Given, $A = \begin{bmatrix$}[r] 3 & 4 \ 4 & -3 \end{bmatrix} \text{ and B} = \begin{bmatrix}[r] 24 \ 7 \end{bmatrix}; find the matrix X such that AX = B.

Let order of matrix X = m × n

Given,

AX = B

$\begin{bmatrix$}[r] 3 & 4 \ 4 & -3 \end{bmatrix}{2 \times 2}X{m \times n} = \begin{bmatrix}[r] 24 \ 7 \end{bmatrix}_{2 \times 1}

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

∴ m = 2

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

∴ n = 1

Order of matrix X = m × n = 2 × 1.

Let matrix X = $\begin{bmatrix$}[r] a \ b \end{bmatrix}

Substituting matrix in AX = B we get,

$\Rightarrow \begin{bmatrix$}[r] 3 & 4 \ 4 & -3 \end{bmatrix}\begin{bmatrix}[r] a \ b \end{bmatrix} = \begin{bmatrix}[r] 24 \ 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 3 \times a + 4 \times b \ 4 \times a + (-3) \times b \end{bmatrix} = \begin{bmatrix}[r] 24 \ 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 3a + 4b \ 4a - 3b \end{bmatrix} = \begin{bmatrix}[r] 24 \ 7 \end{bmatrix}.

∴ 3a + 4b = 24 ……..(1)

4a - 3b = 7 ……..(2)

⇒ 3a + 4b = 24

⇒ 3a = 24 - 4b

⇒ a = $\dfrac{24 - 4b}{3}$ ……..(3)

Substituting value of a from equation (3) in equation (2), we get :

$\Rightarrow 4 \times \dfrac{24 - 4b}{3} - 3b = 7 \\[1em] \Rightarrow \dfrac{96 - 16b}{3} - 3b = 7 \\[1em] \Rightarrow \dfrac{96 - 16b - 9b}{3} = 7 \\[1em] \Rightarrow 96 - 25b = 21 \\[1em] \Rightarrow 25b = 96 - 21 \\[1em] \Rightarrow 25b = 75 \\[1em] \Rightarrow b = \dfrac{75}{25} = 3.$

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

a = $\dfrac{24 - 4b}{3} = \dfrac{24 - 4 \times 3}{3} = \dfrac{12}{3}$ = 4.

Hence, X = $\begin{bmatrix$}[r] 4 \ 3 \end{bmatrix}