Let A = begin{bmatrix} 2 & -5 3 & 1 end{bmat | vedclass

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(C) Given $f(x) = x^3 - 2x^2 - 5$. Therefore,$f(A) = A^3 - 2A^2 - 5I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.

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$\begin{bmatrix} -50 & 70 \ -42 & -36 \end{bmatrix}$

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(C) Given $f(x) = x^3 - 2x^2 - 5$. Therefore,$f(A) = A^3 - 2A^2 - 5I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.First,calculate $A^2 = A \cdot A = \begin{bmatrix} 2 & -5 \ 3 & 1 \end{bmatrix} \begin{bmatrix} 2 & -5 \ 3 & 1 \end{bmatrix} = \begin{bmatrix} 4-15 & -10-5 \ 6+3 & -15+1 \end{bmatrix} = \begin{bmatrix} -11 & -15 \ 9 & -14 \end{bmatrix}$.Next,calculate $A^3 = A \cdot A^2 = \begin{bmatrix} 2 & -5 \ 3 & 1 \end{bmatrix} \begin{bmatrix} -11 & -15 \ 9 & -14 \end{bmatrix} = \begin{bmatrix} -22-45 & -30+70 \ -33+9 & -45-14 \end{bmatrix} = \begin{bmatrix} -67 & 40 \ -24 & -59 \end{bmatrix}$.Now,substitute these into the expression for $f(A)$:$f(A) = \begin{bmatrix} -67 & 40 \ -24 & -59 \end{bmatrix} - 2 \begin{bmatrix} -11 & -15 \ 9 & -14 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$$f(A) = \begin{bmatrix} -67 & 40 \ -24 & -59 \end{bmatrix} - \begin{bmatrix} -22 & -30 \ 18 & -28 \end{bmatrix} - \begin{bmatrix} 5 & 0 \ 0 & 5 \end{bmatrix}$$f(A) = \begin{bmatrix} -67 - (-22) - 5 & 40 - (-30) - 0 \ -24 - 18 - 0 & -59 - (-28) - 5 \end{bmatrix} = \begin{bmatrix} -50 & 70 \ -42 & -36 \end{bmatrix}$.

Let $A = \begin{bmatrix} 2 & -5 \\ 3 & 1 \end{bmatrix}$. What is $f(A)$ if $f(x) = x^3 - 2x^2 - 5$?

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