Find the value of x, y and z from the followi | vedclass

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(D) Given the matrix equation: $\begin{bmatrix} x+y & 2 \ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \ 5 & 8 \end{bmatrix}$.Since the matrices a

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$x=4, y=2, z=0$ or $x=2, y=4, z=0$

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(D) Given the matrix equation: $\begin{bmatrix} x+y & 2 \ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \ 5 & 8 \end{bmatrix}$.Since the matrices are equal,their corresponding elements must be equal.Comparing the elements,we get:$1) x+y = 6$$2) xy = 8$$3) 5+z = 5$From equation $(3)$,$z = 5-5 = 0$.Now,we have $x+y = 6$ and $xy = 8$. We can use the identity $(x-y)^2 = (x+y)^2 - 4xy$.$(x-y)^2 = (6)^2 - 4(8) = 36 - 32 = 4$.Taking the square root,$x-y = \pm 2$.Case $1$: If $x-y = 2$ and $x+y = 6$,adding the equations gives $2x = 8 \Rightarrow x = 4$. Substituting $x=4$ into $x+y=6$ gives $y = 2$.Case $2$: If $x-y = -2$ and $x+y = 6$,adding the equations gives $2x = 4 \Rightarrow x = 2$. Substituting $x=2$ into $x+y=6$ gives $y = 4$.Thus,the values are $(x=4, y=2, z=0)$ or $(x=2, y=4, z=0)$.

Find the value of $x, y$ and $z$ from the following equation: $\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$

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