If A = begin{bmatrix} 2 & -1 0 & 2 end{bmatr | vedclass

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(B) Step $1$: Calculate $A^2 = A 	imes A = \begin{bmatrix} 2 & -1 \ 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \ 0 & 2 \end{bmatrix} = \begin{bmatr

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$x = -4, y = 4$

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(B) Step $1$: Calculate $A^2 = A 	imes A = \begin{bmatrix} 2 & -1 \ 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} 4+0 & -2-2 \ 0+0 & 0+4 \end{bmatrix} = \begin{bmatrix} 4 & -4 \ 0 & 4 \end{bmatrix}$.Step $2$: Substitute $A^2$, $A$, and $I_2$ into the equation $A^2 + xA + yI_2 = O_2$.$\begin{bmatrix} 4 & -4 \ 0 & 4 \end{bmatrix} + x \begin{bmatrix} 2 & -1 \ 0 & 2 \end{bmatrix} + y \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Step $3$: Combine the matrices: $\begin{bmatrix} 4 + 2x + y & -4 - x \ 0 & 4 + 2x + y \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Step $4$: Equate corresponding elements: $-4 - x = 0 \implies x = -4$.Step $5$: Substitute $x = -4$ into $4 + 2x + y = 0$: $4 + 2(-4) + y = 0 \implies 4 - 8 + y = 0 \implies y = 4$.

If $A = \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix}$ and $A^2 + xA + yI_2 = O_2$, where $I_2$ and $O_2$ are the identity matrix and null matrix of order $2$ respectively, then:

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