Answer
$\begin{bmatrix} -1 & -1 & 2 \\3 & 2 &-4\\ -1/2 & 0 & 1/2 \end{bmatrix} $
Work Step by Step
Step 1. Write the original matrix together with an identity matrix.
$\begin{array}(\\B=A|I= \\ \\ \end{array} \begin{bmatrix} 2 & 1 & 0 & | &1 & 0 & 0\\1 & 1 & 4 & | &0 & 1 & 0\\2 & 1 & 2 & | &0 & 0 & 1 \end{bmatrix} \begin{array}( \\2R_2-R_1\to R_2 \\R_3-R_1\to R_3 \\ \end{array}$
Step 2. Use row operations on the right side of the matrix to transform the left half into a reduced row-echelon form towards $B=IA^{-1}$.
$\begin{bmatrix} 2 & 1 & 0 & | &1 & 0 & 0\\0 & 1 & 8 & | &-1 & 2 & 0\\0 & 0 & 2 & | &-1 & 0 & 1 \end{bmatrix} \begin{array}( R_1/2\\ \\R_3/2 \\ \end{array}$
$\begin{bmatrix} 1 & 1/2 & 0 & | &1/2 & 0 & 0\\0 & 1 & 8 & | &-1 & 2 & 0\\0 & 0 & 1 & | &-1/2 & 0 & 1/2 \end{bmatrix} \begin{array}( \\R_2-8R_3\to R_2 \\ \\ \end{array}$
$\begin{bmatrix} 1 & 1/2 & 0 & | &1/2 & 0 & 0\\0 & 1 & 0 & | &3 & 2 &-4\\0 & 0 & 1 & | &-1/2 & 0 & 1/2 \end{bmatrix} \begin{array}( R_1-R_2/2 \\ \\ \\ \end{array}$
$\begin{bmatrix} 1 &0 & 0 & | &-1 & -1 & 2\\0 & 1 & 0 & | &3 & 2 &-4\\0 & 0 & 1 & | &-1/2 & 0 & 1/2 \end{bmatrix} $
Step 3. Conclusion: the inverse of the original matrix is
$\begin{bmatrix} -1 & -1 & 2 \\3 & 2 &-4\\ -1/2 & 0 & 1/2 \end{bmatrix} $
