Answer
$A^{-1}=\left[\begin{array}{rr} {1}&{-1}\\
{-1}&{2}\end{array}\right]$
Work Step by Step
In order to calculate the inverse of an $n$ by $n$ non-singular matrix $A$, we will proceed with the following steps:
Step -1. Transform the given matrix $A$ into this form $\left[A|I_{n}\right]$ as follows:
$\left[A|I_{n}\right]$ = $\left[\begin{array}{ll|ll}
{2}&{1}&{1}&{0}\\
{1}&{1}&{0}&{1}\end{array}\right]$
Step-2. Transform the matrix $\left[A|I_{n}\right]$ into reduced row-echelon form by using the row operations: $r_{1}\leftrightarrow r_{2}$ and $R_{2}=r_{2}-2r_{1}$
$\left[\begin{array}{rr|rr}
{1}&{1}&{0}&{1}\\
{2}&{1}&{1}&{0}\end{array}\right]$
Multiply the above matrix by $-1$:
$\left[\begin{array}{ll|ll}
{1}&{1}&{0}&{1}\\
{0}&{-1}&{1}&{-2}\end{array}\right]$
Now, use the row operations: $R_{1}=r_{1}-r_{2}$
$\rightarrow\left[\begin{array}{cc|cc}{1}&{0}&{1}&{-1}\\{0}&{1}&{-1}&{2}\end{array}\right]$
Step-3: The reduced row-echelon form of $\left[A|I_{n}\right]$ will be represented as the identity matrix $I_{n}$ on the left of the vertical line and the $n$ by $n$ matrix on the right of the vertical line is the inverse of $A$.
Thus, we have:
$A^{-1}=\left[\begin{array}{rr} {1}&{-1}\\
{-1}&{2}\end{array}\right]$
(Note that if the identity matrix does not occur on the left, then the matrix $A$ does not have an inverse).