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

Solution:

$A=\mid A$

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

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

$R_{2} \rightarrow R_{2}-2 R_{1}$

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

$R_{3} \rightarrow R_{3}+2 R_{1}$

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

$R_{2} \rightarrow R_{2}-R_{3}$

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

$R_{1} \rightarrow R_{1}-3 R_{3}$

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

$R_{1} \rightarrow R_{1}-2 R_{2}$

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

Therefore,

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