If A = begin{bmatrix} 0 & 3 0 & 0 end{bmatri | vedclass

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(C) Given $A = \begin{bmatrix} 0 & 3 \ 0 & 0 \end{bmatrix}$.First, calculate $A^2$:$A^2 = \begin{bmatrix} 0 & 3 \ 0 & 0 \end{bmatrix} \begin{bmatrix

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$\begin{bmatrix} 1 & 3 \ 0 & 1 \end{bmatrix}$

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(C) Given $A = \begin{bmatrix} 0 & 3 \ 0 & 0 \end{bmatrix}$.First, calculate $A^2$:$A^2 = \begin{bmatrix} 0 & 3 \ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 3 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Since $A^2 = 0$, all higher powers $A^n = 0$ for $n \geq 2$.Given $f(x) = x + x^2 + x^3 + \ldots + x^{2023}$, we have $f(A) = A + A^2 + A^3 + \ldots + A^{2023}$.Substituting $A^n = 0$ for $n \geq 2$, we get $f(A) = A + 0 + 0 + \ldots + 0 = A$.Therefore, $f(A) + I = A + I$.$f(A) + I = \begin{bmatrix} 0 & 3 \ 0 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \ 0 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix}$ and $f(x) = x + x^2 + x^3 + \ldots + x^{2023}$, then $f(A) + I = $

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