Example 3 (to $7^{\circ}$ ). Find $A^{-1}$, if

$$
A=\left(\begin{array}{rrr}
3 & -1 & 3 \\
2 & -1 & 4 \\
1 & 2 & -3
\end{array}\right)
$$
Solution. In the left part of the first block of the Gaussian table, we enter the elements of matrix $A$. In the right part of the block, we write the identity matrix of the third order. The transition from one block to the next is carried out using the Jordan-Gauss formulas. The leading coefficients are outlined. The working table has the following appearance:

| 3 | -1 | 3 | 1 | 0 | 0 |
| ---: | ---: | ---: | ---: | ---: | ---: |
| 2 | -1 | 4 | 0 | 1 | 0 |
| 1 | 2 | -3 | 0 | 0 | 1 |
| 1 | $-1 / 3$ | 1 | $1 / 3$ | 0 | 0 |
| 0 | $-1 / 3$ | 2 | $-2 / 3$ | 1 | 0 |
| 0 | $7 / 3$ | -4 | $-1 / 3$ | 0 | 1 |
| 1 | 0 | -1 | 1 | -1 | 0 |
| 0 | 1 | -6 | 2 | -3 | 0 |
| 0 | 0 | $(10$ | -5 | 7 | 1 |
| 1 | 0 | 0 | $1 / 2$ | $-3 / 10$ | $1 / 10$ |
| 0 | 1 | 0 | -1 | $6 / 5$ | $3 / 5$ |
| 0 | 0 | 1 | $-1 / 2$ | $7 / 10$ | $1 / 10$ |

We obtained $A^{-1}=\left(\begin{array}{rrr}1 / 2 & -3 / 10 & 1 / 10 \\ -1 & 6 / 5 & 3 / 5 \\ -1 / 2 & 7 / 10 & 1 / 10\end{array}\right)$

## Exercises

Find the inverse matrices for the given matrices.

1. $\left(\begin{array}{rrr}1 & 2 & 3 \\ 2 & -3 & 1 \\ 3 & 3 & -4\end{array}\right)$.
2. $\left(\begin{array}{rr}3 & 1 \\ 5 & -3\end{array}\right)$.
3. $\left(\begin{array}{rrr}1 & -1 & 1 \\ 1 & -3 & -2 \\ 2 & 3 & 2\end{array}\right)$.
4. $\left(\begin{array}{rrr}3 & -1 & 3 \\ 2 & -1 & 4 \\ 1 & 2 & -3\end{array}\right)$.
5. $\left(\begin{array}{rrrr}2 & 3 & 2 & 1 \\ 8 & -3 & -1 & 1 \\ 2 & 1 & 1 & 1 \\ 4 & -1 & -1 & 1\end{array}\right)$
6. $\left(\begin{array}{rrrr}3 & -5 & -2 & 2 \\ -4 & 7 & 4 & 4 \\ 4 & -9 & -3 & 7 \\ 2 & 6 & -3 & 2\end{array}\right)$.
7. $\left(\begin{array}{rrr}5 & 2 & 1 \\ 3 & 7 & -11 \\ 8 & 1 & 6\end{array}\right)$
8. $\left(\begin{array}{rrrr}2 & 6 & 2 & 2 \\ 3 & 1 & 3 & 4 \\ 5 & 2 & -1 & 7 \\ 0 & 5 & 6 & -1\end{array}\right)$.

Solve the systems using the matrix method.
9. $\left\{\begin{aligned} 2 x_{1}+3 x_{2}-5 x_{3} & =0, \\ -3 x_{1}+2 x_{2}+4 x_{3} & =3, \\ -x_{1}+2 x_{2}+3 x_{3} & =2 .\end{aligned} \quad\right.$ 10. $\left\{\begin{aligned}-x_{1}+3 x_{2}+2 x_{3} & =4, \\ 2 x_{1}+x_{2}+3 x_{3} & =6, \\ 2 x_{2}+x_{3} & =3 .\end{aligned}\right.$
11. $\left\{\begin{aligned} x_{2}+2 x_{3}-x_{4} & =4, \\ 3 x_{1}-x_{2}+2 x_{4} & =9, \\ 2 x_{2}+2 x_{3} & =-2, \\ 2 x_{1}+3 x_{3}+x_{4} & =15 .\end{aligned}\right.$

## Answers

1. $\frac{1}{76}\left(\begin{array}{rrr}9 & 17 & 11 \\ 11 & -13 & 5 \\ 15 & 3 & -7\end{array}\right) \