Using elementary row transformations, find the inverse of each of the following matrices: $\left[\begin{array}{ccc} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{array}\right]$

Solution:

We have $A=\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{array}\right)$
To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:
$(A \mid I) \rightarrow\left(I \mid A^{-1}\right)$
Let’s proceed
$\begin{array}{l} \left(\begin{array}{ccc|ccc} 3 & 0 & -1 & 1 & 0 & 0 \\ 2 & 3 & 0 & 0 & 1 & 0 \\ 0 & 4 & 1 & 0 & 0 & 1 \end{array}\right) \stackrel{R_{1} \rightarrow R_{1}-R_{2}}{\longrightarrow}\left(\begin{array}{ccc|ccc} 1 & -3 & -1 & 1 & -1 & 0 \\ 2 & 3 & 0 & 0 & 1 & 0 \\ 0 & 4 & 1 & 0 & 0 & 1 \end{array}\right) \\ \stackrel{R_{2} \rightarrow R_{2}-2 R_{1}}{\longrightarrow}\left(\begin{array}{ccc|ccc} 1 & -3 & -1 & 1 & -1 & 0 \\ 0 & 9 & 2 & -2 & 3 & 0 \\ 0 & 4 & 1 & 0 & 0 & 1 \end{array}\right) \end{array}$
$\begin{array}{l} \underline{R_{2} \rightarrow R_{2}-2 R_{3}}\left(\begin{array}{ccc|ccc} 1 & -3 & -1 & 1 & -1 & 0 \\ 0 & 1 & 0 & -2 & 3 & -2 \\ 0 & 4 & 1 & 0 & 0 & 1 \end{array}\right) \\ \stackrel{R_{3} \rightarrow R_{3}-4 R_{2}}{\longrightarrow}\left(\begin{array}{ccc|ccc} 1 & -3 & -1 & 1 & -1 & 0 \\ 0 & 1 & 0 & -2 & 3 & -2 \\ 0 & 0 & 1 & 8 & -12 & 9 \end{array}\right) \end{array}$
$\begin{array}{l} \stackrel{R_{1} \rightarrow R_{1}+R_{3}}{\longrightarrow}\left(\begin{array}{ccc|ccc} 1 & -3 & 0 & 9 & -13 & 9 \\ 0 & 1 & 0 & -2 & 3 & -2 \\ 0 & 0 & 1 & 8 & -12 & 9 \end{array}\right) \\ \stackrel{R_{1} \rightarrow R_{1}+3 R_{2}}{\longrightarrow}\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 3 & -4 & 3 \\ 0 & 1 & 0 & -2 & 3 & -2 \\ 0 & 0 & 1 & 8 & -12 & 9 \end{array}\right) \end{array}$
Therefore we have the inverse of the matrix as
$A^{-1}=\left(\begin{array}{ccc} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{array}\right)$