Answer
$$A=\left[ \begin {array}{cc} \frac{3}{42}&\frac{1}{42}\\ -\frac{1}{21}&\frac{2}{21}
\end {array} \right]
.$$
Work Step by Step
Since
$$((3A)^{-1})^{-1}=3A,$$
then by calculating the inverse of the matrix $ \left[ \begin {array}{cc}4&-1\\2&3 \end {array} \right]
$ we have
$$3A=\left[ \begin {array}{cc} \frac{3}{14}&\frac{1}{14}\\ -\frac{1}{7}&\frac{2}{7}
\end {array} \right]
.$$
Hence, $A$ is given by
$$A=\frac{1}{3}\left[ \begin {array}{cc} \frac{3}{14}&\frac{1}{14}\\ -\frac{1}{7}&\frac{2}{7}
\end {array} \right]=\left[ \begin {array}{cc} \frac{3}{42}&\frac{1}{42}\\ -\frac{1}{21}&\frac{2}{21}
\end {array} \right]
.$$
