Question

Find the inverse of each of the following matrices by using elementary row transformations:

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

Answer 1

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

$$\begin{gathered} \left[\begin{array}{ccc}1& 2& 3\\ 2& 5& 7\\ -2& -4& -5\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A \\ {R}_2\to R_2-2R_1 \\ \left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 1\\ -2& -4& -5\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 0& 0& 1\end{array}\right] \\ {R}_3\to R_3+2R_1 \\ \left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 2& 0& 1\end{array}\right] \\ {R}_2\to R_2-R_3 \\ \left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -4& 1& -1\\ 2& 0& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} R_1\to R_1-3R_3 \\ \left[\begin{array}{ccc}1& 2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-5& 0& -3\\ -4& 1& -1\\ 2& 0& 1\end{array}\right] \\ {R}_1\to R_1-2R_2 \\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}3& -2& -1\\ -4& 1& -1\\ 2& 0& 1\end{array}\right] \end{gathered}$$

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