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

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

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

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(B) Given that,$A=\begin{bmatrix} 1 & -2 \ 4 & 5 \end{bmatrix}$ 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 $f(A) = A^2 - 3A + 7I$:$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}$.Finally,add the given matrix:$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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