91Ó°ÊÓ

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

We have to find the inverse of the given matrix.

$\left[Aâˆ\pounds I_3\right]=\left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 3& 2& -1& 0& 1& 0\\ 3& 1& 2& 0& 0& 1\end{array}\right]$

In order to transform the matrix $\left[A|I_3\right]$into reduced row echelon form first perform the row operations $R_2=r_2-3r_1$and then $R_3=r_3-3r_1$.

$$\begin{gathered} \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 3& 2& -1& 0& 1& 0\\ 3& 1& 2& 0& 0& 1\end{array}\right]→\left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& -1& -4& -3& 1& 0\\ 3& 1& 2& 0& 0& 1\end{array}\right] \\ →\left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& -1& -4& -3& 1& 0\\ 0& -2& -1& -3& 0& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& -1& -4& -3& 1& 0\\ 0& -2& -1& -3& 0& 1\end{array}\right]→\left[\begin{array}{cccccc}1& 1& 1& 1& 0& 0\\ 0& 1& 4& 3& -1& 0\\ 0& -2& -1& -3& 0& 1\end{array}\right] \\ →\left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& -2& -1& -3& 0& 1\end{array}\right] \end{gathered}$$

Perform $R_3=r_3+2r_2$and then $R_3=\frac{1}{7}{r}_3$.

$$\begin{gathered} \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& -2& -1& -3& 0& 1\end{array}\right]→\left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 7& 3& -2& 1\end{array}\right] \\ →\left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right] \end{gathered}$$

$$\begin{gathered} \left[\begin{array}{cccccc}1& 0& -3& -2& 1& 0\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]→\left[\begin{array}{cccccc}1& 0& 0& -\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ 0& 1& 4& 3& -1& 0\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right] \\ →\left[\begin{array}{cccccc}1& 0& 0& -\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ 0& 1& 0& \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ 0& 0& 1& \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right] \end{gathered}$$

We can see that the reduced row echelon form of $\left[A|I_3\right]$contains the identity matrix $I_3$on the left of the vertical bar.

The $3$by $3$matrix on the right side of the vertical bar is the inverse of $A$.

role="math" localid="1647445785318" $A^{-1}=\left[\begin{array}{ccc}-\frac{5}{7}& \frac{1}{7}& \frac{3}{7}\\ \frac{9}{7}& \frac{1}{7}& -\frac{4}{7}\\ \frac{3}{7}& -\frac{2}{7}& \frac{1}{7}\end{array}\right]$.