In Problems 17–19, find the inverse, if there is one, of each matrix. If there is not an inverse, say that the matrix is singular.

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

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

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

Transform the matrix $[A\mid I_3]$into reduced row echelon form.

Perform the operations $R_2=R_2-R_1$and then $R_3=R_3-R_1$,

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

Now, perform the operations $R_2=-1·R_2$and then $R_1=R_1-3R_2$,

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

Now, perform the operations $R_3=R_3+4R_2$and then $R_3=\frac{1}{7}{R}_3$,

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

Now, perform the operations $R_1=R_1+3R_3$and then $R_2=R_2-2R_3$,

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

Now, we can see that the identity matrix is on the left of the vertical bar.

Then the matrix on the right of the vertical bar is the inverse of $A$.

Therefore, the inverse of the given matrix is$\left[\begin{array}{ccc}-\frac{5}{7}& \frac{9}{7}& \frac{3}{7}\\ \frac{1}{7}& \frac{1}{7}& -\frac{2}{7}\\ \frac{3}{7}& -\frac{4}{7}& \frac{1}{7}\end{array}\right]$