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

$x = 4, y = 2, z = -6, w = 4$

The correct answer is Option (2) → $x = 4, y = 2, z = -6, w = 4$ ##

We have,

$\begin{bmatrix} xy & 4 \\ z+6 & x+y \end{bmatrix} = \begin{bmatrix} 8 & w \\ 0 & 6 \end{bmatrix}$$

By equality of matrix,

$x + y = 6 \quad \dots(i)$

$xy = 8 \quad \dots(ii)$

From Eq. (i), we get $x = 6 - y \quad \dots(iii)$

On putting $x = 6 - y$ in Eq. (ii), we get

$(6 - y) \cdot y = 8$

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

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

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

$\Rightarrow y = 2 \text{ or } y = 4$

$∴x = 6 - 2 = 4 \quad [∵x = 6 - y]$

or $x = 6 - 4 = 2$

Also, $z + 6 = 0$

$\Rightarrow z = -6 \text{ and } w = 4$

$∴x = 2, y = 4 \text{ or } x = 4, y = 2, z = -6 \text{ and } w = 4$