ y = 18 - 8

⇒ y = 10.

∴ 3x - 8 = 7

⇒ 3x = 7 + 8

⇒ 3x = 15

⇒ x = $\frac{15}{3}$

⇒ x = 5.

Hence, x = 5, y = 10.

Given,

$$2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}$$

Solving,

$$\Rightarrow \begin{bmatrix} 6 & 8 \\ 10 & 2x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 + 1 & 8 + y \\ 10 + 0 & 2x + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & 8 + y \\ 10 & 2x + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}.$$

∴ 8 + y = 0

⇒ y = -8

∴ 2x + 1 = 5

⇒ 2x = 5 - 1

⇒ 2x = 4

⇒ x = $\frac{4}{2}$

⇒ x = 2.

Hence, x = 2, y = -8.

Given,

$$2 \begin{bmatrix} x & 7 \\ 9 & y - 5 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}.$$

Solving,

$$\Rightarrow \begin{bmatrix} 2x & 14 \\ 18 & 2y - 10 \end{bmatrix} + \begin{bmatrix} 6 & -7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x + 6 & 14 - 7 \\ 18 + 4 & 2y - 10 + 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x + 6 & 7 \\ 22 & 2y - 5 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}.$$

∴ 2x + 6 = 10

⇒ 2x = 10 - 6

⇒ 2x = 4

⇒ x = $\frac{4}{2}$

⇒ x = 2.

∴ 2y - 5 = 15

⇒ 2y = 15 + 5

⇒ 2y = 20

⇒ y = $\frac{20}{2}$

⇒ y = 10.

Hence, x = 2, y = 10.

(i) Given,

$$A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix}$$ and $$B = \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix}$$

Solving,

$$AB = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(2) + (-2)(3) + (1)(1) & (1)(1) + (-2)(2) + (1)(1) \\ (2)(2) + (1)(3) + (3)(1) & (2)(1) + (1)(2) + (3)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 2 - 6 + 1 & 1 - 4 + 1 \\ 4 + 3 + 3 & 2 + 2 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}.$$

Hence, AB = $$\begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix}$$.

(ii) Yes, it is possible.

The number of columns in B equals the number of rows in A . The resulting matrix BA will be a 3×3 matrix.

$$BA = \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix} (2)(1) + (1)(2) & (2)(-2) + (1)(1) & (2)(1) + (1)(3) \\ (3)(1) + (2)(2) & (3)(-2) + (2)(1) & (3)(1) + (2)(3) \\ (1)(1) + (1)(2) & (1)(-2) + (1)(1) & (1)(1) + (1)(3) \end{bmatrix} \\[1em] = \begin{bmatrix} 2 + 2 & -4 + 1 & 2 + 3 \\ 3 + 4 & -6 + 2 & 3 + 6 \\ 1 + 2 & -2 + 1 & 1 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & -3 & 5 \\ 7 & -4 & 9 \\ 3 & -1 & 4 \end{bmatrix}.$$

Hence, BA = $$\begin{bmatrix} 4 & -3 & 5 \\ 7 & -4 & 9 \\ 3 & -1 & 4 \end{bmatrix}$$.

(i) Given,

A = $$\begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix}$$ (order 2 × 2)

B = $$\begin{bmatrix} 2 \\ -3 \end{bmatrix}$$ (order 2 × 1)

For matrix multiplication :

The number of columns in the first matrix (A) must equal the number of rows in the second matrix(B).

Since the number of columns in A (2) equals the number of rows in B (2), the product AB exists.

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

Hence, order of matrix AB 2 × 1.

(ii) $$AB = \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} \times \begin{bmatrix} 2 \\ -3 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(2) + (3)(-3) \\ (2)(2) + (-1)(-3) \end{bmatrix} \\[1em] = \begin{bmatrix} 2 - 9 \\ 4 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix} -7 \\ 7 \end{bmatrix}.$$

Hence, AB = $$\begin{bmatrix} -7 \\ 7 \end{bmatrix}.$$

(i) Given,

For matrix multiplication :

Number of columns in the first matrix must be equal to number of rows in the second matrix.

We know that,

The number of rows in resultant matrix equals to number of rows in first matrix and number of columns in resultant matrix equals to number of columns in second matrix.

$M_{1 \times 2} \times N_{2 \times 2} = MN_{1 \times 2}$

Hence, MN exists and order of MN = 1 × 2.

(ii) $$MN = \begin{bmatrix} 1 & -2 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(2) + (-2)(-1) & (1)(1) + (-2)(2) \end{bmatrix} \\[1em] = \begin{bmatrix} 2 + 2 & 1 - 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & -3 \end{bmatrix}.$$

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

(iii)

For matrix multiplication :

Number of columns in the second matrix must be equal to number of rows in the first matrix.

$N_{2 \times 2} \times M_{1 \times 2}$

Not possible because Number of columns in the second matrix ≠Number of rows in the first matrix.

Hence, NM does not exists.

Given,

A = $$\begin{bmatrix} 2 & 3 \end{bmatrix}$$ B = $$\begin{bmatrix} -2 \\ 1 \end{bmatrix}$$

For matrix multiplication :

Number of columns in the first matrix (A) = Number of rows in the second matrix(B).

Since the number of columns in A (2) equals the number of rows in B (2), the product AB exists.

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

$A_{1 \times 2} \times B_{2 \times 1} = AB_{1 \times 1}$

∴ Order of matrix AB will be 1 × 1.

Since the number of columns in B (1) equals the number of rows in A (1), the product AB exists.

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

∴ m = 2 and n = 2.

Hence, order of matrix AB 1 × 1 and BA 2 × 2.

(i) $$AB = \begin{bmatrix} 2 & 3 \end{bmatrix} \times \begin{bmatrix} -2 \\ 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (2)(-2) + (3)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} -4 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 \end{bmatrix}.$$

Hence, AB = $$\begin{bmatrix} -1 \end{bmatrix}$$.

(ii) For matrix multiplication :

Number of columns in the first matrix (B) = Number of rows in the second matrix(A).

Since the number of columns in B (1) equals the number of rows in A (1), the product BA exists.

$AB_{2 \times 1} \times B_{1 \times 2} = BA_{2 \times 2}$

$$BA = \begin{bmatrix} -2 \\ 1 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix} (-2)(2) & (-2)(3) \\ (1)(2) & (1)(3) \end{bmatrix} \\[1em] = \begin{bmatrix} -4 & -6 \\ 2 & 3 \end{bmatrix} \\[1em]$$

Hence, BA = $$\begin{bmatrix} -4 & -6 \\ 2 & 3 \end{bmatrix}$$.

(i) Solving AB,

$$\Rightarrow \begin{bmatrix} 1 & 3 \\ 2 & -1 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 \times 2 + 3 \times 0 & 1 \times 1 + 3 \times 3 \\ 2 \times 2 + (-1) \times 0 & 2 \times 1 + (-1) \times 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 + 0 & 1 + 9 \\ 4 + 0 & 2 + (-3) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 & 10 \\ 4 & -1 \end{bmatrix}.$$

Hence, AB = $$\begin{bmatrix} 2 & 10 \\ 4 & -1 \end{bmatrix}$$.

(ii) Solving BA,

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

Hence, BA = $$\begin{bmatrix} 4 & 5 \\ 6 & -3 \end{bmatrix}$$.

(iii) As calculated above,

AB ≠ BA

Hence, AB ≠ BA.

Solving AB,

$$\Rightarrow \begin{bmatrix} 5 & -1 \\ 6 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 \times 2 + (-1) \times (-3) & 5 \times 1 + (-1) \times 4 \\ 6 \times 2 + 2 \times (-3) & 6 \times 1 + 2 \times 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 10 + 3 & 5 + (-4) \\ 12 + (-6) & 6 + 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 13 & 1 \\ 6 & 14 \end{bmatrix}.$$

AB = $$\begin{bmatrix} 13 & 1 \\ 6 & 14 \end{bmatrix}$$

Solving BA,

$$\Rightarrow \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} \times \begin{bmatrix} 5 & -1 \\ 6 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 5 + 1 \times 6 & 2 \times (-1) + 1 \times 2 \\ -3 \times 5 + 4 \times 6 & -3 \times (-1) + 4 \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 10 + 6 & -2 + 2 \\ -15 + 24 & 3 + 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 16 & 0 \\ 9 & 11 \end{bmatrix}.$$

BA = $$\begin{bmatrix} 16 & 0 \\ 9 & 11 \end{bmatrix}$$

AB ≠ BA.

Hence, proved that AB ≠ BA.

Solving AB,

$$\Rightarrow \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \times \begin{bmatrix} 3 & -2 \\ -4 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -1 \times 3 + 2 \times (-4) & -1 \times (-2) + 2 \times 5 \\ 3 \times 3 + 4 \times (-4) & 3 \times (-2) + 4 \times 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3 + (-8) & 2 + 10 \\ 9 + (-16) & (-6) + 20 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -11 & 12 \\ -7 & 14 \end{bmatrix}.$$

AB = $$\begin{bmatrix} -11 & 12 \\ -7 & 14 \end{bmatrix}$$

Solving BA,

$$\Rightarrow \begin{bmatrix} 3 & -2 \\ -4 & 5 \end{bmatrix} \times \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 \times (-1) + (-2) \times 3 & 3 \times 2 + (-2) \times 4 \\ -4 \times (-1) + 5 \times 3 & -4 \times 2 + 5 \times 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3 + (-6) & 6 + (-8) \\ 4 + 15 & -8 + 20 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -9 & -2 \\ 19 & 12 \end{bmatrix}.$$

BA = $$\begin{bmatrix} -9 & -2 \\ 19 & 12 \end{bmatrix}$$

Now, AB + BA

$$\Rightarrow \begin{bmatrix} -11 & 12 \\ -7 & 14 \end{bmatrix} + \begin{bmatrix} -9 & -2 \\ 19 & 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -11 + (-9) & 12 + (-2) \\ -7 + 19 & 14 + 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -20 & 10 \\ 12 & 26 \end{bmatrix}.$$

Hence, (AB + BA) = $$\begin{bmatrix} -20 & 10 \\ 12 & 26 \end{bmatrix}$$.

Let's find AB,

$$\Rightarrow \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 \times 2 + 2 \times 3 & 1 \times 1 + 2 \times 2 \\ 2 \times 2 + 4 \times 3 & 2 \times 1 + 4 \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 + 6 & 1 + 4 \\ 4 + 12 & 2 + 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 & 5 \\ 16 & 10 \end{bmatrix}.$$

Let's find AC,

$$\Rightarrow \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} -2 & 7 \\ 5 & -1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 \times (-2) + 2 \times 5 & 1 \times 7 + 2 \times (-1) \\ 2 \times (-2) + 4 \times 5 & 2 \times 7 + 4 \times (-1) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -2 + 10 & 7 + (-2) \\ -4 + 20 & 14 + (-4) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 & 5 \\ 16 & 10 \end{bmatrix}.$$

Hence, proved that AB = AC.

Solving (AB)C,

$$\Rightarrow \Big(\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\ -3 & 0 \end{bmatrix}\Big) \times \begin{bmatrix} -3 & -1 \\ -2 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 \times 2 + 1 \times (-3) & 4 \times 1 + 1 \times 0 \\ 2 \times 2 + 3 \times (-3) & 2 \times 1 + 3 \times 0 \end{bmatrix} \times \begin{bmatrix} -3 & -1 \\ -2 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 + (-3) & 4 + 0 \\ 4 + (-9) & 2 + 0 \end{bmatrix} \times \begin{bmatrix} -3 & -1 \\ -2 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 & 4 \\ -5 & 2 \end{bmatrix} \times \begin{bmatrix} -3 & -1 \\ -2 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 \times (-3) + 4 \times (-2) & 5 \times (-1) + 4 \times 2 \\ -5 \times (-3) + 2 \times (-2) & -5 \times (-1) + 2 \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -15 - 8 & -5 + 8 \\ 15 - 4 & 5 + 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -23 & 3 \\ 11 & 9 \end{bmatrix}. \\[1em] \therefore (AB)C = \begin{bmatrix} -23 & 3 \\ 11 & 9 \end{bmatrix}$$

Solving A(BC),

$$\Rightarrow \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \times \Big(\begin{bmatrix} 2 & 1 \\ -3 & 0 \end{bmatrix} \times \begin{bmatrix} -3 & -1 \\ -2 & 2 \end{bmatrix}\Big) \[1em] \Rightarrow \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} 2 \times (-3) + 1 \times (-2) & 2 \times (-1) + 1 \times 2 \\ -3 \times (-3) + 0 \times (-2) & -3 \times (-1) + 0 \times 2 \end{bmatrix} \[1em] \Rightarrow \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} -6 - 2 & -2 + 2 \\ 9 + 0 & 3 + 0 \end{bmatrix} \[1em] \Rightarrow \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} -8 & 0 \\ 9 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 \times (-8) + 1 \times 9 & 4 \times 0 + 1 \times 3 \\ 2 \times (-8) + 3 \times 9 & 2 \times 0 + 3 \times 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -32 + 9 & 0 + 3 \\ -16 + 27 & 0 + 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -23 & 3 \\ 11 & 9 \end{bmatrix}. \\[1em] \therefore A(BC) = \begin{bmatrix} -23 & 3 \\ 11 & 9 \end{bmatrix}$$

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

(i) A(B + C)

$$\Rightarrow \begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix} \times \Big(\begin{bmatrix} 5 & 0 \\ 2 & -3 \end{bmatrix} + \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}\Big) \\[1em] \Rightarrow \begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 5 + 2 & 0 - 1 \\ 2 + 4 & -3 + 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 7 & -1 \\ 6 & -3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 \times 7 + 4 \times 6 & 3 \times (-1) + 4 \times (-3) \\ -1 \times 7 + 2 \times 6 & -1 \times (-1) + 2 \times (-3) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 21 + 24 & -3 + (-12) \\ -7 + 12 & 1 + (-6) \end{bmatrix} \\[1em] = \begin{bmatrix} 45 & -15 \\ 5 & -5 \end{bmatrix}.$$

Hence, A(B + C) = $$\begin{bmatrix} 45 & -15 \\ 5 & -5 \end{bmatrix}$$

(ii) (AB + AC)

$$\Rightarrow AB = \begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 5 & 0 \\ 2 & -3 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 \times 5 + 4 \times 2 & 3 \times 0 + 4 \times (-3) \\ -1 \times 5 + 2 \times 2 & -1 \times 0 + 2\times(-3) \end{bmatrix} \\[1em] = \begin{bmatrix} 15 + 8 & 0 + (-12) \\ -5 + 4 & 0 + (-6) \end{bmatrix}\\[1em] = \begin{bmatrix} 23 & -12 \\ -1 & -6 \end{bmatrix} \\[1em] \Rightarrow AC = \begin{bmatrix} 3 & 4 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 \times 2 + 4 \times 4 & 3\times(-1) + 4 \times 0 \\ -1 \times 2 + 2 \times 4 & -1\times(-1) + 2 \times 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 6 + 16 & -3 + 0 \\ -2 + 8 & 1 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 22 & -3 \\ 6 & 1 \end{bmatrix} \\[1em] \Rightarrow AB + AC = \begin{bmatrix} 23 & -12 \\ -1 & -6 \end{bmatrix} + \begin{bmatrix} 22 & -3 \\ 6 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 23 + 22 & -12 - 3 \\ -1 + 6 & -6 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 45 & -15 \\ 5 & -5 \end{bmatrix}.$$

Hence, (AB + AC) = $$\begin{bmatrix} 45 & -15 \\ 5 & -5 \end{bmatrix}$$

(iii) As calculated,

$$A(B + C) = (AB + AC)= \begin{bmatrix} 45 & -15 \\ 5 & -5 \end{bmatrix}.$$

Hence, A(B + C) = (AB + AC).

(i) (A - B)C

$$\Rightarrow \Big(\begin{bmatrix} 5 & 7 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -2 \\ -4 & 0 \end{bmatrix}\Big) \times \begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 - 3 & 7 - (-2) \\ 3 - (-4) & 2 - 0 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 & 9 \\ 7 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (2)(2) + (9)(1) & (2)(-3) + (9)(4) \\ (7)(2) + (2)(1) & (7)(-3) + (2)(4) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 + 9 & -6 + 36 \\ 14 + 2 & -21 + 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 13 & 30 \\ 16 & -13 \end{bmatrix}.$$

Hence, (A - B)C =$$\begin{bmatrix} 13 & 30 \\ 16 & -13 \end{bmatrix}.$$

(ii) AC - BC

$$\Rightarrow AC = \begin{bmatrix} 5 & 7 \\ 3 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} (5)(2) + (7)(1) & (5)(-3) + (7)(4) \\ (3)(2) + (2)(1) & (3)(-3) + (2)(4) \end{bmatrix} \\[1em] = \begin{bmatrix} 10 + 7 & -15 + 28 \\ 6 + 2 & -9 + 8 \end{bmatrix} \\[1em] = \begin{bmatrix} 17 & 13 \\ 8 & -1 \end{bmatrix} \\[1em] \Rightarrow BC = \begin{bmatrix} 3 & -2 \\ -4 & 0 \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ 1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} (3)(2) + (-2)(1) & (3)(-3) + (-2)(4) \\ (-4)(2) + (0)(1) & (-4)(-3) + (0)(4) \end{bmatrix} \\[1em] = \begin{bmatrix} 6 - 2 & -9 - 8 \\ -8 + 0 & 12 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & -17 \\ -8 & 12 \end{bmatrix} \\[1em] \Rightarrow AC - BC = \begin{bmatrix} 17 & 13 \\ 8 & -1 \end{bmatrix} - \begin{bmatrix} 4 & -17 \\ -8 & 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 17 - 4 & 13 - (-17) \\ 8 - (-8) & -1 - 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 13 & 30 \\ 16 & -13 \end{bmatrix}.$$

Hence, AC - BC =$$\begin{bmatrix} 13 & 30 \\ 16 & -13 \end{bmatrix}.$$

(ii) Yes, (A - B)C = AC - BC = $$\begin{bmatrix} 13 & 30 \\ 16 & -13 \end{bmatrix}.$$

Hence, (A - B)C = AC - BC .

Solving for AB,

$$\Rightarrow AB = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix} \times \begin{bmatrix} 6 \\ 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (3)(6) + (-2)(1) \\ (-1)(6) + (4)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 18 - 2 \\ -6 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 16 \\ -2 \end{bmatrix}.$$

Substituting values in AB + 2C – 4D,we get :

$$\Rightarrow \begin{bmatrix} 16 \\ -2 \end{bmatrix} + 2 \times \begin{bmatrix} -4 \\ 5 \end{bmatrix} - 4 \times \begin{bmatrix} 2 \\ 2 \end{bmatrix} \\[1em] = \begin{bmatrix} 16 \\ -2 \end{bmatrix} + \begin{bmatrix} -8 \\ 10 \end{bmatrix} - \begin{bmatrix} 8 \\ 8 \end{bmatrix} \\[1em] = \begin{bmatrix} 16 + (-8) \\ -2 + 10 \end{bmatrix} - \begin{bmatrix} 8 \\ 8 \end{bmatrix} \\[1em] = \begin{bmatrix} 8 \\ 8 \end{bmatrix} - \begin{bmatrix} 8 \\ 8 \end{bmatrix} \\[1em] = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$

Hence, AB + 2C – 4D = $$\begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$

Solving A²,

$$\Rightarrow A^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(1) + (0)(2) & (1)(0) + (0)(1) \\ (2)(1) + (1)(2) & (2)(0) + (1)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 0 & 0 + 0 \\ 2 + 2 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}.$$

Solving AB,

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

Solving B²,

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

A² + AB + B²

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

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

(i) A² – A + BC

Solving A²,

$$\Rightarrow A^2 = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \times \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4\times4 + (-2)\times6 & 4\times(-2) + (-2)\times(-3) \\ 6\times4 + (-3)\times6 & 6\times(-2) + (-3)\times(-3) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 16 - 12 & -8 + 6 \\ 24 - 18 & -12 + 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}.$$

Solving BC,

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

Solving A² – A + BC

$$\Rightarrow \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} - \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} + \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix}.$$

Hence, (A² – A + BC) = $$\begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix}$$

Solving AC,

$$\Rightarrow AC = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 1 + 3 \times (-1) & 2 \times 0 + 3 \times 4 \\ 5 \times 1 + 7 \times (-1) & 5 \times 0 + 7 \times 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 - 3 & 0 + 12 \\ 5 - 7 & 0 + 28 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix}.$$

AC = $$\begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix}$$

Solving B²,

$$\Rightarrow B^2 = \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \times \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 \times 0 + 4 \times (-1) & 0 \times 4 + 4 \times 7 \\ -1 \times 0 + 7 \times (-1) & -1 \times 4 + 7 \times 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -4 & 28 \\ -7 & -4 + 49 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -4 & 28 \\ -7 & 45 \end{bmatrix}.$$

Solving 10C,

$$\Rightarrow 10C = 10 \times \begin{bmatrix} 1 & 0 \\ -1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix}.$$

Now, AC + B² – 10C

$$\Rightarrow \begin{bmatrix} -1 & 12 \\ -2 & 28 \end{bmatrix} + \begin{bmatrix} -4 & 28 \\ -7 & 45 \end{bmatrix} - \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -5 & 40 \\ -9 & 73 \end{bmatrix} - \begin{bmatrix} 10 & 0 \\ -10 & 40 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -5 - 10 & 40 - 0 \\ -9 - (-10) & 73 - 40 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -15 & 40 \\ 1 & 33 \end{bmatrix}.$$

Hence, (AC + B² – 10C) = $$\begin{bmatrix} -15 & 40 \\ 1 & 33 \end{bmatrix}$$

Given,

A = $$\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$$

$$A^T = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$$

Solving A^TB,

$$\Rightarrow \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} \times \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 4 + 1 \times (-1) & 2 \times (-2) + 1 \times 3 \\ 5 \times 4 + 3 \times (-1) & 5 \times (-2) + 3 \times 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix}.$$

Solving BI,

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

Now, A^TB + BI

$$\Rightarrow \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 + 4 & -1 + (-2) \\ 17 + (-1) & -1 + 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix}.$$

Hence, (A^TB + BI) = $$\begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix}$$

Solving A²,

$$\Rightarrow A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 \times 3 + 1 \times (-1) & 3 \times 1 + 1 \times 2 \\ -1 \times 3 + 2 \times (-1) & -1 \times 1 + 2 \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}.$$

Solving 5A,

$$\Rightarrow 5A = 5 \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}.$$

Solving 7I,

$$\Rightarrow 7I = 7 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}.$$

Now, A² – 5A + 7I

$$\Rightarrow \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 - (15) + 7 & 5 - (5) + 0 \\ -5 - 5 + 0 & 3 - (10) + 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$

Hence, A² – 5A + 7I = $$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$.

Solving A²,

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

Solving 6A,

$$\Rightarrow 6A = 6 \times \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 24 & 6 \\ -6 & 12 \end{bmatrix}.$$

Now, 6A – A²

$$\Rightarrow \begin{bmatrix} 24 & 6 \\ -6 & 12 \end{bmatrix} - \begin{bmatrix} 15 & 6 \\ -6 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 24 - 15 & 6 - 6 \\ -6 - (-6) & 12 - 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \\[1em] \Rightarrow 9\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 9I.$$

Hence, proved that 6A – A² = 9I.

Solving A(B + C),

$$\Rightarrow \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \times \Big(\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} + \begin{bmatrix} 4 & 1 \\ 1 & 5 \end{bmatrix}\Big) \\[1em] \Rightarrow \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} 1+4 & 2+1 \\ 2+1 & 4+5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} 5 & 3 \\ 3 & 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 \times 5 + 3 \times 3 & 1 \times 3 + 3 \times 9 \\ 2 \times 5 + 4 \times 3 & 2 \times 3 + 4 \times 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 + 9 & 3 + 27 \\ 10 + 12 & 6 + 36 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 14 & 30 \\ 22 & 42 \end{bmatrix}.$$

Solving 14I,

$$\Rightarrow 14 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}.$$

Now, A(B + C) – 14I

$$\Rightarrow \begin{bmatrix} 14 & 30 \\ 22 & 42 \end{bmatrix} - \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 14 - 14 & 30 - 0 \\ 22 - 0 & 42 - 14 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 30 \\ 22 & 28 \end{bmatrix}.$$

Hence, A(B + C) – 14I = $$\begin{bmatrix} 0 & 30 \\ 22 & 28 \end{bmatrix}$$.

Given,

X² = 4X + 5I

Solving for L.H.S.,

$$X^2 = \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 \times 1 + 1 \times 8 & 1 \times 1 + 1\times 3 \\ 8 \times 1 + 3 \times 8 & 8 \times 1 + 3 \times 3 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 8 & 1 + 3 \\ 8 + 24 & 8 + 9 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix}.$$

Solving for R.H.S.,

$$4X + 5I = 4\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} + 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 + 5 & 4 + 0 \\ 32 + 0 & 12 + 5 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix}.$$

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

Hence, proved that X² = 4X + 5I.

Solving A²,

$$\Rightarrow A^2 = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \times \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} ab \times ab + b^2 \times (-a^2) & ab \times b^2 + b^2 \times (-ab) \\ (-a^2) \times ab + (-ab) \times (-a^2) & (-a^2) \times b^2 + (-ab) \times (-ab) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} a^2 b^2 -a^2 b^2 & a b^3 -a b^3 \\ -a^3 b +a^3 b & -a^2 b^2 +a^2 b^2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$

Hence, proved that A² = 0.

Solving A²,

$$\Rightarrow A^2 = \begin{bmatrix} 2 & -3 \\ a & b \end{bmatrix} \times \begin{bmatrix} 2 & -3 \\ a & b \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 2 + (-3) \times a & 2 \times (-3) + (-3) \times b \\ a \times 2 + b \times a & a \times (-3) + b \times b \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 - 3a & -6 - 3b \\ 2a + ab & -3a + b^2 \end{bmatrix}.$$

Given,

A² = I,

$$\begin{bmatrix} 4 - 3a & -6 - 3b \\ 2a + ab & -3a + b^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$

∴ 4 - 3a = 1,

⇒ -3a = -3

⇒ a = 1.

∴ -6 - 3b = 0,

⇒ -3b = 6

⇒ b = -2.

Hence, a = 1 and b = -2.

Let's find A²,

$$\Rightarrow A^2 = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 2 + 12 \times 0 & 2 \times 12 + 12 \times 1 \\ 0 \times 2 + 1 \times 0 & 0 \times 12 + 1 \times 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 24 + 12 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix}.$$

Since, A² = B,

$$\Rightarrow \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix}$$

∴ x = 36

Hence, x = 36.

Let B = $$\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$$ Then, AB = $$\begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}$$

Since AB exists, we have:

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

Number of rows of A = Number of rows in AB = 2

Order of A is 2 × 2

Let A = $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$.

Then,

$$\Rightarrow \begin{bmatrix} a & b \\ c & d \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2a + 4b & 3a + 5b \\ 2c + 4d & 3c + 5d \end{bmatrix} =\begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}.$$

Solving for a and b:

∴ 2a + 4b = 0

⇒ 2(a + 2b) = 0

⇒ a + 2b = 0

⇒ a = -2b ......(1)

∴ 3a + 5b = -4

Substituting value of a from equation(1) in 3a + 5b = -4, we get:

⇒ 3(-2b) + 5b = -4

⇒ -6b + 5b = -4

⇒ -b = -4

⇒ b = 4.

Substituting value of b in equation 1, we get,

⇒ a = -2(4)

⇒ a = -8.

Solving for c and d:

∴ 2c + 4d = 10

⇒ 2(c + 2d) = 10

⇒ c + 2d = 5

⇒ c = 5 - 2d .......(2)

∴ 3c + 5d = 3

Substituting value of c from equation (2) in 3c + 5d = 3, we get :

⇒ 3c + 5d = 3

⇒ 3(5 - 2d) + 5d = 3

⇒ 15 - 6d + 5d = 3

⇒ 15 - d = 3

⇒ d = 15 - 3

⇒ d = 12.

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

⇒ c = 5 - 2(12)

⇒ c = 5 - 24

⇒ c = -19.

$$\therefore A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} -8 & 4 \\ -19 & 12 \end{bmatrix}$$.

Hence, A = $$\begin{bmatrix} -8 & 4 \\ -19 & 12 \end{bmatrix}$$.

Let Y = $$\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$$.

Then, YX = $$\begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix}$$

Since YX exists, we have:

Number of rows of X = Number of columns in Y = 2

Number of columns of X = Number of columns in YX = 2

Order of X is 2 × 2.

Let X = $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$.

Then,

$$\Rightarrow \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix}= \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5a - 7c & 5b - 7d \\ -2a + 3c & -2b + 3d \end{bmatrix} = \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix} \\[1em]$$

Solving for a and c :

∴ -2a + 3c = 7

⇒ 2a = 3c - 7

⇒ a = $\frac{3c - 7}{2}$ ...(1)

∴ 5a - 7c = -16 .......(2)

Substituting value of a from equation (1) in (2), we get :

$$\begin{aligned}\Rightarrow 5\Big(\frac{3c - 7}{2}\Big) - 7c = -16 \\[1em] \Rightarrow \Big(\frac{15c - 35}{2}\Big) - 7c = -16 \\[1em] \Rightarrow \frac{15c - 35}{2} - \frac{7c \times 2}{2} = -16 \\[1em] \Rightarrow \frac{15c - 35 - 14c}{2} = -16 \\[1em] \Rightarrow \frac{c - 35}{2} = -16 \\[1em] \Rightarrow c - 35 = -16 \times 2 \\[1em] \Rightarrow c - 35 = -32 \\[1em] \Rightarrow c = -32 + 35 \\[1em] \Rightarrow c = 3.\end{aligned}$$

Substituting value of c in equation (1), we get:

⇒ a = $\frac{3(3) - 7}{2}$

⇒ a = $\frac{9 - 7}{2}$

⇒ a = $\frac{2}{2}$

⇒ a = 1.

Solving for b and d:

∴ -2b + 3d = 2

⇒ 2b = 3d - 2

⇒ b = $\frac{3d - 2}{2}$ ....(3)

∴ 5b - 7d = -6 ......(4)

Substituting value of b from equation (3) in (4), we get:

$$\begin{aligned}\Rightarrow 5\Big(\frac{3d - 2}{2}\Big) - 7d = -6 \\[1em] \Rightarrow \Big(\frac{15d - 10}{2}\Big) - 7d = -6 \\[1em] \Rightarrow \frac{15d - 10}{2} - \frac{7d \times 2}{2} = -6 \\[1em] \Rightarrow \frac{15d - 10 - 14d}{2} = -6 \\[1em] \Rightarrow \frac{d - 10}{2} = -6 \\[1em] \Rightarrow d - 10 = -6 \times 2 \\[1em] \Rightarrow d - 10 = -12 \\[1em] \Rightarrow d = -12 + 10 \\[1em] \Rightarrow d = -2.\end{aligned}$$

Substituting value of d in equation (3), we get:

⇒ b = $\frac{3(-2) - 2}{2}$

⇒ b = $\frac{-6 - 2}{2}$

⇒ b = $\frac{-8}{2}$

⇒ b = -4.

$$\therefore X = \begin{bmatrix} 1 & -4 \\ 3 & -2 \end{bmatrix}$$.

Hence, X = $$\begin{bmatrix} 1 & -4 \\ 3 & -2 \end{bmatrix}$$.

(i) Let B = $$\begin{bmatrix} 3 & 2 \\ -1 & 5 \end{bmatrix}$$ Then, AB = $$\begin{bmatrix} 9 & -11 \end{bmatrix}$$

Order of B = 2 × 2

Order of AB = 1 × 2

Since AB exists, we have:

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

Number of rows of A = Number of rows in AB = 1

Order of A is 1 × 2.

$A_{1 \times 2} \times B_{2 \times 2} = AB_{1 \times 2}$

Hence, order of A is 1 × 2.

(ii) Let A = $$\begin{bmatrix} a & b \end{bmatrix}$$.

Then,

$$\Rightarrow \begin{bmatrix} a & b \end{bmatrix} \times \begin{bmatrix} 3 & 2 \\ -1 & 5 \end{bmatrix} = \begin{bmatrix} 9 & -11 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3a - b & 2a + 5b \end{bmatrix} = \begin{bmatrix} 9 & -11 \end{bmatrix}.$$

Solving for a and b:

∴ 3a - b = 9

⇒ b = 3a - 9 ....(1)

∴ 2a + 5b = -11 ...(2)

Substituting value of b from equation (1) in (2), we get :

⇒ 2a + 5(3a - 9) = -11

⇒ 2a + 15a - 45 = -11

⇒ 17a = -11 + 45

⇒ 17a = 34

⇒ a = $\frac{34}{17}$

⇒ a = 2.

Substituting value of a in equation (1), we get :

⇒ b = 3(2) - 9

⇒ b = 6 - 9

⇒ b = -3.

$$\therefore A = \begin{bmatrix} 2 & -3 \end{bmatrix}$$.

Hence, A = $$\begin{bmatrix} 2 & -3 \end{bmatrix}$$.

(i) Let N = $$\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$$

Given, NM = 6I

NM = $$\begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$$

Order of N = 2 × 2

Order of NM = 2 × 2

Since NM exists, we have:

Number of rows of M = Number of columns in N = 2

Number of columns of M = Number of columns in NM = 2

Order of M is 2 × 2.

$M_{2 \times 2} \times N_{2 \times 2} = MN_{2 \times 2}$

Hence, order of M is 2 × 2.

(ii) Let M = $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$.

Then,

$$\Rightarrow \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix} = 6\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4a + 2c & 4b + 2d \\ -a + c & -b + d \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}.$$

Solving for a and c :

∴ -a + c = 0

⇒ c = a....(1)

∴ 4a + 2c = 6

⇒ 2(2a + c = 3)

⇒ 2a + c = 3...(2)

Substituting value of c from equation (1) in 2a + c = 3, we get:

⇒ 2a + a = 3

⇒ 3a = 3

⇒ a = $\frac{3}{3}$

⇒ a = 1

Hence, a = c = 1

Solving for b and d:

∴ -b + d = 6

⇒ d = b + 6...(3)

∴ 4b + 2d = 0

⇒ 2(2b + d) = 0

⇒ 2b + d = 0...(4)

Substituting value of d from equation (3) in 2b + d = 0, we get:

⇒ 2b + b + 6 = 0

⇒ 3b + 6 = 0

⇒ 3b = -6

⇒ b = $\frac{-6}{3}$

⇒ b = -2

Substituting value of b in equation(3), we get:

⇒ d = -2 + 6

⇒ d = 4

Hence, M = $$\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}.$$

(i) Let B = $$\begin{bmatrix} 5 & -2 \\ 1 & 3 \end{bmatrix}$$ Then, BA = $$\begin{bmatrix} 4 \\ 11 \end{bmatrix}$$

Order of B = 2 × 2

Order of BA = 2 × 1

Since BA exists, we have:

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

Number of columns of A = Number of columns in BA = 1

Order of A is 2 × 1.

$B_{2 \times 2} \times A_{2 \times 1} = BA_{2 \times 1}$

Hence, order of A is 2 × 1.

(ii) Let A = $$\begin{bmatrix} x \\ y \end{bmatrix}$$.

Then,

$$\Rightarrow \begin{bmatrix} 5 & -2 \\ 1 & 3 \end{bmatrix} \times \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 11 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5x - 2y \\ x + 3y \end{bmatrix} = \begin{bmatrix} 4 \\ 11 \end{bmatrix}.$$

Solving for x and y:

∴ 5x - 2y = 4... (1)

∴ x + 3y = 11

⇒ x = 11 - 3y....(2)

Substituting value of d from equation (2) in 5x - 2y = 4, we get:

⇒ 5(11 - 3y) - 2y = 4

⇒ 55 - 15y - 2y = 4

⇒ 55 - 17y = 4

⇒ 17y = 55 - 4

⇒ 17y = 51

⇒ y = $\frac{51}{17}$

⇒ y = 3

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

⇒ x = 11 - 3(3)

⇒ x = 11 - 9

⇒ x = 2

Hence, A = $$\begin{bmatrix} 2 \\ 3 \end{bmatrix}.$$

A = $$\begin{bmatrix} 2 & -1 \\ -3 & 4 \end{bmatrix}$$ and B = $$\begin{bmatrix} 8 \\ -17 \end{bmatrix}$$

Given,

AC = B

Order of A = 2 × 2

Order of AC = Order of B = 2 × 1

Since AC exists, we have :

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

Number of columns of C = Number of columns in B = 1

Order of C is 2 × 1.

Let C = $$\begin{bmatrix} x \\ y \end{bmatrix}$$

AC = B

$$\Rightarrow \begin{bmatrix} 2 & -1 \\ -3 & 4 \end{bmatrix} \times \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ -17 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x - y \\ -3x + 4y \end{bmatrix} = \begin{bmatrix} 8 \\ -17 \end{bmatrix}.$$

∴ 2x - y = 8

⇒ y = 2x - 8 .......(1)

∴ -3x + 4y = -17

Substituting value of y from equation(1) in -3x + 4y = -17, we get:

⇒ -3x + 4(2x - 8) = -17

⇒ -3x + 8x - 32 = -17

⇒ 5x - 32 = -17

⇒ 5x = -17 + 32

⇒ 5x = 15

⇒ x = $\frac{15}{5}$

⇒ x = 3.

Substituting value of x in equation (1), we get :

⇒ y = 2(3) - 8

⇒ y = 6 - 8

⇒ y = -2.

Hence, C = $$\begin{bmatrix} 3 \\ -2 \end{bmatrix}.$$

A = $$\begin{bmatrix} 3 & 2 \\ -1 & 1 \end{bmatrix}$$ and B = $$\begin{bmatrix} 14 & 3 \\ 2 & 4 \end{bmatrix}$$

Given,

AC = B

Order of A = 2 × 2

Order of AC = Order of B = 2 × 2

Since AC exists, we have:

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

Number of columns of C = Number of columns in B = 2

Order of C is 2 × 2.

Let C = $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

AC = B

$$\Rightarrow \begin{bmatrix} 3 & 2 \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 14 & 3 \\ 2 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3a + 2c & 3b + 2d \\ -a + c & -b + d \end{bmatrix} = \begin{bmatrix} 14 & 3 \\ 2 & 4 \end{bmatrix}.$$

Solving for a and c:

∴ -a + c = 2

⇒ c = a + 2 ...(1)

∴ 3a + 2c = 14 ......(2)

Substituting value of c from equation(1) in (2), we get :

⇒ 3a + 2(a + 2) = 14

⇒ 3a + 2a + 4 = 14

⇒ 5a = 14 - 4

⇒ 5a = 10

⇒ a = $\frac{10}{5}$

⇒ a = 2.

Substituting value of a in equation (1), we get :

⇒ c = 2 + 2

⇒ c = 4.

Solving for b and d :

∴ -b + d = 4

⇒ d = b + 4 .......(3)

∴ 3b + 2d = 3 .......(4)

Substituting value of d from equation (3) in (4), we get:

⇒ 3b + 2(b + 4) = 3

⇒ 3b + 2b + 8 = 3

⇒ 5b + 8 = 3

⇒ 5b = 3 - 8

⇒ 5b = -5

⇒ b = $\frac{-5}{5}$

⇒ b = -1.

Substituting value of b in equation (3), we get :

⇒ d = -1 + 4

⇒ d = 3.

Hence, C = $$\begin{bmatrix} 2 & -1\ 4 & 3 \end{bmatrix}.$$

Given,

A = $$\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$$ and B = $$\begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}$$

Solving for A²:

$$\Rightarrow A^2 = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(1) + (3)(2) & (1)(3) + (3)(4) \\ (2)(1) + (4)(2) & (2)(3) + (4)(4) \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 6 & 3 + 12 \\ 2 + 8 & 6 + 16 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 & 15 \\ 10 & 22 \end{bmatrix}.$$

Solving for 5B²:

$$\Rightarrow 5B^2 = 5\Big(\begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}\Big) \\[1em] = 5\Big(\begin{bmatrix} (-2)(-2) + (1)(-3) & (-2)(1) + (1)(2) \\ (-3)(-2) + (2)(-3) & (-3)(1) + (2)(2) \end{bmatrix}\Big) \\[1em] = 5\begin{bmatrix} 4 - 3 & -2 + 2 \\ 6 - 6 & -3 + 4 \end{bmatrix} \\[1em] = 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}.$$

Solving for A² – 5B² = 5C:

$$\Rightarrow \begin{bmatrix} 7 & 15 \\ 10 & 22 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = 5C \\[1em] \Rightarrow \begin{bmatrix} 7 - 5 & 15 - 0 \\ 10 - 0 & 22 - 5 \end{bmatrix} = 5C \\[1em] \Rightarrow \frac{1}{5} \begin{bmatrix} 2 & 15 \\ 15 & 17 \end{bmatrix} = C \\[1em] \Rightarrow \begin{bmatrix} \frac{2}{5} & 3 \\ 3 & \frac{17}{5} \end{bmatrix} = C$$

Hence, C = $$\begin{bmatrix} \frac{2}{5} & 3 \\ 3 & \frac{17}{5} \end{bmatrix}$$

Given,

B = $$\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix}$$

X = B² - 4B

Solving for B²:

$$\Rightarrow B^2 = \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(1) + (1)(8) & (1)(1) + (1)(3) \\ (8)(1) + (3)(8) & (8)(1) + (3)(3) \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 8 & 1 + 3 \\ 8 + 24 & 8 + 9 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix}.$$

Solving for 4B:

$$\Rightarrow 4B = 4\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix}.$$

Now X = B² - 4B:

$$\Rightarrow X = \begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 - 4 & 4 - 4 \\ 32 - 32 & 17 - 12 \end{bmatrix} \\[1em] = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}.$$

Hence, X =$$\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}.$$

Now,

$$\Rightarrow X \times \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \times \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5a + 0 \\ 0a + 5b \end{bmatrix} = \begin{bmatrix} 5 \\ 50 \end{bmatrix}.$$

Solving for a and b:

∴ 5a = 5

a = $\frac{5}{5}$

a = 1.

∴ 5b = 50

b = $\frac{50}{5}$

b = 10.

Hence, a = 1 and b = 10.

Solving,

$$\Rightarrow \begin{bmatrix} 2 \cos 60^\circ & -2 \sin 30^\circ \\ -\tan 45^\circ & \cos 0^\circ \end{bmatrix} \times \begin{bmatrix} \cot 45^\circ & \cosec 30^\circ \\ \sec 60^\circ & \sin 90^\circ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times \frac{1}{2} & -2 \times \frac{1}{2} \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (1)(1) + (-1)(2) & (1)(2) + (-1)(1) \\ (-1)(1) + (1)(2) & (-1)(2) + (1)(1) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 - 2 & 2 - 1 \\ -1 + 2 & -2 + 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}.$$

Hence, $$\begin{bmatrix} 2 \cos 60^\circ & -2 \sin 30^\circ \\ -\tan 45^\circ & \cos 0^\circ \end{bmatrix} \times \begin{bmatrix} \cot 45^\circ & \cosec 30^\circ \\ \sec 60^\circ & \sin 90^\circ \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}$$

Given,

A = $$\begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix}$$, B = $$\begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix}$$

AB = 0

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

Solve for x and y:

∴ 6 + 6y = 0

⇒ 6y = -6

⇒ y = $\frac{-6}{6}$

⇒ y = -1.

∴ 2x + 12 = 0

⇒ 2x = -12

⇒ x = $\frac{-12}{2}$

⇒ x = -6.

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

$$\phantom{\Rightarrow} AB = C \\[1em] \Rightarrow \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 4 & 0 \\ y & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ x & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x \times 4 + 0 \times y & x \times 0 + 0 \times 1 \\ 1 \times 4 + 1 \times y & 1 \times 0 + 1 \times 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ x & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4x + 0 & 0 + 0 \\ 4 + y & 0 + 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ x & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4x & 0 \\ 4 + y & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ x & 1 \end{bmatrix} \\[1em] \Rightarrow 4x = 4 \text{ and } 4 + y = x \\[1em] \Rightarrow x = \frac{4}{4} \text{ and } 4 + y = x \\[1em] \Rightarrow x = 1 \text{ and } 4 + y = 1 \\[1em] \Rightarrow x = 1 \text{ and } y = 1 - 4 = -3.$$

Hence, x = 1 and y = -3.

Given,

$$\begin{bmatrix} -3 & 2 \\ 0 & -5 \end{bmatrix}$$ $$\begin{bmatrix} x \\ 2 \end{bmatrix}$$ = $$\begin{bmatrix} -5 \\ y \end{bmatrix}$$

Solving:

$$\Rightarrow \begin{bmatrix} (-3)(x) + (2)(2) \\ (0)(x) + (-5)(2) \end{bmatrix} = \begin{bmatrix} -5 \\ y \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3x + 4 \\ -10 \end{bmatrix} = \begin{bmatrix} -5 \\ y \end{bmatrix}.$$

∴ y = -10

∴ -3x + 4 = -5

⇒ -3x = -5 - 4

⇒ -3x = -9

⇒ x = $\frac{-9}{-3}$

⇒ x = 3.

Hence, x = 3 and y = -10.

$$\Rightarrow \begin{bmatrix} 1 & 2 \\ 3 & 3 \end{bmatrix} \times \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix} = \begin{bmatrix} x & 0 \\ 9 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (1)(x) + (2)(0) & (1)(0) + (2)(y) \\ (3)(x) + (3)(0) & (3)(0) + (3)(y) \end{bmatrix} = \begin{bmatrix} x & 0 \\ 9 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x + 2 & 2y \\ 3x & 3y \end{bmatrix} = \begin{bmatrix} x & 0 \\ 9 & 0 \end{bmatrix}$$

Solving for x and y:

∴ 3x = 9

⇒ x = $\frac{9}{3}$

⇒ x = 3.

∴ 2y = 0

⇒ y = 0.

Hence, x = 3 and y = 0.

Solving,

$$\Rightarrow \begin{bmatrix} 2x & x \\ y & 3y \end{bmatrix} \begin{bmatrix} 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (2x)(3) + (x)(2) \\ (y)(3) + (3y)(2) \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6x + 2x \\ 3y + 6y \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8x \\ 9y \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix}.$$

Solving for x and y:

∴ 8x = 16

⇒ x = $\frac{16}{8}$

⇒ x = 2.

∴ 9y = 9

⇒ y = $\frac{9}{9}$

⇒ y = 1.

Hence, x = 2 and y = 1.

Solving AB:

$$\Rightarrow \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} \times \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (3)(a) + (0)(0) & (3)(b) + (0)(c) \\ (0)(a) + (2)(0) & (0)(b) + (2)(c) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3a & 3b \\ 0 & 2c \end{bmatrix}.$$

Solving A + B:

$$\Rightarrow \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 + a & b \\ 0 & 2 + c \end{bmatrix}.$$

Given,

AB = A + B

$$\Rightarrow \begin{bmatrix} 3a & 3b \\ 0 & 2c \end{bmatrix} = \begin{bmatrix} 3 + a & b \\ 0 & 2 + c \end{bmatrix}$$

∴ 3a = 3 + a

⇒ 3a - a = 3

⇒ 2a = 3

⇒ a = $\frac{3}{2}$

∴ 3b = b

⇒ 3b - b = 0

⇒ 2b = 0

⇒ b = 0

∴ 2c = 2 + c

⇒ 2c - c = 2

⇒ c = 2.

Hence, values of a = $\frac{3}{2}$, b = 0 c = 2.

Solving,

$$\Rightarrow \begin{bmatrix} 2x & x \\ y & 3y \end{bmatrix} \begin{bmatrix} 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (2x)(3) + (x)(2) \\ (y)(3) + (3y)(2) \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6x + 2x \\ 3y + 6y \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8x \\ 9y \end{bmatrix} = \begin{bmatrix} 16 \\ 9 \end{bmatrix}.$$

∴ 8x = 16

⇒ x = $\frac{16}{8}$

⇒ x = 2.

∴ 9y = 9

⇒ y = $\frac{9}{9}$

⇒ y = 1.

Hence, x = 2 and y = 1.

Given,

A = $$\begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix}$$ and I = $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

$$\Rightarrow A^2 = \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \times \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \\[1em] = \begin{bmatrix} (2)(2) + (0)(-1) & (2)(0) + (0)(7) \\ (-1)(2) + (7)(-1) & (-1)(0) + (7)(7) \end{bmatrix} \\[1em] = \begin{bmatrix} 4 + 0 & 0 + 0 \\ -2 - 7 & 0 + 49 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix}.$$

Solving for A² = 9A + mI:

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

∴ m = -14.

Hence, m = -14.

Given,

A = $$\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix}$$ and B = $$\begin{bmatrix} 9 & -16 \\ 0 & -y \end{bmatrix}$$

Solving A²:

$$\Rightarrow A^2 = \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (3)(3) + (x)(0) & (3)(x) + (x)(1) \\ (0)(3) + (1)(0) & (0)(x) + (1)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 9 + 0 & 3x + x \\ 0 + 0 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix}.$$

Solving for x and y:

A² = B

$$\Rightarrow \begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & -16 \\ 0 & -y \end{bmatrix}$$

∴ 4x = -16

⇒ x = $\frac{-16}{4}$

⇒ x = -4.

∴ -y = 1

⇒ y = -1.

Hence, x = -4 and y = -1.

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

∴ 2y = -4

⇒ y = $\frac{-4}{2}$

⇒ y = -2.

∴ 2x = 6

x = $\frac{6}{2}$

⇒ x = 3.

Hence, x = 3 and y = -2.

Given,

A = $$\begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix}$$ and B = $$\begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix}$$

Solving for A²:

$$\Rightarrow A^2 = \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} \times \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (3)(3) + (0)(5) & (3)(0) + (0)(1) \\ (5)(3) + (1)(5) & (5)(0) + (1)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 9 + 0 & 0 + 0 \\ 15 + 5 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 & 0 \\ 20 & 1 \end{bmatrix}.$$

Solving for B²:

$$\Rightarrow B^2 = \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} \times \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} (-4)(-4) + (2)(1) & (-4)(2) + (2)(0) \\ (1)(-4) + (0)(1) & (1)(2) + (0)(0) \end{bmatrix} \\[1em] = \begin{bmatrix} 16 + 2 & -8 + 0 \\ -4 + 0 & 2 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix}.$$

Solving for 2AB:

$$\Rightarrow 2AB = 2 \Big(\begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} \times \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix}\Big) \\[1em] = 2\begin{bmatrix} (3)(-4) + (0)(1) & (3)(2) + (0)(0) \\ (5)(-4) + (1)(1) & (5)(2) + (1)(0) \end{bmatrix} \\[1em] = 2\begin{bmatrix} -12 + 0 & 6 + 0 \\ -20 + 1 & 10 + 0 \end{bmatrix} \\[1em] = 2\begin{bmatrix} -12 & 6 \\ -19 & 10 \end{bmatrix} \\[1em] = \begin{bmatrix} -24 & 12 \\ -38 & 20 \end{bmatrix}.$$

A² – 2AB + B²

$$\Rightarrow \begin{bmatrix} 9 & 0 \\ 20 & 1 \end{bmatrix} - \begin{bmatrix} -24 & 12 \\ -38 & 20 \end{bmatrix} + \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 - (-24) & 0 - 12 \\ 20 - (-38) & 1 - 20 \end{bmatrix} + \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 33 & -12 \\ 58 & -19 \end{bmatrix} + \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 33 + 18 & -12 - 8 \\ 58 - 4 & -19 + 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 51 & -20 \\ 54 & -17 \end{bmatrix}.$$

Hence, A² – 2AB + B² = $$\begin{bmatrix} 51 & -20 \\ 54 & -17 \end{bmatrix}$$

The order of a matrix is written as "rows by columns."

If a matrix has x rows and y columns, its order is written as x × y.

Hence, option 3 is the correct option.

The order of an m × n matrix indicates that it has m rows and n columns. To find the total number of elements, you multiply the number of rows by the number of columns = mn.

Hence, option 3 is the correct option.

i = Row

j = Column

In a₂₃, i = 2 (Second row) j = 3 (Third column)

In given matrix:

A = $$\begin{bmatrix} 3 & 4 & -8 \\ 2 & 0 & -2 \\ 1 & 7 & -9 \end{bmatrix}$$

The element in the second row and third column is −2.

Hence, option 4 is the correct option.

Given,

A = $$\begin{bmatrix} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -7 \end{bmatrix}$$

A diagonal matrix is a square matrix in which all the elements outside the main diagonal are zero.

Thus, A is a diagonal matrix.

Hence, option 3 is the correct option.

a_ij = (2i + j)

⇒ a₁₁ = 2(1) + 1 = 3

⇒ a₂₁ = (2)2 + 1 = 5

⇒ a₃₁ = (2)3 + 1 = 7

Matrix A = $$\begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix}$$

Hence, option 2 is the correct option.

a_ij = i + j.

∴ a₁₁ = 1 + 1 = 2, a₁₂ = 1 + 2 = 3.

a₂₁ = 2 + 1 = 3, a₂₂ = 2 + 2 = 4.

a₃₁ = 3 + 1 = 4, a₃₂ = 3 + 2 = 5.

Matrix A = $$\begin{bmatrix} 2 & 3 \\ 3 & 4 \\ 4 & 5 \end{bmatrix}$$

Hence, option 3 is the correct option.

If a matrix has 12 elements the possible orders are, 1 x 12, 12 x 1, 2 x 6, 6 x 2, 3 x 4, 4 x 3.

There are 6 possible orders.

Hence, option 2 is the correct option.

Given,

a_ij = $\frac{(i + 2j)^2}{2}$

$$\begin{aligned}a_{11} = \frac{[1 + 2(1)]^2}{2} = \frac{3^2}{2} = \frac{9}{2}, a_{12} = \frac{[1 + 2(2)]^2}{2} = \frac{5^2}{2} = \frac{25}{2} \\[1em] a_{21} = \frac{[2 + 2]^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8, a_{22} = \frac{[2 + 2(2)]^2}{2} = \frac{6^2}{2} = \frac{36}{2} = 18.\end{aligned}$$

Matrix A = $$\begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix}$$

Hence, option 4 is the correct option.

Given,

$$\begin{bmatrix} 5 & 3 \\ x & 7 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 7 \end{bmatrix}$$

Solving for x, y and z:

∴ x = 1

∴ y = 5

∴ z = 3

⇒ x + y + z = 9

Hence, option 1 is the correct option.

Given,

$$\Rightarrow \begin{bmatrix} x - y & 2x + z \\ 2x - y & 3z + w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$$

Solving for x and y:

∴ 2x - y = 0

⇒ y = 2x ...(1)

∴ x - y = -1 ...(2)

Substituting value of y from equation (1) in x - y = -1, we get:

⇒ x - 2x = -1

⇒ -x = -1

⇒ x = 1.

Substituting value of x in equation(1), we get:

⇒ y = 2(1)

⇒ y = 2.

Solving for w and z:

∴ 2x + z = 5

⇒ 2(1) + z = 5

⇒ 2 + z = 5

⇒ z = 5 - 2

⇒ z = 3.

∴ 3z + w = 13

⇒ 3(3) + w = 13

⇒ 9 + w = 13

⇒ w = 13 - 9

⇒ w = 4.

∴ x + y + z + w = 1 + 2 + 3 + 4 = 10.

Hence, option 3 is the correct option.

If A = m × n then A^T = n × m

The total number of elements is calculated by multiplying the dimensions: m × n. Since m × n = n × m

Therefore, total number of elements in A is exactly the same as the total number of elements in A^T

Hence, option 4 is the correct option.

Given,

$$\begin{bmatrix} x + y \\ x - y \end{bmatrix} = \begin{bmatrix} 8 \\ 4 \end{bmatrix}$$

∴ x + y = 8 .....(1)

∴ x - y = 4 .....(2)

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

⇒ x + y + x - y = 8 + 4

⇒ 2x = 12

⇒ x = $\frac{12}{2}$

⇒ x = 6.

Substituting value of x in equation (1) :

⇒ x + y = 8

⇒ 6 + y = 8

⇒ y = 8 - 6

⇒ y = 2.

∴ xy = 6 × 2 = 12.

Hence, option 4 is the correct option.

$$\begin{bmatrix} p - q & 2q \\ 2q + r & p + q \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 9 & 5 \end{bmatrix}$$

Solving for p, q and r:

∴ 2q = 4...(1)

⇒ q = $\frac{4}{2}$

⇒ q = 2.

∴ p - q = 1

⇒ p - 2 = 1

⇒ p = 1 + 2

⇒ p = 3.

∴ 2q + r = 9

⇒ 2(2) + r = 9

⇒ 4 + r = 9

⇒ r = 9 - 4

⇒ r = 5.

∴ p + q + r = 3 + 2 + 5 = 10.

Hence, option 2 is the correct option.

Given,

$$\begin{bmatrix} x + 3 & 4 \\ y - 4 & x + y \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix}$$

Solving for x and y :

∴ x + 3 = 5

⇒ x = 5 - 3

⇒ x = 2.

∴ y - 4 = 3

⇒ y = 3 + 4

⇒ y = 7.

Hence, option 1 is the correct option.

Given matrix has :

3 rows and 5 columns

Therefore, the order of the matrix is 3 × 5.

Hence, option 3 is the correct option.

Given,

A of order 2 × 1

B of order 2 × 2

Since their orders are different, thus matrix A and B cannot be added.

Thus, A + B is not possible.

Hence, option 4 is the correct option.

Given,

A = $$\begin{bmatrix} -5 & 4 \\ 3 & -8 \end{bmatrix}$$ and B = $$\begin{bmatrix} 2 & -3 \\ -1 & 4 \end{bmatrix}$$

Solving for A - B:

$$\Rightarrow \begin{bmatrix} -5 - 2 & 4 - (-3) \\ 3 - (-1) & -8 - 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -7 & 7 \\ 4 & -12 \end{bmatrix}.$$

Hence, option 1 is the correct option.

Given,

$$\begin{bmatrix} 4 & 2 \\ -3 & -9 \end{bmatrix}$$

Solving for -3A:

$$\Rightarrow -3\begin{bmatrix} 4 & 2 \\ -3 & -9 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -12 & -6 \\ 9 & 27 \end{bmatrix}$$

Hence, option 3 is the correct option.

Given,

A = $$\begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix}$$ and B = $$\begin{bmatrix} 5 & 0 \\ 3 & 2 \end{bmatrix}$$

Solving for 3A + 2B:

$$\Rightarrow 3\begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix} + 2\begin{bmatrix} 5 & 0 \\ 3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 & 6 \\ 12 & 3 \end{bmatrix} + \begin{bmatrix} 10 & 0 \\ 6 & 4 \end{bmatrix} \\[1em] \Rightarrow 3\begin{bmatrix} 3 & 2 \\ 4 & 1 \end{bmatrix} + 2\begin{bmatrix} 5 & 0 \\ 3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 + 10 & 6 + 0 \\ 12 + 6 & 3 + 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 19 & 6 \\ 18 & 7 \end{bmatrix}$$

Hence, option 1 is the correct option.

Given,

A + $$\begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix}$$ = $$\begin{bmatrix} 3 & -6 \\ 5 & -8 \end{bmatrix}$$

Solving for A:

$$\Rightarrow A = \begin{bmatrix} 3 & -6 \\ 5 & -8 \end{bmatrix} - \begin{bmatrix} 4 & 6 \\ -3 & 7 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 - 4 & -6 - 6 \\ 5 - (-3) & -8 - 7 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 & -12 \\ 8 & -15 \end{bmatrix}$$

Hence, option 2 is the correct option.

Given,

$$\cosθ \cdot \begin{bmatrix} \cosθ & \sinθ \\ -\sinθ & \cosθ \end{bmatrix} + \sinθ \cdot \begin{bmatrix} \sinθ & -\cosθ \\ \cosθ & \sinθ \end{bmatrix}$$

Solving:

$$\Rightarrow \begin{bmatrix} \cos^2θ & \cosθ\sinθ \\ -\sinθ\cosθ & \cos^2θ \end{bmatrix} + \cdot \begin{bmatrix} \sin^2θ & -\sinθ\cosθ \\ \sinθ\cosθ & \sin^2θ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \cos^2θ + \sin^2θ & \cosθ\sinθ - \sinθ\cosθ\ -\sinθ\cosθ + \sinθ\cosθ & \cos^2θ + \sin^2θ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$$

Hence, option 3 is the correct option.

Given,

$$2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}$$

Solving:

$$\Rightarrow \begin{bmatrix} 6 & 8 \\ 10 & 2x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 + 1 & 8 + y \\ 10 + 0 & 2x + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & 8 + y \\ 10 & 2x + 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}.$$

∴ 8 + y = 0

⇒ y = -8

∴ 2x + 1 = 5

⇒ 2x = 5 - 1

⇒ 2x = 4

⇒ x = $\frac{4}{2}$

⇒ x = 2

∴ x - y = 2 - (-8) = 10

Hence, option 1 is the correct option.

Given,

$$2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$$

Solving:

$$\Rightarrow \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}.$$

∴ 2 + y = 5

⇒ y = 5 - 2

⇒ y = 3

∴ 2x + 2 = 8

⇒ 2x = 8 - 2

⇒ 2x = 6

⇒ x = $\frac{6}{2}$ = 3.

∴ x + y = 3 + 3 = 6.

Hence, option 3 is the correct option.

Given,

$$\begin{bmatrix} a + b & 2 \\ 5 & ab \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$$

∴ a + b = 6

⇒ b = 6 - a...(1)

∴ ab = 8...(2)

Substituting value of b from equation(1) in ab = 8, we get:

⇒ a (6 - a) = 8

⇒ 6a - a² = 8

⇒ a² - 6a + 8 = 0

⇒ a² - 4a - 2a + 8 = 0

⇒ a(a - 4) -2(a - 4) = 0

⇒ (a - 2)(a - 4) = 0

(a - 2)= 0 or (a - 4) = 0 [Using zero product rule]

⇒ a = 2 or a = 4

If a = 2, then b = 6 − 2 = 4.

If a = 4, then b = 6 − 4 = 2.

Hence, option 3 is the correct option.

Given,

A = $$\begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{bmatrix}$$

A′ = $$\begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 2 & 3 & 0 \end{bmatrix}$$

(A + A′)

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

Hence, option 1 is the correct option.

$$\Rightarrow \begin{bmatrix} x & y + 2 & z - 3 \end{bmatrix} + \begin{bmatrix} y & 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x + y & y + 2 + 4 & z - 3 + 5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x + y & y + 6 & z + 2 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix} \\[1em]$$

Solving for y and z:

∴ y + 6 = 9

⇒ y = 9 - 6

⇒ y = 3.

∴ x + y = 4

⇒ x + 3 = 4

⇒ x = 4 - 3

⇒ x = 1

∴ z + 2 = 12

⇒ z = 12 - 2

⇒ z = 10.

∴ xyz = (1)(3)(10) = 30.

Hence, option 4 is the correct option.

$$\Rightarrow x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x \\ x \end{bmatrix} + \begin{bmatrix} 3y \\ 5y \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x + 3y - 8 \\ x + 5y - 11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$

∴ 2x + 3y - 8 = 0

⇒ 2x + 3y = 8 ....(1)

∴ x + 5y - 11 = 0

⇒ x = 11 - 5y ....(2)

Substituting value of x from equation (2) in (1), we get :

⇒ 2(11 - 5y) + 3y = 8

⇒ 22 - 10y + 3y = 8

⇒ -7y = 8 - 22

⇒ -7y = -14

⇒ y = $\frac{-14}{-7}$

⇒ y = 2.

Substituting value of y in equation (2) we get:

⇒ x = 11 - 5(2)

⇒ x = 11 - 10

⇒ x = 1.

∴ 2x - 3y = 2(1) - 3(2) = 2 - 6 = -4.

Hence, option 2 is the correct option.

Given,

$$\Rightarrow \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} + 2A = 4 \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \\[1em] \Rightarrow 2A = 4 \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} \\[1em] \Rightarrow A = \frac{1}{2} \Big(4\begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} \Big) \\[1em] \Rightarrow A = \frac{1}{2} \Big(\begin{bmatrix} -4 & 8 \\ 12 & 16 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ -4 & 1 \end{bmatrix} \Big) \\[1em] \Rightarrow A = \frac{1}{2} \Big(\begin{bmatrix} -4 - 2 & 8 - 3 \\ 12 - (-4) & 16 - 1 \end{bmatrix}\Big) \\[1em] \Rightarrow \frac{1}{2}\begin{bmatrix} -6 & 5 \\ 16 & 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -3 & \frac{5}{2} \\ 8 & \frac{15}{2} \end{bmatrix}.$$

Hence, option 3 is the correct option.

For matrix multiplication A × B to be defined, the inner dimensions must match.

If matrix A has the order m × n.And matrix B has the order p × q.The product AB is only possible if n = p.

The number of columns of A is equal to the number of rows of B.

Hence, option 2 is the correct option.

Given,

Matrix A: m × n

Matrix B: p × q

The product BA is defined if number of columns in B = number of rows in A, m = q .

Hence, option 2 is the correct option.

Given,

Matrix P: 3 × n

Matrix Q: n × p

The product PQ is defined if number of columns in P × number of rows in Q, 3 × p .

Hence, option 1 is the correct option.

Let A = $$\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$$

Let B = $$\begin{bmatrix} 6 \\ 8 \end{bmatrix}$$

Let order of P be m × n:

Thus, AP = B

For matrix multiplication:

The number of columns in A must equal the number of rows in P.

⇒ m = 2

The order of the product B = Rows of A × Columns of P.

⇒ 2 × 1 = 2 × n

⇒ n = 1.

Order of matrix P = 2 × 1.

Hence, option 2 is the correct option.

A = $$\begin{bmatrix} 3 & -6 \end{bmatrix}$$

Order of matrix A = 1 × 2

B = $$\begin{bmatrix} 3 & -1 \\ 4 & 3 \end{bmatrix}$$

Order of matrix B = 2 × 2

The resulting matrix AB will have the number of rows from the first matrix A and the number of columns from the second matrix B.

Order of matrix AB = 1 × 2

Hence, option 1 is the correct option.

Given,

$$\begin{bmatrix} 2 & 0 \\ 0 & 4 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$$ = $$\begin{bmatrix} 2 \\ -8 \end{bmatrix}$$

Solving,

$$\Rightarrow \begin{bmatrix} 2(x) + 0(y) \\ 0(x) + 4(y) \end{bmatrix}= \begin{bmatrix} 2 \\ -8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x \\ 4y \end{bmatrix}= \begin{bmatrix} 2 \\ -8 \end{bmatrix}$$

∴ 2x = 2

⇒ x = $\frac{2}{2}$

⇒ x = 1.

∴ 4y = -8

⇒ y = $\frac{-8}{4}$

⇒ y = -2.

Hence, option 1 is the correct option.

Given,

$$\begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix} = 0$$

Solving,

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

∴ x - 2 = 0

⇒ x = 2

Hence, option 3 is the correct option.

Calculating,

$$\Rightarrow AB = \begin{bmatrix} -1 & 2 \\ \end{bmatrix}\begin{bmatrix} 3 \\ 4 \end{bmatrix} \\[1em] = \begin{bmatrix} -1 \times 3 + 2 \times 4 \\ \end{bmatrix} \\[1em] = \begin{bmatrix} -3 + 8 \\ \end{bmatrix} \\[1em] = \begin{bmatrix} 5 \end{bmatrix}.$$

Hence, option 3 is the correct option.

Solving for x and y:

$$\Rightarrow x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x - y \\ 3x + y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \\[1em]$$

∴ 2x - y = 10 ....(1)

∴ 3x + y = 5 .....(2)

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

⇒ 2x - y + 3x + y = 10 + 5

⇒ 2x + 3x = 15

⇒ 5x = 15

⇒ x = $\frac{15}{5}$

⇒ x = 3.

Substituting value of x in 2x - y = 10, we get :

⇒ 2(3) - y = 10

⇒ 6 - y = 10

⇒ y = 6 - 10

⇒ y = -4.

∴ xy = (3)(-4) = -12.

Hence, option 2 is the correct option.

Given,

A = $$\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}$$ and B = $$\begin{bmatrix} -4 \\ 3 \end{bmatrix}$$

Solving for AB:

$$\Rightarrow \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix} \times \begin{bmatrix} -4 \\ 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (4)(-4) + (3)(3) \\ (1)(-4) + (2)(3) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -16 + 9 \\ -4 + 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -7 \\ 2 \end{bmatrix}.$$

Hence, option 1 is the correct option.

Order of matrix A = 1 × 2

Order of matrix B = 2 × 1

Since, no. of columns in A is equal to the no. of rows in B and no. of columns in B is equal to the no. of rows in A.

∴ AB and BA are possible.

$$AB = \begin{bmatrix} a \times c + b \times d \end{bmatrix} \\[1em] = \begin{bmatrix} ac + bd \end{bmatrix}. \\[1em] BA = \begin{bmatrix} c \times a & c \times b \\ d \times a & d \times b \end{bmatrix} \\[1em] = \begin{bmatrix} ca & cb \\ da & db \end{bmatrix}.$$

∴ AB ≠ BA.

Hence, Option 3 is the correct option.

Calculating,

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

Hence, option 3 is the correct option.

If A is of order m × n:

For matrix multiplication AB to exist, the number of columns of A must equal the number of rows of B.

Number of rows in B = n

For matrix multiplication BA to exist, the number of columns of B must be equal to the number of rows of A.

Number of columns in B = m.

B is of order = n × m.

Hence, option 2 is the correct option.

Given,

A = $$\begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix}$$ and I = $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Solving for AI:

$$AI = \begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 5(1) + 10(0) & 5(0) + 10(1) \\ 3(1) + (-4)(0) & 3(0) + (-4)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 5 + 0 & 0 + 10 \\ 3 + 0 & 0 + (-4) \end{bmatrix} \\[1em] = \begin{bmatrix} 5 & 10 \\ 3 & -4 \end{bmatrix}.$$

Hence, option 3 is the correct option.

Given,

$$A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

Substituting values we get :

$$A^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} . \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\\[1em] = \begin{bmatrix} 0 \times 0 + 1 \times 1 & 0 \times 1 + 1 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 1 + 0 \times 0 \end{bmatrix}\\[1em] = \begin{bmatrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{bmatrix}\\[1em] = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$

Hence, option 3 is the correct option.

Order of matrix A = 2 × 2

Order of matrix B = 2 × 1

Order of matrix M (let) = a × b

We know that,

Two matrix can be multiplied if the no. of columns of the first matrix is equal to the no. of rows of the second matrix and the resultant matrix has the no. of rows of first matrix and no. of columns of second matrix.

Given,

⇒ AM = B

⇒ A_{2 × 2} × M_{a × b} = B_{2 × 1}

⇒ a = 2 and b = 1.

Order of matrix M = 2 × 1.

Hence, Option 2 is the correct option.

$$\phantom{\Rightarrow} A^2 = \begin{bmatrix} 2 & 2 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & 2 \\ 0 & 2 \end{bmatrix} \\[1em] \Rightarrow A^2 = \begin{bmatrix} 2 \times 2 + 2\times 0 & 2 \times 2 + 2 \times 2 \\ 0 \times 2 + 2 \times 0 & 0 \times 2 + 2 \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & x \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 4 + 0 & 4 + 4 \\ 0 + 0 & 0 + 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & x \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 0 & 4 \end{bmatrix} \\[1em] \Rightarrow x = 8.$$

Hence, Option 3 is the correct option.

Given,

$$A^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$.

$$\therefore \begin{bmatrix} 6 & 9 \\ -4 & k \end{bmatrix}\begin{bmatrix} 6 & 9 \\ -4 & k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 6 \times 6 + 9 \times (-4) & 6 \times 9 + 9 \times k \\ -4 \times 6 + k \times (-4) & (-4) \times 9 + k \times k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 36 + (-36) & 54 + 9k \\ -24 - 4k & -36 + k^2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 54 + 9k \\ -24 - 4k & -36 + k^2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow 54 + 9k = 0 \\[1em] \Rightarrow 9k = -54 \\[1em] \Rightarrow k = -\frac{54}{9} = -6.$$

Hence, Option 2 is the correct option.

The product of a row matrix and a column matrix is possible.

A row matrix has order 1 × n and a column matrix has order m × 1.

If n = m, so the product is defined and gives a 1 × 1 matrix.

So, assertion (A) is true.

Two matrices of the same order cannot be multiplied.

If the matrix are of the same order and not square, example 2 × 3 and 2 × 3, then (2 × 3) × (2 × 3) is not defined because the number of columns of the first matrix is not equal to the number of rows of the second matrix.

So the statement “Two matrices of the same order can be multiplied” is false.

Reason (R) is false.

A is true, R is false.

Hence, option 3 is the correct option.

In a rectangular matrix number of rows ≠ number of columns.

A diagonal matrix is defined only for square matrices, where all non-diagonal elements are 0 and diagonal elements may be nonzero.

Hence, a rectangular matrix can never be diagonal, because diagonal position exists only when rows = columns.

So, assertion (A) is false.

A square matrix in which every non-diagonal element is 0 is called a diagonal matrix.

This is the correct definition of a diagonal matrix.

So, reason (R) is true.

A is false, R is true.

Hence, option 4 is the correct option.

Matrix addition follows the commutative law — but only when A and B are of the same order.

If that is the condition, A + B = B + A always holds true.

So, Assertion (A) is true.

Matrix addition is defined only when both matrices have the same order.

So, Reason (R) is true.

Both A and R are true, but R is not the correct explanation of A.

Hence, option 2 is the correct option.

If A and B are square matrices of the same order (say n × n), then :

Thus, for AB, no. of columns in A = no. of rows in B.

Thus, AB is possible.

For BA, no. of columns in B = no. of rows in A.

Thus, BA is possible.

Assertion (A) is true.

The rule for matrix multiplication is : AB exists if the number of columns of A = number of rows of B.

Reason (R) is false.

A is true, R is false.

Hence, option 3 is the correct option.

Order of matrix A = 1 × 2

Order of matrix B = 2 × 2

Since, no. of columns in A is equal to number of rows in B.

∴ AB is possible.

Both A and R are true, and R is the correct explanation of A.

Hence, option 1 is the correct option.

Given,

$$\Rightarrow M = \frac{1}{2}A - 2B + 5l \\[1em] = \frac{1}{2}\begin{bmatrix} -6 & 0 \\ 4 & 2 \end{bmatrix} - 2\begin{bmatrix} 1 & 0 \\ 1 & 3 \end{bmatrix} + 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} -3 & 0 \\ 2 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 2 & 6 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \\[1em] = \begin{bmatrix} -3 - 2 + 5 & 0 - 0 + 0 \\ 2 - 2 + 0 & 1 - 6 + 5 \end{bmatrix} \\[1em] = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$

Hence, M = $$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$

(a) Given,

AB is a null matrix.

Let AB = X, where X is a null matrix of order a × b.

⇒ AB = X

⇒ A_{2 × 2} × B_{2 × 1} = X_{a × b}

We know that,

The resultant matrix has no. of rows equal to the rows in the first matrix and no. of columns equal to the no. of columns in the second matrix.

∴ a = 2 and b = 1.

Hence, the order of null matrix is 2 × 1.

(b) Calculating,

$$\therefore \begin{bmatrix} x & 1 \\ y & 2 \end{bmatrix}_{2 \times 2} \begin{bmatrix} x \\ x - 2 \end{bmatrix}_{2 \times 1} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x^2 + 1(x - 2) \\ xy + 2(x - 2) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} x^2 + x - 2 \\ xy + 2x - 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \text{ ………….(1)}$$

From equation (1) :

⇒ x² + x - 2 = 0

⇒ x² + 2x - x - 2 = 0

⇒ x(x + 2) - 1(x + 2) = 0

⇒ (x - 1)(x + 2) = 0

⇒ x - 1 = 0 or x + 2 = 0

⇒ x = 1 or x = -2.

From equation (1) :

⇒ xy + 2x - 4 = 0

Substituting x = 1, we get :

⇒ 1.y + 2.1 - 4 = 0

⇒ y + 2 - 4 = 0

⇒ y - 2 = 0

⇒ y = 2.

Substituting x = -2, we get :

⇒ (-2).y + 2.(-2) - 4 = 0

⇒ -2y - 4 - 4 = 0

⇒ -2y - 8 = 0

⇒ -2y = 8

⇒ y = $\frac{8}{-2}$ = -4.

Hence, x = 1, y = 2 or x = -2, y = -4.

(a) Order of matrix A = 2 × 2, Order of matrix C = 1 × 2

The product AC is not possible as the no. of columns in A is not equal to the no. of rows in C.

Hence, product AC is not possible.

(b) Given,

$$\Rightarrow X = AB + B^2 - DC \\[1em] \Rightarrow X = \begin{bmatrix} -1 & 3 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} - \begin{bmatrix} 4 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -1 \times 1 + 3 \times 0 & -1 \times -2 + 3 \times 3 \\ 2 \times 1 + 0 \times 0 & 2 \times -2 + 0 \times 3 \end{bmatrix} + \begin{bmatrix} 1 \times 1 + -2 \times 0 & 1 \times -2 + -2 \times 3 \\ 0 \times 1 + 3 \times 0 & 0 \times -2 + 3 \times 3 \end{bmatrix} - \begin{bmatrix} 4 \times 1 & 4 \times -4 \\ 1 \times 1 & 1 \times -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -1 + 0 & 2 + 9 \\ 2 + 0 & -4 + 0 \end{bmatrix} + \begin{bmatrix} 1 + 0 & -2 - 6 \\ 0 + 0 & 0 + 9 \end{bmatrix} - \begin{bmatrix} 4 & -16 \\ 1 & -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -1 & 11 \\ 2 & -4 \end{bmatrix} + \begin{bmatrix} 1 & -8 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 4 & -16 \\ 1 & -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -1 + 1 - 4 & 11 + (-8) - (-16) \\ 2 + 0 - 1 & (-4) + 9 - (-4) \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -4 & 11 - 8 + 16 \\ 1 & -4 + 9 + 4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix} -4 & 19 \\ 1 & 9 \end{bmatrix}$$

Hence, X = $$\begin{bmatrix} -4 & 19 \\ 1 & 9 \end{bmatrix}$$.