Finding inverse of a $3\times3$ matrix

Question

I am given a $3 \times3$ matrix and am asked to find the inverse using elementary row operations. I know how they work, but have no idea of which steps to apply first, followed by which steps.

First, the matrices:

$$\begin{pmatrix} 1 & 1 & -3\\ 2 & 1 & -3\\ 2 & 2 & 1 \end{pmatrix}$$

All I know thus far is that, if there is a series of operations (pre-multipliers)

$E_nE_{n-1}...E_2E_1A$ that reduces to the identity matrix, the same sequence $ E_nE_{n-1}...E_2E_1I$ reduces to the inverse of $A$, $A^{-1}$.

Any help? If not, I will use another method already because this is not working thus far.

UPDATE

Thanks to the community, I got the final answer:

$$\begin{pmatrix} -1 & 1 & 0\\ \frac8 7 & -1 & \frac 3 7\\ \frac{-2}{7} & 0 & \frac 1 7 \end{pmatrix}$$

Answer 1

I was taught to augment the matrix with the identity, then apply the row operations:

$$\begin{pmatrix} 1 & 1 & -3 & 1 & 0 & 0\\ 2 & 1 & -3&0&1&0\\ 2 & 2 & 1&0&0&1 \end{pmatrix}$$

Subtract twice row 1 from row 2 and twice row 1 from row 3 (yes, this is two operations)

$$\begin{pmatrix} 1 & 1 & -3 & 1 & 0 & 0\\ 0 & -1 & 3&-2&1&0\\ 0 & 0 & 7&-2&0&1 \end{pmatrix}$$

Multiply row 2 by -1 and row 3 by $\frac 17$

$$\begin{pmatrix} 1 & 1 & -3 & 1 & 0 & 0\\ 0 & 1 & -3&2&-1&0\\ 0 & 0 & 1&-2/7&0&1/7 \end{pmatrix}$$

Subtract row 2 from row 1, then add three times the third to the second and you are there. The right three columns will be your inverse.

Answer 2

That way of calculating ineverses (which was very slow imo), was to write it this way:

$$\begin{pmatrix}1 & 1 & -3 & | & 1 & 0 & 0\\ 2 & 1 & -3 & | & 0 & 1 & 0\\ 2 & 2 & 1 & | & 0 & 0 & 1 \end{pmatrix}$$

Ten make those elementary operations on the first one to reduce to the identity, while making the same operations on the right one. When you get the identity on the left one, what you have on the right one will be the inverse.

Answer 3

$$\mathbf{A}^{-1} = \begin{bmatrix} a & b & c\\ d & e & f \\ g & h & k\\ \end{bmatrix}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{bmatrix} \, A & \, B & \,C \\ \, D & \, E & \,F \\ \, G & \,H & \, K\\ \end{bmatrix}^T = \frac{1}{\det(\mathbf{A})} \begin{bmatrix} \, A & \, D & \,G \\ \, B & \, E & \,H \\ \, C & \,F & \, K\\ \end{bmatrix}$$ If the determinant is non-zero, the matrix is invertible, with the elements of the above matrix on the right side given by $$\begin{matrix} A = (ek-fh) & D = (ch-bk) & G = (bf-ce) \\ B = (fg-dk) & E = (ak-cg) & H = (cd-af) \\ C = (dh-eg) & F = (gb-ah) & K = (ae-bd). \\ \end{matrix}$$