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