Find the value of $x,y$ and $z$ from the following equation:
(i) $\left[\begin{array}{cc}4& 3\\ x& 5\end{array}\right]=\left[\begin{array}{cc}y& z\\ 1& 5\end{array}\right]$

(ii) $\left[\begin{array}{cc}x+y& 2\\ 5+z& xy\end{array}\right]=\left[\begin{array}{cc}6& 2\\ 5& 8\end{array}\right]$

(iii) $\left[\begin{array}{c}x+y+z\\ x+z\\ y+z\end{array}\right]=\left[\begin{array}{c}9\\ 5\\ 7\end{array}\right]$

(i) $\left[\begin{array}{cc}4& 3\\ x& 5\end{array}\right]=\left[\begin{array}{cc}y& z\\ 1& 5\end{array}\right]$

As the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get: $x=1,y=4,$ and $z=3$

(ii) $\left[\begin{array}{cc}x+y& 2\\ 5+z& xy\end{array}\right]=\left[\begin{array}{cc}6& 2\\ 5& 8\end{array}\right]$

As the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:

$x+y=6,xy=8,5+z=5$

Now, $5+z=5\Rightarrow z=0$

We know that:
${\left(x-y\right)}^2={\left(x+y\right)}^2-4xy$
$\Rightarrow {\left(x-y\right)}^2=36-32=4$
$\Rightarrow x-y=\pm 2$

Now, when $x-y=2$ and $x+y=6$, we get $x=4$ and $y=2$
When $x-y=-2$ and $x+y=6$, we get $x=2$ and $y=4$
$\therefore x=4,y=2$ and $z=0orx=2,y=4$ and  $z=0$

(iii)  $\left[\begin{array}{c}x+y+z\\ x+z\\ y+z\end{array}\right]=\left[\begin{array}{c}9\\ 5\\ 7\end{array}\right]$

As the two matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:

$x+y+z=9$        ....(1)
$x+z=5$            ....(2)
$y+z=7$            ....(3)

From (1) and (2), we have:
$y+5=9$
$\Rightarrow y=4$

Then, from (3), we have: $4+z=7$
$\Rightarrow z=3$
$\therefore x+z=5$
$\Rightarrow x=2$
$\therefore x=2,y=4$ and $z=3$