Answer
$A^{-1}=\left[\begin{array}{rr}
1 & 2\\
-1/2 & 2/3
\end{array}\right]$
Work Step by Step
$A=\displaystyle \left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right] \Rightarrow A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{ll}
d & -b\\
-c & a
\end{array}\right]$
($A^{-1}$ exists if and only if $ad-bc\neq 0)$
----
$\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
0.4 & -1.2\\
0.3 & 0.6
\end{array}\right]$
$ad-bc=0.4(0.6)-(-1.2)(0.3)=0.6$
$\displaystyle \frac{1}{ad-bc}=\frac{1}{0.6}=\frac{5}{3}$
$A^{-1}=\displaystyle \frac{1}{0.6}\left[\begin{array}{ll}
0.6 & 1.2\\
-0.3 & 0.4
\end{array}\right]$
$A^{-1}=\left[\begin{array}{ll}
1 & 2\\
-1/2 & 2/3
\end{array}\right]$
