Answer
$A^{-1}=\left[\begin{array}{rr}
{1}&{-5/2}\\
{-1}&{3}\end{array}\right]$
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
$\left[A|I_{n}\right] =\left[\begin{array}{cc|cc}{6}&{5}&{1}&{0}\\{2}&{2}&{0}&{1}\end{array}\right]\qquad \left(\begin{array}{l}
r_{1}\leftrightarrow r_{2}.\\
.
\end{array}\right)$
STEP 2
$\left[\begin{array}{cc|cc}{2}&{2}&{0}&{1}\\{6}&{5}&{1}&{0}\end{array}\right]\qquad \left(\begin{array}{l}
.\\
R_{2}=r_{2}-3r_{1}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rr|rr}
{2}&{2}&{0}&{1}\\
{0}&{-1}&{1}&{-3}\end{array}\right]\qquad \left(\begin{array}{l}
R_{1}=r_{1}+2r_{2}.\\
\times(-1).
\end{array}\right)$
$\rightarrow\left[\begin{array}{rr|rr}
{2}&{0}& {2}&{-5}\\
{0}&{1}&{-1}&{ 3}\end{array}\right]\qquad \left(\begin{array}{l}
\div 2.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rr|rr}
{1}&{0}& {1}&{-5/2}\\
{0}&{1}&{-1}&{ 3}\end{array}\right]$
STEP 3
$A^{-1}=\left[\begin{array}{rr}
{1}&{-5/2}\\
{-1}&{3}\end{array}\right]$
