If $2\begin{bmatrix$}[r] 3 & x \ 0 & 1 \end{bmatrix} + 3\begin{bmatrix}[r] 1 & 3 \ y & 2 \end{bmatrix} = \begin{bmatrix}[r] z & -7 \ 15 & 8 \end{bmatrix}; find the values of x, y and z.

Given,

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

By definition of equality of matrices we get,

z = 9,

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

3y = 15
⇒ y = 5.

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