Answer
$A^{-1}=\left[\begin{array}{rrr}
{-\displaystyle \frac{5}{7}}&{\displaystyle \frac{1}{7}}&{\displaystyle \frac{3}{7}}\\\\
{\displaystyle \frac{9}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{4}{7}}\\\\
{\displaystyle \frac{3}{7}}&{-\displaystyle \frac{2}{7}}&{\displaystyle \frac{1}{7}}\end{array}\right]$
Work Step by Step
To find the inverse of an $n$ by $n$ nonsingular matrix $A,$ proceed as follows:
STEP 1: Form the matrix $\left[A|I_{n}\right]$
STEP 2: Transform the matrix $\left[A|I_{n}\right]$ into reduced row echelon form.
STEP 3: The reduced row echelon form of $\left[A|I_{n}\right]$ will contain
the identity matrix $I_{n}$ on the left of the vertical bar;
the $n$ by $n$ matrix on the right of the vertical bar is the inverse of $A.$
If the identity matrix can not be obtained on the left, A has no inverse.
----
STEP 1
$\left[A|I_{n}\right] = \left[\begin{array}{rrr|rrr}
{1}&{1}&{1}&{1}&{0}&{0}\\
{3}&{2}&{-1}&{0}&{1}&{0}\\
{3}&{1}&{2}&{0}&{0}&{1}\end{array}\right] \qquad \left(\begin{array}{l}
.\\
R_{2}=r_{2}-3r_{1}.\\
R_{3}=r_{3}-3r_{1}.
\end{array}\right)$
STEP 2
$\rightarrow\left[\begin{array}{rrr|rrr}
{1}&{1}&{1}&{1}&{0}&{0}\\
{0}&{-1}&{-4}&{-3}&{1}&{0}\\
{0}&{-2}&{-1}&{-3}&{0}&{1}\end{array}\right]\qquad \left(\begin{array}{l}
R_{1}=r_{1}+r_{2}.\\
R_{2}=-r_{2}.\\
R_{3}=r_{3}-2r_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrr|rrr}
{1}&{0} &{-3}&{-2}&{1}&{0}\\
{0}&{1} &{4} & {3}&{-1}&{0}\\
{0}&{0} &{7} & {3} &{-2}&{1}\end{array}\right]\qquad\left(\begin{array}{l}
.\\
.\\
\div 7.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrr|rrr}
{1}&{0}&{-3}&{-2}&{1}&{0}\\
{0}&{1}&{4}&{3}&{-1}&{0}\\
{0}&{0}&{1}&{\displaystyle \frac{3}{7}}&{-\displaystyle \frac{2}{7}}&{\displaystyle \frac{1}{7}}\end{array}\right] \left(\begin{array}{l}
R_{1}=r_{1}+3r_{3}.\\
R_{2}=-3r_{3}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{ccc|ccc}
{1}&{0}&{0}&{-\displaystyle \frac{5}{7}}&{\displaystyle \frac{1}{7}}&{\displaystyle \frac{3}{7}}\\\\
{0}&{1}&{0}&{\displaystyle \frac{9}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{4}{7}}\\\\
{0}&{0}&{1}&{\displaystyle \frac{3}{7}}&{-\displaystyle \frac{2}{7}}&{\displaystyle \frac{1}{7}}\end{array}\right]$
STEP 3
$A^{-1}=\left[\begin{array}{rrr}
{-\displaystyle \frac{5}{7}}&{\displaystyle \frac{1}{7}}&{\displaystyle \frac{3}{7}}\\
{\displaystyle \frac{9}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{4}{7}}\\
{\displaystyle \frac{3}{7}}&{-\displaystyle \frac{2}{7}}&{\displaystyle \frac{1}{7}}\end{array}\right]$
