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If $A=\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]$ and $f(x)=x+x^2+x^3+\ldots . .+x^{2023}$ then $\mathrm{f}(\mathrm{A})+\mathrm{I}=$
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$\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$
$A=\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]$
$\begin{aligned} & \mathrm{A}^2=\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \\ & \therefore \mathrm{A}^n=0, n \geq 2 \\ & f(\mathrm{~A})=\mathrm{A}+\mathrm{A}^2+\mathrm{A}^3+\ldots .+\mathrm{A}^{2023} \\ & f(\mathrm{~A})=\mathrm{A} \\ & f(\mathrm{~A})+\mathrm{I}=\mathrm{A}+\mathrm{I} \\ & =\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]\end{aligned}$
$\begin{aligned} & \mathrm{A}^2=\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \\ & \therefore \mathrm{A}^n=0, n \geq 2 \\ & f(\mathrm{~A})=\mathrm{A}+\mathrm{A}^2+\mathrm{A}^3+\ldots .+\mathrm{A}^{2023} \\ & f(\mathrm{~A})=\mathrm{A} \\ & f(\mathrm{~A})+\mathrm{I}=\mathrm{A}+\mathrm{I} \\ & =\left[\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]\end{aligned}$
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