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