#### Answer

$A^{-1}=\left[\begin{array}{rr} {1}&{-5/2}\\
{-1}&{3}\end{array}\right]$

#### 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: $\left[A|I_{n}\right]$ = $\left[\begin{array}{ll|ll}
{6}&{5}&{1}&{0}\\
{2}&{2}&{0}&{1}\end{array}\right]$
Step-2. Transform the matrix $\left[A|I_{n}\right]$ into reduced row echelon form by using Row operations: $r_{1}\leftrightarrow r_{2}$ and $R_{1}=r_{2}-3r_{1}$
$\left[\begin{array}{rr|rr}
{2}&{2}&{0}&{1}\\
{0}&{-1}&{1}&{-3}\end{array}\right]$
Now, use Row operations: $R_{1}=r_{1}+2r_{2}$ and then multiply the Row-2 by $-1$ to obtain:
$\left[\begin{array}{cc|cc}{2}&{0}&{2}&{-5}\\{0}&{1}&{-1}&{3}\end{array}\right]$
Finally divide Row-1 by $2$ to obtain: $\left[\begin{array}{cc|cc}{1}&{0}&{1}&{-5/2}\\{0}&{1}&{-1}&{3}\end{array}\right]$
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: $A^{-1}=\left[\begin{array}{rr} {1}&{-5/2}\\
{-1}&{3}\end{array}\right]$
(Note that if the identity matrix does not obtain on the left, then the matrix $A$ does have no inverse).