Question

Using elementary row transformations, find the inverse of the matrix $A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 5& 7\\ -2& -4& -5\end{array}\right]$

Answer 1

Let $A=IA$
$\left[\begin{array}{ccc}1& 2& 3\\ 2& 5& 7\\ -2& -4& -5\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A$

Operate $R_2\to R_2-2R_1$ and $R_3\to R_3+2R_1$

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

Operate $R_2\to R_2-R_3$

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

Operate $R_1\to R_1-3R_3$

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

Operate $R_1\to R_1-2R_2$

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

Which is of the form
$I=A^{-1}A$

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