Question

If $A=\left[\begin{array}{ccc}1& 0& -1\\ 2& 1& -1\\ 2& 3& 2\end{array}\right]$, then find $A^{-1}$ by using elementery row transformations and hence find matrix $B$ such that $A^2+A+I=BA$.

Answer 1

Lets take the given matrix
$A=\left[\begin{array}{ccc}1& 0& -1\\ 2& 1& -1\\ 2& 3& 2\end{array}\right]$

We know that $A{A}^{-1}=I$

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

Perform $R_2\to R_2-2R_1$ and $R_3\to R_3-2R_1$

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

Perform $R_3\to R_3-3R_2$

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

Perform $R_1\to R_1+R_3$ and $R_2\to R_2-R_3$

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}5& -3& 1\\ -6& 4& -1\\ 4& -3& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}I\end{array}\right]\left[A^{-1}\right]=\left[\begin{array}{ccc}5& -3& 1\\ -6& 4& -1\\ 4& -3& 1\end{array}\right]$

$\therefore A^{-1}=\left[\begin{array}{ccc}5& -3& 1\\ -6& 4& -1\\ 4& -3& 1\end{array}\right]$

Now, to find matrix $B$, multiply $A^2+A+I=BA$ by $A^{-1}$ .
$\Rightarrow A(A.A^{-1})+A.A^{-1}+I.A^{-1}=B.A.A^{-1}$
$\Rightarrow A+I+A^{-1}=B(\because A{A}^{-1}=I)$
$\Rightarrow B=\left[\begin{array}{ccc}1& 0& -1\\ 2& 1& -1\\ 2& 3& 2\end{array}\right]+\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]+\left[\begin{array}{ccc}5& -3& 1\\ -6& 4& -1\\ 4& -3& 1\end{array}\right]$
$\Rightarrow B=\left[\begin{array}{ccc}7& -3& 0\\ -4& 6& -2\\ 6& 0& 4\end{array}\right]$