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