If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, then the value of $A^2 - 5A + 6I$ is

$A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

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$A^2 = A \times A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

Position (1,1): $(3)(3) + (1)(-1) = 9 - 1 = 8$

Position (1,2): $(3)(1) + (1)(2) = 3 + 2 = 5$

Position (2,1): $(-1)(3) + (2)(-1) = -3 - 2 = -5$

Position (2,2): $(-1)(1) + (2)(2) = -1 + 4 = 3$

$A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

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$5A = 5 \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$

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$6I = 6 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$6I = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$

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$A^2 - 5A + 6I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$

Position (1,1): $8 - 15 + 6 = -1$

Position (1,2): $5 - 5 + 0 = 0$

Position (2,1): $-5 - (-5) + 0 = 0$

Position (2,2): $3 - 10 + 6 = -1$

$A^2 - 5A + 6I = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$