# Solve this

Question:

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

Solution:

We have, $A=\left[\begin{array}{lll}-1 & 2 & 0\end{array}\right.$

$\begin{array}{lll}-1 & 1 & 1\end{array}$

$\left.\begin{array}{lll}0 & 1 & 0\end{array}\right]$

Now,

$A^{2}=\left[\begin{array}{lll}-1 & 2 & 0\end{array}\right.$

$\begin{array}{lll}-1 & 1 & 1\end{array}$

$\left.\begin{array}{lll}0 & 1 & 0\end{array}\right]\left[\begin{array}{lll}-1 & 2 & 0\end{array}\right.$

$-1 \quad 1 \quad 1$

$0 \quad 1 \quad 0]=\left[\begin{array}{lll}1-2+0 & -2+2+0 & 0+2+0\end{array}\right.$

$\begin{array}{lll}1-1+0 & -2+1+1 & 0+1+0\end{array}$

$0-1+0 \quad 0+1+0 \quad 0+1+0]=\left[\begin{array}{lll}-1 & 0 & 2\end{array}\right.$

$\begin{array}{lll}0 & 0 & 1\end{array}$

$\left.\begin{array}{lll}-1 & 1 & 1\end{array}\right]$

and $A^{2} \times A=\left[\begin{array}{lll}-1 & 0 & 2\end{array}\right.$

$\begin{array}{lll}0 & 0 & 1\end{array}$

$\left.\begin{array}{lll}-1 & 1 & 1\end{array}\right]\left[\begin{array}{lll}-1 & 2 & 0\end{array}\right.$

$\begin{array}{lll}-1 & 1 & 1\end{array}$

$\left.\begin{array}{lll}0 & 1 & 0\end{array}\right]=\left[\begin{array}{ll}1+0+0 & -2+0+2 & 0\end{array}\right.$

$\begin{array}{lll}0+0+0 & 0+0+1 & 0\end{array}$

$0 \quad-2+1+1 \quad 1]=\left[\begin{array}{lll}1 & 0 & 0\end{array}\right.$

$\begin{array}{lll}0 & 1 & 0\end{array}$

$\left.\begin{array}{rll}0 & 0 & 1\end{array}\right]=I_{3} \quad$ [Identity matrix of order 3]

$\Rightarrow A^{2} \times A=I_{3}$

$\Rightarrow A^{2}=A^{-1}$