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If A=[31−12]A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, then A2−5A+7I=A^2 - 5A + 7I =

Solution

✅ Correct Option: 1

A2=[85−53]A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}, 5A=[155−510]5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}, 7I=[7007]7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}. A2−5A+7I=[0000]=OA^2 - 5A + 7I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O.

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