Question
If $\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y}\end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6\end{bmatrix},$ then find the values of $x, y, z$ and $w.$

Answer

$\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y}\end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6\end{bmatrix}$
The corresponding entries of the two equal matrices are equal,
$\Rightarrow xy = 8 ...(1),$
$w = 4 ...(2),$
$z + 6 = 0 ...(3),$
And $x + y = 6 ...(4)$
From equation $(2)$ and equation $(3)$ we get $z = -6$ and $w = 4.$
From equation $(4)$ we have,
$x + y = 6$
$\Rightarrow x = 6 - y,$
Subsituting value of $x$ in equation $(1)$ we get,
$\Rightarrow (6 - y)y = 8$
$\Rightarrow y^2 - 6y + 8 = 0$
$\Rightarrow (y - 2)(y - 4) = 0,$
$\Rightarrow y = 2, 4$
Subsituting the value of $y$ in equation $(1)$ we get,
$\Rightarrow x = 4, 2$
Therefore, value of $x, y, z, w$ are $2, 4, -6, 4$ or $4, 2, -6, 4.$

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