Question

$\left(\begin{array}{ccc}1& 3& -2\\ -3& 0& 1\\ 2& 1& 0\end{array}\right)$
Find Inverse Elementary transformation

Answer 1

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

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

swap matrices R1 and R2

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

$R_2\to R_2+\frac{1}{3}{R}_1$

$=\left[\begin{array}{ccccccc}-3& 0& 1& \mid & 0& 1& 0\\ 0& 3& -\frac{5}{3}& \mid & 1& \frac{1}{3}& 0\\ 2& 1& 0& \mid & 0& 0& 1\end{array}\right]$

$R_3\leftarrow R_3+\frac{2}{3}\cdot R_1$

$=\left[\begin{array}{ccccccc}-3& 0& 1& \mid & 0& 1& 0\\ 0& 3& -\frac{5}{3}& \mid & 1& \frac{1}{3}& 0\\ 0& 1& \frac{2}{3}& \mid & 0& \frac{2}{3}& 1\end{array}\right]$

$R_3\leftarrow R_3-\frac{1}{3}\cdot R_2$

$=\left[\begin{array}{ccccccc}-3& 0& 1& \mid & 0& 1& 0\\ 0& 3& -\frac{5}{3}& \mid & 1& \frac{1}{3}& 0\\ 0& 0& \frac{11}{9}& \mid & -\frac{1}{3}& \frac{5}{9}& 1\end{array}\right]$

$R_3\leftarrow \frac{9}{11}\cdot R_3$

$=\left[\begin{array}{ccccccc}-3& 0& 1& \mid & 0& 1& 0\\ 0& 3& -\frac{5}{3}& \mid & 1& \frac{1}{3}& 0\\ 0& 0& 1& \mid & -\frac{3}{11}& \frac{5}{11}& \frac{9}{11}\end{array}\right]$

$R_2\leftarrow R_2+\frac{5}{3}\cdot R_3$

$=\left[\begin{array}{ccccccc}-3& 0& 1& \mid & 0& 1& 0\\ 0& 3& 0& \mid & \frac{6}{11}& \frac{12}{11}& \frac{15}{11}\\ 0& 0& 1& \mid & -\frac{3}{11}& \frac{5}{11}& \frac{9}{11}\end{array}\right]$

$R_1\leftarrow R_1-1\cdot R_3$

$=\left[\begin{array}{ccccccc}-3& 0& 0& \mid & \frac{3}{11}& \frac{6}{11}& -\frac{9}{11}\\ 0& 3& 0& \mid & \frac{6}{11}& \frac{12}{11}& \frac{15}{11}\\ 0& 0& 1& \mid & -\frac{3}{11}& \frac{5}{11}& \frac{9}{11}\end{array}\right]$

$R_2\leftarrow \frac{1}{3}\cdot R_2$

$=\left[\begin{array}{ccccccc}-3& 0& 0& \mid & \frac{3}{11}& \frac{6}{11}& -\frac{9}{11}\\ 0& 1& 0& \mid & \frac{2}{11}& \frac{4}{11}& \frac{5}{11}\\ 0& 0& 1& \mid & -\frac{3}{11}& \frac{5}{11}& \frac{9}{11}\end{array}\right]$

$R_1\leftarrow -\frac{1}{3}\cdot R_1$

$=\left[\begin{array}{ccccccc}1& 0& 0& \mid & -\frac{1}{11}& -\frac{2}{11}& \frac{3}{11}\\ 0& 1& 0& \mid & \frac{2}{11}& \frac{4}{11}& \frac{5}{11}\\ 0& 0& 1& \mid & -\frac{3}{11}& \frac{5}{11}& \frac{9}{11}\end{array}\right]$

$=\left(\begin{array}{ccc}-\frac{1}{11}& -\frac{2}{11}& \frac{3}{11}\\ \frac{2}{11}& \frac{4}{11}& \frac{5}{11}\\ -\frac{3}{11}& \frac{5}{11}& \frac{9}{11}\end{array}\right)$