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