If A = begin{bmatrix} 0 & 5 0 & 0 end{bmatri | vedclass

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(D) Given $A = \begin{bmatrix} 0 & 5 \ 0 & 0 \end{bmatrix}$.First, calculate $A^2$:$A^2 = \begin{bmatrix} 0 & 5 \ 0 & 0 \end{bmatrix} \begin{bmatrix

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$\begin{bmatrix} 1 & 5 \ 0 & 1 \end{bmatrix}$

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(D) Given $A = \begin{bmatrix} 0 & 5 \ 0 & 0 \end{bmatrix}$.First, calculate $A^2$:$A^2 = \begin{bmatrix} 0 & 5 \ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 5 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Since $A^2 = O$ (the zero matrix), all higher powers of $A$ will also be zero matrices, i.e.,$A^n = O$ for all $n \geq 2$.Given $f(x) = x + x^2 + \dots + x^{2018}$, we have:$f(A) = A + A^2 + A^3 + \dots + A^{2018}$.Substituting the powers of $A$:$f(A) = A + O + O + \dots + O = A$.Now, calculate $f(A) + I$:$f(A) + I = A + I = \begin{bmatrix} 0 & 5 \ 0 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 5 \ 0 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix}$ and $f(x) = x + x^2 + \dots + x^{2018}$, then $f(A) + I =$

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