How to find the inverse of a 3x3 matrix using row operations?

The Gauss-Jordan method: augment $A$ with the identity matrix $[A \mid I]$ and row-reduce until the left side becomes $I$. The right side will be $A^{-1}$. $$[A \mid I] = \begin{pmatrix} 1 & 2 & 3 & | & 1 & 0 & 0 \\ 0 & 1 & 4 & | & 0 & 1 & 0 \\ 5 & 6 & 0 & | & 0 & 0 & 1 \end{pmatrix}$$ **Step 1:** $R_3 \leftarrow R_3 - 5R_1$ $$\begin{pmatrix} 1 & 2 & 3 & | & 1 & 0 & 0 \\ 0 & 1 & 4 & | & 0 & 1 & 0 \\ 0 & -4 & -15 & | & -5 & 0 & 1 \end{pmatrix}$$ **Step 2:** $R_3 \leftarrow R_3 + 4R_2$ $$\begin{pmatrix} 1 & 2 & 3 & | & 1 & 0 & 0 \\ 0 & 1 & 4 & | & 0 & 1 & 0 \\ 0 & 0 & 1 & | & -5 & 4 & 1 \end{pmatrix}$$ **Step 3:** $R_2 \leftarrow R_2 - 4R_3$ $$\begin{pmatrix} 1 & 2 & 3 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & 20 & -15 & -4 \\ 0 & 0 & 1 & | & -5 & 4 & 1 \end{pmatrix}$$ **Step 4:** $R_1 \leftarrow R_1 - 2R_2 - 3R_3$ $$\begin{pmatrix} 1 & 0 & 0 & | & -24 & 18 & 5 \\ 0 & 1 & 0 & | & 20 & -15 & -4 \\ 0 & 0 & 1 & | & -5 & 4 & 1 \end{pmatrix}$$ **Therefore:** $$A^{-1} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}$$ **Verification:** $A \cdot A^{-1}$ should equal $I$.