If $\left[ {\begin{array}{cccccccccccccccccccc}{xy}&4\\{z + 6}&{x + y}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}8&w\\0&6\end{array}} \right]$, then find the values of $x,y,z$ and $w$.

We have, $\left[ {\begin{array}{cccccccccccccccccccc}{xy}&4\\{z + 6}&{x + y}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}8&w\\0&6\end{array}} \right]$

By equality of matrix, $x + y = 6$ and $xy = 8$

$\Rightarrow$ $x = 6 - y$ and $(6 - y) \cdot y = 8$

$\Rightarrow$ ${y^2} - 6y + 8 = 0$

$\Rightarrow$ ${y^2} - 4y - 2y + 8 = 0$

$\Rightarrow$ $(y - 2)(y - 4) = 0$

$\Rightarrow$ $y = 2$ or $y = 4$

$\therefore$ $x = 6 - 2 = 4$
or $x = 6 - 4 = 2$

Hence, $z + 6 = 0$
$\Rightarrow$ $z = - 6$ and $w = 4$

$\therefore$ $x = 2,y = 4$ or $x = 4,y = 2,z = - 6$ and $w = 4$