Question

Find the value of x, y and z from the following equations:
$\left(i\right)\left[\begin{array}{cc}4& 3\\ x& 6\end{array}\right]$

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

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

Answer 1

Given, $\left[\begin{array}{cc}4& 3\\ x& 5\end{array}\right]=\left[\begin{array}{cc}y& z\\ 1& 5\end{array}\right]$, By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get 4=y,3 =z and $x=1\Rightarrow x=1$, y=4 and z=3

Given, $\left[\begin{array}{cc}x+y& 2\\ 5+z& xy\end{array}\right]=\left[\begin{array}{cc}6& 2\\ 5& 8\end{array}\right]$. By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get
x+y=6 .....(i)
5+z=5 .....(ii)
and xy=8 ....(iii)
From Eq. (ii), we get z=0,
From Eq. (i), y =6-x ....(iv)
Substituting the value of y in Eq. (iii), we obtain
$x(6-x)=8\Rightarrow x^2-6x+8=0\Rightarrow (x-2)(x-4)=0\Rightarrow x=2$ or x=4
When x=2, then from Eq.(iv), y=6-2=4 and when x=4, then from Eq.(iv), y =6-4=2.
So, either x=2, y=4 and z=0
or x=4, y=2 and z=0

Given,$\left[\begin{array}{c}x+y+z\\ x+z\\ y+z\end{array}\right]=\left[\begin{array}{c}9\\ 5\\ 7\end{array}\right]$. By definition of equality of matrix as the given matrices are equal, their corresponding elements are equal. Comparing the corresponding elements, we get
x+y+z=9 ....(i)
x+z=5 ....(ii)
y+z=7 ....(iii)
Subtracting Eq. (ii) from Eq. (i), we get y =4
Subtracting Eq. (iii) from Eq. (i), we get x=2
Subtracting y=4 in Eq. (iii), we get
$4+z=7\Rightarrow z=7-4=3$
Hence, x=2, y=4, z=3