Find x and y if :

(i) $3\begin{bmatrix$}[r] 4 & x \end{bmatrix} + 2\begin{bmatrix}[r] y & -3 \end{bmatrix} = \begin{bmatrix}[r] 10 & 0 \end{bmatrix}

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

(i) Given,

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

By equality of matrices we get,

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

3x - 6 = 0
⇒ 3x = 6
⇒ x = 2

Hence, x = 2 and y = -1.

(ii) Given,

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

By definition of equality of matrices we get,

-x + 8 = 7 ……..(i)

2x - 4y = -8 ……(ii)

Solving eq. (i) we get,

⇒ x = 8 - 7 = 1.

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

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

Hence, x = 1 and y = 2.5