Answer
$$A^{-1}= \left[ \begin {array}{cccccc} {\frac {3}{20}}&{\frac {3}{20}}&\frac{1}{10}
\\ \frac{3}{10}&-\frac{1}{30}&-\frac{2}{15}\\ -\frac{1}{5}&-\frac{1}{5}&\frac{1}{5}\end {array} \right]
.$$
Work Step by Step
Let $A$ be a matrix given by
$$A=\left[ \begin {array}{cccccc} 2&3&1 \\ 2&-3&-3
\\ 4&0&3 \end {array} \right].$$
To find $A^{-1}$, we have
$$\left[ A \ \ I \right]=\left[ \begin {array}{cccccc} 2&3&1&1&0&0\\ 2&-3&-3
&0&1&0\\ 4&0&3&0&0&1\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[I \ \ A^{-1} \right]= \left[ \begin {array}{cccccc} 1&0&0&{\frac {3}{20}}&{\frac {3}{20}}&\frac{1}{10}
\\ 0&1&0&\frac{3}{10}&-\frac{1}{30}&-\frac{2}{15}\\ 0&0
&1&-\frac{1}{5}&-\frac{1}{5}&\frac{1}{5}\end {array} \right]
.
$$
Then $A^{-1}$ is given by
$$A^{-1}= \left[ \begin {array}{cccccc} {\frac {3}{20}}&{\frac {3}{20}}&\frac{1}{10}
\\ \frac{3}{10}&-\frac{1}{30}&-\frac{2}{15}\\ -\frac{1}{5}&-\frac{1}{5}&\frac{1}{5}\end {array} \right]
.$$
