Answer
$\begin{bmatrix}3 & -2 & -5\\-1 & 1 & 1 \\-3 & 2 & 6 \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} 4 & 2 & 3 & | &1 & 0 & 0\\3 & 3 & 2 & | &0 & 1 & 0\\1 & 0 & 1 & | &0 & 0 & 1 \end{bmatrix} \begin{array}( R_1/4 \\(R_2-3R_3)/3\to R_2 \\ \\ \end{array}$
Step 2. Use row operations given on the right side of the matrix to transform the left half into a reduced row-echelon form towards $B=IA^{-1}$
$\begin{bmatrix} 1 & 1/2 & 3/4 & | &1/4 & 0 & 0\\0 & 1 & -1/3 & | &0 & 1/3 & -1 \\1 & 0 & 1 & | &0 & 0 & 1 \end{bmatrix} \begin{array}( \\ \\2(R_1-R_3)\to R_3 \\ \end{array}$
$\begin{bmatrix} 1 & 1/2 & 3/4 & | &1/4 & 0 & 0\\0 & 1 & -1/3 & | &0 & 1/3 & -1 \\0 & 1 & -1/2 & | &1/2 & 0 & -2 \end{bmatrix} \begin{array}( \\ \\6(R_2-R_3)\to R_3 \\ \end{array}$
$\begin{bmatrix} 1 & 1/2 & 3/4 & | &1/4 & 0 & 0\\0 & 1 & 0 & | &0 & 1/3 & -1 \\0 & 0 & 1 & | &-3 & 2 & 6 \end{bmatrix} \begin{array}( \\ R_2+R_3/3\to R_2 \\ \\ \end{array}$
$\begin{bmatrix} 1 & 1/2 & 3/4 & | &1/4 & 0 & 0\\0 & 1 & 0 & | &-1 & 1 & 1 \\0 & 0 & 1 & | &-3 & 2 & 6 \end{bmatrix} \begin{array}( R_1-R_2/2-3R_3/4\to R_1\\ \\ \\ \end{array}$
$\begin{bmatrix} 1 & 0 &0 & | &3 & -2 & -5\\0 & 1 & 0 & | &-1 & 1 & 1 \\0 & 0 & 1 & | &-3 & 2 & 6 \end{bmatrix} \begin{array}( \\ \\ \\ \end{array}$
Step 3. Conclusion: the inverse of the original matrix is
$\begin{bmatrix}3 & -2 & -5\\-1 & 1 & 1 \\-3 & 2 & 6 \end{bmatrix} $
