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If $A=\left[\begin{array}{ll}0 & 5 \\ 0 & 0\end{array}\right]$ and $f(x)=x+x^2+\ldots+x^{2018}$, then $f(A)+I=$
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Verified Answer
The correct answer is:
$\left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]$
We have,
$$
\begin{aligned}
A & =\left[\begin{array}{ll}
0 & 5 \\
0 & 0
\end{array}\right] \\
A^2 & =\left[\begin{array}{ll}
0 & 5 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 5 \\
0 & 0
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$$
Similarly. $A^3=A^4=\ldots . A^n=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$$
\begin{aligned}
\therefore \quad f(A) & =A+A^2+A^3+\ldots+A^{2018} \\
f(A) & =A+0+0+\ldots+0=A \\
f(A)+I & =\left[\begin{array}{ll}
0 & 5 \\
0 & 0
\end{array}\right]+\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 5 \\
0 & 1
\end{array}\right]
\end{aligned}
$$
$$
\begin{aligned}
A & =\left[\begin{array}{ll}
0 & 5 \\
0 & 0
\end{array}\right] \\
A^2 & =\left[\begin{array}{ll}
0 & 5 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 5 \\
0 & 0
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$$
Similarly. $A^3=A^4=\ldots . A^n=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$$
\begin{aligned}
\therefore \quad f(A) & =A+A^2+A^3+\ldots+A^{2018} \\
f(A) & =A+0+0+\ldots+0=A \\
f(A)+I & =\left[\begin{array}{ll}
0 & 5 \\
0 & 0
\end{array}\right]+\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 5 \\
0 & 1
\end{array}\right]
\end{aligned}
$$
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