Answer
$\begin{bmatrix} 1 & 0 & -1\\1 &0 & -3/2\\-6 & 1 & 13/2\end{bmatrix} $
Work Step by Step
Step 1. Write the original matrix together with an identity matrix.
$\begin{array}(\\B=A|I= \\ \\ \end{array} \begin{bmatrix} 3 & -2 & 0 & | &1 & 0 & 0\\5 & 1 & 1 & | &0 & 1 & 0\\2 & -2 & 0 & | &0 & 0 & 1 \end{bmatrix} \begin{array}( 10R_1\to R_1 \\6R_2\to R_2 \\15R_3\to R_3 \\ \end{array}$
Step 2. Use row operations on the right side of the matrix to transform the left half into a reduced row-echelon form towards $B=IA^{-1}$.
$\begin{bmatrix} 30 & -20 & 0 & | &10 & 0 & 0\\30 & 6 & 6 & | &0 &6 & 0\\30 & -30 & 0 & | &0 & 0 & 15 \end{bmatrix} \begin{array}( \\R_2-R_1\to R_2 \\R_1-R_3\to R_3 \\ \end{array}$
$\begin{bmatrix} 30 & -20 & 0 & | &10 & 0 & 0\\0 & 26 & 6 & | &-10 &6 & 0\\0 & 10 & 0 & | &10 & 0 & -15 \end{bmatrix} \begin{array}( R_1+2R_3\to R_1\\5R_2\to R_2 \\13R_3\to R_3 \\ \end{array}$
$\begin{bmatrix} 30 & 0 & 0 & | &30 & 0 & -30\\0 & 130 & 30 & | &-50 &30 & 0\\0 & 130 & 0 & | &130 & 0 & -195 \end{bmatrix} \begin{array}( \\ \\R_2-R_3\to R_3 \\ \end{array}$
$\begin{bmatrix} 30 & 0 & 0 & | &30 & 0 & -30\\0 & 130 & 30 & | &-50 &30 & 0\\0 & 0 & 30 & | &-180 & 30 & 195 \end{bmatrix} \begin{array}( \\R_2-R_3\to R_2 \\\ \end{array}$
$\begin{bmatrix} 30 & 0 & 0 & | &30 & 0 & -30\\0 & 130 & 0 & | &130 &0 & -195\\0 & 0 & 30 & | &-180 & 30 & 195 \end{bmatrix} \begin{array}( R_1/30\to R_1 \\R_2/130\to R_2 \\R_3/30\to R_3 \end{array}$
$\begin{bmatrix} 1 & 0 & 0 & | &1 & 0 & -1\\0 & 1 & 0 & | &1 &0 & -3/2\\0 & 0 & 1 & | &-6 & 1 & 13/2\end{bmatrix} $
Step 3. Conclusion: the inverse of the original matrix is
$\begin{bmatrix} 1 & 0 & -1\\1 &0 & -3/2\\-6 & 1 & 13/2\end{bmatrix} $
