Solve this

Solution:

To verify: $A^{2}+A=A(A+I)$,

Given: $A=\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$

$A^{2}=\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$

$=\left[\begin{array}{ccc}1+0+0 & 0+0-3 & -3+0-3 \\ 2+2+0 & 0+1+3 & -6+3+3 \\ 0+2+0 & 0+1+1 & 0+3+1\end{array}\right]$

$=\left[\begin{array}{ccc}1 & -3 & -6 \\ 4 & 4 & 0 \\ 2 & 2 & 4\end{array}\right]$

LHS:

$A^{2}+A=\left[\begin{array}{ccc}1 & -3 & -6 \\ 4 & 4 & 0 \\ 2 & 2 & 4\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$

$=\left[\begin{array}{ccc}1+1 & -3+0 & -6-3 \\ 4+2 & 4+1 & 0+3 \\ 2+0 & 2+1 & 4+1\end{array}\right]$

$=\left[\begin{array}{ccc}2 & -3 & -9 \\ 6 & 5 & 3 \\ 2 & 3 & 5\end{array}\right]$

$\mathrm{RHS}$.

$A+I=\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$=\left[\begin{array}{ccc}1+1 & 0+0 & -3+0 \\ 2+0 & 1+1 & 3+0 \\ 0+0 & 1+0 & 1+1\end{array}\right]$

$=\left[\begin{array}{ccc}2 & 0 & -3 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{array}\right]$

$A(A+I)=\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}2 & 0 & -3 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{array}\right]$

$\begin{aligned}=&\left[\begin{array}{ccc}2+0+0 & 0+0-3 & -3+0-6 \\ 4+2+0 & 0+2+3 & -6+3+6 \\ 0+2+0 & 0+2+1 & 0+3+2\end{array}\right] \\=&\left[\begin{array}{ccc}2 & -3 & -9 \\ 6 & 5 & 3 \\ 2 & 3 & 5\end{array}\right] \end{aligned}$

Therefore, LHS = RHS.

Hence, $A^{2}+A=A(A+l)$ is verified.