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If $\mathrm{A}=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$, then verify that $\mathrm{A}^2+\mathrm{A}$ $=\mathrm{A}(\mathrm{A}+\mathrm{I})$, where $\mathrm{I}$ is $3 \times 3$ unit matrix.
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Verified Answer
We have, $\mathrm{A}=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$
$\therefore \mathrm{A}^2=\mathrm{A} \cdot \mathrm{A}$
$=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4\end{array}\right]$
$\therefore \quad \mathrm{A}^2+\mathrm{A}=\left[\begin{array}{ccc}2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5\end{array}\right]$
Now, $\mathrm{A}+\mathrm{I}=\left[\begin{array}{ccc}2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{array}\right]$
$$
\text { and } \begin{aligned}
A(A+I) &=\left[\begin{array}{ccc}
1 & 0 & -1 \\
2 & 1 & 3 \\
0 & 1 & 1
\end{array}\right] \cdot\left[\begin{array}{ccc}
2 & 0 & -1 \\
2 & 2 & 3 \\
0 & 1 & 2
\end{array}\right] \\
&=\left[\begin{array}{ccc}
2 & -1 & -3 \\
6 & 5 & 7 \\
2 & 3 & 5
\end{array}\right]
\end{aligned}
$$
Thus, we see that $A^2+A=A(A+I)$
[using Eqs. (i) and (ii)]
$\therefore \mathrm{A}^2=\mathrm{A} \cdot \mathrm{A}$
$=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4\end{array}\right]$
$\therefore \quad \mathrm{A}^2+\mathrm{A}=\left[\begin{array}{ccc}2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5\end{array}\right]$
Now, $\mathrm{A}+\mathrm{I}=\left[\begin{array}{ccc}2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{array}\right]$
$$
\text { and } \begin{aligned}
A(A+I) &=\left[\begin{array}{ccc}
1 & 0 & -1 \\
2 & 1 & 3 \\
0 & 1 & 1
\end{array}\right] \cdot\left[\begin{array}{ccc}
2 & 0 & -1 \\
2 & 2 & 3 \\
0 & 1 & 2
\end{array}\right] \\
&=\left[\begin{array}{ccc}
2 & -1 & -3 \\
6 & 5 & 7 \\
2 & 3 & 5
\end{array}\right]
\end{aligned}
$$
Thus, we see that $A^2+A=A(A+I)$
[using Eqs. (i) and (ii)]
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