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

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(A) Given $A = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix}$.First,we calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmat

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$4A - 3I$

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(A) Given $A = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix}$.First,we calculate $A^2 = A 	imes A$:$A^2 = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} (2)(2) + (-1)(-1) & (2)(-1) + (-1)(2) \ (-1)(2) + (2)(-1) & (-1)(-1) + (2)(2) \end{bmatrix} = \begin{bmatrix} 5 & -4 \ -4 & 5 \end{bmatrix}$.Now,we evaluate the options. Let's check option $A$ $(4A - 3I)$:$4A - 3I = 4 \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 8 & -4 \ -4 & 8 \end{bmatrix} - \begin{bmatrix} 3 & 0 \ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -4 \ -4 & 5 \end{bmatrix}$.Since $A^2 = 4A - 3I$,the correct option is $A$.

If $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$ and $I$ is the unit matrix of order $2$,then $A^2$ equals

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