Question

If $A=\left[\begin{array}{ccc}1& 0& -1\\ 2& 1& 3\\ 0& 1& 1\end{array}\right]$, then verify that $A^2+A=A(A+I)$, where I is $3\times 3$ unit matrix.

Answer 1

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

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

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

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

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

$\Rightarrow A^2+A=\left[\begin{array}{ccc}2& -1& -3\\ 6& 5& 7\\ 2& 3& 5\end{array}\right]$ ⋯(i)

Finding $A(A+I)$, where I is $3\times 3$ unit matrix.

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

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

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

$\Rightarrow A\cdot (A+I)=\left[\begin{array}{ccc}2& -1& -3\\ 6& 5& 7\\ 2& 3& 5\end{array}\right]$ ⋯(ii)

From equation (i) and (ii), we get

$A^2+A=A(A+I)$

Hence, verified.