Given $\begin{bmatrix$}[r] 2 & 1 \ -3 & 4 \end{bmatrix}X = \begin{bmatrix}[r] 7 \ 6 \end{bmatrix}, write :

(i) the order of the matrix X

(ii) the matrix X.

(i) Since,

$\begin{bmatrix$}[r] 2 & 1 \ -3 & 4 \end{bmatrix}X = \begin{bmatrix}[r] 7 \ 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2 & 1 \ -3 & 4 \end{bmatrix}X \text{ is a } 2 \times 1 \text{ matrix, but} \begin{bmatrix}[r] 2 & 1 \ -3 & 4 \end{bmatrix} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \Rightarrow \text{X is a } 2 \times 1 \text{ matrix}.

The order of the matrix is 2 × 1.

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

Given,

$\begin{bmatrix$}[r] 2 & 1 \ -3 & 4 \end{bmatrix}X = \begin{bmatrix}[r] 7 \ 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2 & 1 \ -3 & 4 \end{bmatrix} \begin{bmatrix}[r] x \ y \end{bmatrix} = \begin{bmatrix}[r] 7 \ 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2 \times x + 1 \times y \ -3 \times x + 4 \times y \end{bmatrix} = \begin{bmatrix}[r] 7 \ 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2x + y \ -3x + 4y \end{bmatrix} = \begin{bmatrix}[r] 7 \ 6 \end{bmatrix} \\[1em]

By definition of equality of matrices we get,

2x + y = 7 or y = 7 - 2x     (…Eq 1)

-3x + 4y = 6                      (…Eq 2)

Putting value of y from Eq 1 in Eq 2

⇒ -3x + 4y = 6
⇒ -3x + 4(7 - 2x) = 6
⇒ -3x + 28 - 8x = 6
⇒ -11x = 6 - 28
⇒ -11x = -22
⇒ x = 2.

∴ x = 2 and y = 7 - 2x = 7 - 2(2) = 7 - 4 = 3.

Since, $\text{X} = \begin{bmatrix$}[r] x \ y \end{bmatrix} \\[1em] \therefore \text{X} = \begin{bmatrix}[r] 2 \ 3 \end{bmatrix}

Hence, the matrix $\text{X} = \begin{bmatrix$}[r] 2 \ 3 \end{bmatrix} .