Answer
$A^{-1}=\left[\begin{array}{ll}
{-1}&{-1/2}\\
{-3}&{-2}\end{array}\right]$
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
$\left[A|I_{n}\right] =\left[\begin{array}{rr|rr}
{-4}&{1}&{1}&{0}\\
{6}&{-2}&{0}&{1}\end{array}\right]\left(\begin{array}{l}
R_{1}=r_{1}+r_{2}.\\
.
\end{array}\right)$
STEP 2
$\rightarrow \left[\begin{array}{rr|rr}
{2}& {-1}&{1}&{1}\\
{6}&{-2}& {0}&{1}\end{array}\right] \qquad \left(\begin{array}{l}
.\\
R_{2}=r_{2}-3r_{1}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rr|rr}
{2}& {-1}&{1} &{1}\\
{0}&{ 1}& {-3}&{-2}\end{array}\right] \qquad \left(\begin{array}{l}
R_{1}=r_{1}+r_{2}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rr|rr}
{2}& {0}& {-2} &{-1}\\
{0}&{ 1} & {-3}&{-2}\end{array}\right] \qquad\left(\begin{array}{l}
\div 2.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{cc|cc}
{1}&{0}&{-1}&{-\displaystyle \frac{1}{2}}\\
{0}&{1}&{-3}&{-2}\end{array}\right]$
STEP 3
$A^{-1}=\left[\begin{array}{ll}
{-1}&{-1/2}\\
{-3}&{-2}\end{array}\right]$