Question:
Find $x, y$ and $z$ so that $A=B$, where
$A=\left[\begin{array}{ccc}x-2 & 3 & 2 z \\ 18 z & y+2 & 6 z\end{array}\right], B=\left[\begin{array}{ccc}y & z & 6 \\ 6 y & x & 2 y\end{array}\right]$
Solution:
Since all the corresponding elements of a matrix are equal,
$A=\left[\begin{array}{ccc}x-2 & 3 & 2 z \\ 18 z & y+2 & 6 z\end{array}\right], B=\left[\begin{array}{ccc}y & z & 6 \\ 6 y & x & 2 y\end{array}\right]$
Here,
$x-2=y \quad \ldots(1)$
$z=3 \quad \ldots(2)$
$18 z=6 y \quad \ldots(3)$
Putting the value of $z$ in eq. (3), we get
$18(3)=6 y$
$\Rightarrow 54=6 y$
$\Rightarrow y=\frac{54}{6}=9$
Putting the value of $y$ in eq. (1), we get
$x-2=9$
$\Rightarrow x=9+2$
$\Rightarrow x=11$
$\therefore x=11, y=9$ and $z=3$