Find the values of x, y and z if $\begin{bmatrix$}[r] x + 2 & 6 \ 3 & 5z \end{bmatrix} = \begin{bmatrix}[r] -5 & y^2 + y \ 3 & -20 \end{bmatrix} .

Given $\begin{bmatrix$}[r] x + 2 & 6 \ 3 & 5z \end{bmatrix} = \begin{bmatrix}[r] -5 & y^2 + y \ 3 & -20 \end{bmatrix} .

By definition of equality of matrices, we get

x + 2 = -5 or x = -7,
y² + y = 6       [….Eq 1],
5z = -20 or z = -4.

From Eq 1 we get,

$\Rightarrow y^2 + y - 6 = 0 \\[0.5em] \Rightarrow y^2 + 3y - 2y - 6 = 0 \\[0.5em] \Rightarrow y(y + 3) - 2(y + 3) = 0 \\[0.5em] \Rightarrow (y - 2)(y + 3) = 0 \\[0.5em] \Rightarrow y = 2 \text{ or } y = -3. \\[0.5em]$

∴ x = -7, y = 2 or -3 and z = -4.

Hence, the values are: x = -7, y = 2 and z = -4 OR x = -7, y = -3 and z = -4.