If A = begin{bmatrix} 1 & -2 4 & 5 end{bmatr | vedclass

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(B) Given $A = \begin{bmatrix} 1 & -2 \ 4 & 5 \end{bmatrix}$ and $f(t) = t^2 - 3t + 7$.First, calculate $A^2$:$A^2 = \begin{bmatrix} 1 & -2 \ 4 & 5

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$\begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$

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(B) Given $A = \begin{bmatrix} 1 & -2 \ 4 & 5 \end{bmatrix}$ and $f(t) = t^2 - 3t + 7$.First, calculate $A^2$:$A^2 = \begin{bmatrix} 1 & -2 \ 4 & 5 \end{bmatrix} \begin{bmatrix} 1 & -2 \ 4 & 5 \end{bmatrix} = \begin{bmatrix} 1(1) + (-2)(4) & 1(-2) + (-2)(5) \ 4(1) + 5(4) & 4(-2) + 5(5) \end{bmatrix} = \begin{bmatrix} -7 & -12 \ 24 & 17 \end{bmatrix}$.Now, calculate $f(A) = A^2 - 3A + 7I$, where $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$:$f(A) = \begin{bmatrix} -7 & -12 \ 24 & 17 \end{bmatrix} - 3 \begin{bmatrix} 1 & -2 \ 4 & 5 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$$f(A) = \begin{bmatrix} -7 & -12 \ 24 & 17 \end{bmatrix} - \begin{bmatrix} 3 & -6 \ 12 & 15 \end{bmatrix} + \begin{bmatrix} 7 & 0 \ 0 & 7 \end{bmatrix}$$f(A) = \begin{bmatrix} -7-3+7 & -12+6+0 \ 24-12+0 & 17-15+7 \end{bmatrix} = \begin{bmatrix} -3 & -6 \ 12 & 9 \end{bmatrix}$.Finally, calculate $f(A) + \begin{bmatrix} 3 & 6 \ -12 & -9 \end{bmatrix}$:$\begin{bmatrix} -3 & -6 \ 12 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \ -12 & -9 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & -2 \\ 4 & 5 \end{bmatrix}$ and $f(t) = t^2 - 3t + 7$, then $f(A) + \begin{bmatrix} 3 & 6 \\ -12 & -9 \end{bmatrix}$ is equal to

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