If A = begin{bmatrix} 3 & 1 -1 & 2 end{bmatr | vedclass

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(A) Given $A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$.First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} \begi

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(A) Given $A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$.First,calculate $A^{2} = A \cdot A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$.$A^{2} = \begin{bmatrix} 3(3) + 1(-1) & 3(1) + 1(2) \ -1(3) + 2(-1) & -1(1) + 2(2) \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix}$.Now,evaluate the expression $L.H.S. = A^{2} - 5A + 7I$.$L.H.S. = \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix} - 5 \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.$L.H.S. = \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \ 0 & 7 \end{bmatrix}$.$L.H.S. = \begin{bmatrix} 8-15+7 & 5-5+0 \ -5-(-5)+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} = 0$.Thus,$A^{2} - 5A + 7I = 0$ is proved.

If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$,show that $A^{2} - 5A + 7I = 0$.

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