Answer
$A^{-1}=\left[\begin{array}{rrr}
{3}&{-3}&{1}\\
{-2}&{2}&{-1}\\
{-4}&{5}&{-2}\end{array}\right]$
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
$\left[A|I_{n}\right] = \left[\begin{array}{rrr|rrr}
{1}&{-1}&{1}&{1}&{0}&{0}\\
{0}&{-2}&{1}&{0}&{1}&{0}\\
{-2}&{-3}&{0}&{0}&{0}&{1}\end{array}\right] \qquad \left(\begin{array}{l}
.\\
.\\
R_{3}=r_{3}+2r_{1}.
\end{array}\right)$
STEP 2
$\rightarrow\left[\begin{array}{rrr|rrr}
1 & -1 & 1 & 1 & 0 & 0\\
0 & -2 & 1 & 0 & 1 & 0\\
0 & -5 & 2 & 2 & 0 & 1
\end{array}\right]\qquad \left(\begin{array}{l}
.\\
R_{2}=2r_{2}-r_{3}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrr|rrr}
1 & -1 & 1 & 1 & 0 & 0\\
0 & 1 & 0 & -2 & 2 & -1\\
0 & -5 & 2 & 2 & 0 & 1
\end{array}\right]\qquad \left(\begin{array}{l}
R_{1}=r_{1}+r_{2}.\\
.\\
R_{3}=R_{3}+5r_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrr|rrr}
1 & 0 & 1 & -1 & 2 & -1\\
0 & 1 & 0 & -2 & 2 & -1\\
0 & 0 & 2 & -8 & 10 & -4
\end{array}\right] \qquad \left(\begin{array}{l}
.\\
.\\
\div 2.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrr|rrr}
1 & 0 & 1 & -1 & 2 & -1\\
0 & 1 & 0 & -2 & 2 & -1\\
0 & 0 & 1 & -4 & 5 & -2
\end{array}\right]\qquad \left(\begin{array}{l}
R_{1}=r_{1}-r_{3}.\\
.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 3 & -3 & 1\\
0 & 1 & 0 & -2 & 2 & -1\\
0 & 0 & 1 & -4 & 5 & -2
\end{array}\right]$
STEP 3
$A^{-1}=\left[\begin{array}{rrr}
{3}&{-3}&{1}\\
{-2}&{2}&{-1}\\
{-4}&{5}&{-2}\end{array}\right]$
