Question

Find the values of $x,y$ and $z$ form the following equations
$\left(i\right)$ $\left[\begin{array}{cc}4& 3\\ x& 5\end{array}\right]=\left[\begin{array}{cc}y& z\\ 1& 5\end{array}\right]$
$\left(ii\right)$ $\left[\begin{array}{cc}x+y& 2\\ 5+z& xy\end{array}\right]=\left[\begin{array}{cc}6& 2\\ 5& 8\end{array}\right]$
$\left(iii\right)$ $\left[\begin{array}{c}x+y+z\\ x+z\\ y+z\end{array}\right]=\left[\begin{array}{c}9\\ 5\\ 7\end{array}\right]$

Answer 1

$\left(i\right):$ Given: $\left[\begin{array}{cc}4& 3\\ x& 5\end{array}\right]=\left[\begin{array}{cc}y& z\\ 1& 5\end{array}\right]$
Since matrices are equal,
Comparing corresponding terms,
$\therefore x=1,y=4,z=3$

$\left(ii\right):$ Given: $\left[\begin{array}{cc}x+y& 2\\ 5+z& xy\end{array}\right]=\left[\begin{array}{cc}6& 2\\ 5& 8\end{array}\right]$
Since matrices are equal.
Comparing corresponding terms,
$x+y=6\cdots \left(1\right)$
$xy=8\cdots \left(2\right)$
$5+z=5\cdots \left(3\right)$
$\Rightarrow 5+z=5$
$\Rightarrow z=5-5$
$\Rightarrow z=0$
From equation $\left(1\right),$
$x+y=6$
$\Rightarrow x=6-y$
Put in equation $\left(2\right)$
$\Rightarrow (6-y)y=8$
$\Rightarrow 6y-y^2=8$
$\Rightarrow y^2-6y+8=0$
$\Rightarrow y^2-4y-2y+8=0$
$\Rightarrow (y-2)(y-4)=0$
$\Rightarrow y=2\text{or}y=4$
Putting values of $y$ in equation $\left(1\right)$ i.e. $x+y=6$
When $y=2\to x=4$ or $y=4\to x=2$
Therefore,
$x=2,y=4,z=0$ or $x=4,y=2,z=0$

$\left(iii\right):$ Given: $\left[\begin{array}{c}x+y+z\\ x+z\\ y+z\end{array}\right]=\left[\begin{array}{c}9\\ 5\\ 7\end{array}\right]$
Since matrices are equal,
Comparing corresponding elements,
$x+y+z=9\cdots \left(1\right)$
$x+z=5\cdots \left(2\right)$
$y+z=7\cdots \left(3\right)$
From equation $\left(2\right)$
$x+z=5$
Putting this in $\left(1\right),$ we get
$y=4\cdots \left(4\right)$
From equation $\left(3\right)$
Putting this in $\left(1\right),$ we get
$x=2\cdots \left(5\right)$
Using $\left(4\right)$ and $\left(5\right)$ in $\left(1\right),$ we get
$z=3$
$\therefore x=2,y=4$ and $z=3.$