Question

If $\left[\begin{array}{cc}x-2& x+y\\ z-3& 4a-8\end{array}\right]=\left[\begin{array}{cc}0& 3\\ 4& 12\end{array}\right]$, then the value of $(x+y+z)/a$ is

Answer 1

$\left[\begin{array}{cc}x-2& x+y\\ z-3& 4a-8\end{array}\right]=\left[\begin{array}{cc}0& 3\\ 4& 12\end{array}\right]$

$x-2=3\Rightarrow x=5$

$x+y=3\Rightarrow y=3-5=-2$

$z-3=4\Rightarrow z=7$

$4a-8=12\Rightarrow a=5$

$\therefore \frac{x+y+z}{a}=\frac{5-2+7}{5}=2$