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# Find the values of x,y and z from the following equations:$(ii)\;\begin{bmatrix}x+y & z\\5+z & xy\end{bmatrix}=\begin{bmatrix}6 & 2\\5 & 8\end{bmatrix}\qquad$

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• If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij} = B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
Given $\begin{bmatrix}x+y & 2\\5+z & xy\end{bmatrix}=\begin{bmatrix}6 & 2\\5 & 8\end{bmatrix}.$ Since these matrices are equal, we can an obtain the value of x,y,z by comparing the matrices' corresponding elements.
By comparing the given two matrices of equal order, we can see that:
$x+y = 6$ (i)
$5+z = 5$ (ii)
$xy = 8$ (iii)
From (i) we get $x = 6-y$. Substituting in (iii), we get $(6-y)y = 8$ $\rightarrow$ $y^2-6y-8=0$.
Solving for $y$, $(y-2)(y-4) = 0$ $\rightarrow$ $y=2$ or $y=4$.
Given than $xy=8$, If $y=2$, then $x=4$ and if $y=4$, then $x=2$.
Solving (ii), we get $5+z = 5$. $\rightarrow$ $z = 0$.
Solving for x, y and z we get two solutions $(x,y,z) = (2,4,0)$ or $(4,2,0)$.