If $\left[\begin{array}{cc}x & 3 x-y \\ 2 x+z & 3 y-\omega\end{array}\right]=\left[\begin{array}{ll}3 & 2 \\ 4 & 7\end{array}\right]$, find $x, y, z, \omega$.
Since all the corresponding elements of a matrix are equal,
$\left[\begin{array}{cc}x & 3 x-y \\ 2 x+z & 3 y-w\end{array}\right]=\left[\begin{array}{ll}3 & 2 \\ 4 & 7\end{array}\right]$
$x=3$ .....(1)
$3 x-y=2$ ....(2)
Putting the value of $x$ in eq. (2), we get
$3(3)-y=2$
$\Rightarrow 9-y=2$
$\Rightarrow-y=-7$
$\Rightarrow y=7$
$2 x+z=4$ ....(3)
Putting the value of $x$ in eq. (3), we get
$2(3)+z=4$
$\Rightarrow 6+z=4$
$\Rightarrow z=4-6$
$\Rightarrow z=-2$
$3 y-w=7$ ....(4)
Putting the value of $y$ in eq. (4), we get
$3(7)-w=7$
$\Rightarrow 21-w=7$
$\Rightarrow 21-7=w$
$\Rightarrow w=14$
$\therefore x=3, y=7, z=-2$ and $w=14$