Example. Solve the system of equations

$$
\left\{\begin{array}{r}
x_{1}+2 x_{2}+x_{3}=4 \\
3 x_{1}-5 x_{2}+3 x_{3}=1 \\
2 x_{1}+7 x_{2}-x_{3}=8
\end{array}\right.
$$

using Cramer's rule.
SOLUTION.

1. We calculate the determinant of the system matrix, expanding it along the first row:

$$
\Delta=\left|\begin{array}{rrr}
1 & 2 & 1 \\
3 & -5 & 3 \\
2 & 7 & -1
\end{array}\right|=1 \cdot(-16)-2 \cdot(-9)+1 \cdot 31=33
$$

Since it is not equal to zero, the system of equations has a unique solution.

2. We calculate the determinants

$$
\begin{aligned}
& \Delta_{1}=\left|\begin{array}{rrr}
4 & 2 & 1 \\
1 & -5 & 3 \\
8 & 7 & -1
\end{array}\right|=4 \cdot(-16)-2 \cdot 25+1 \cdot 47=33 \\
& \Delta_{2}=\left|\begin{array}{rrr}
1 & 4 & 1 \\
3 & 1 & 3 \\
2 & 8 & -1
\end{array}\right|=1 \cdot(-25)-4 \cdot(-9)+1 \cdot 22=33 \\
& \Delta_{3}=\left|\begin{array}{rrr}
1 & 2 & 4 \\
3 & -5 & 1 \\
2 & 7 & 8
\end{array}\right|=1 \cdot(-47)-2 \cdot 22+4 \cdot 31=33
\end{aligned}
$$

3. Using Cramer's formulas (1), we find the solution to the system of equations

$$
x_{1}=1, \quad x_{2}=1, \quad x_{3}=1
$$

Answer. $x_{1}=1, x_{2}=1, x_{3}=1$.

Conditions of the Problem. Solve the systems of equations using Cramer's rule.

1. $\left\{\begin{array}{r}x_{1}+3 x_{2}-x_{3}=2, \\ 2 x_{1}-3 x_{2}+2 x_{3}=0, \\ 3 x_{1}-2 x_{2}-x_{3}=4 .\end{array}\right.$
2. $\left\{\begin{array}{c}x_{1}+2 x_{2}+x_{3}=5, \\ 3 x_{1}-5 x_{2}+3 x_{3}=-7, \\ 2 x_{1}+7 x_{2}-x_{3}=13 .\end{array}\right.$
3. $\left\{\begin{array}{l}2 x_{1}+3 x_{2}+x_{3}=1, \\ 3 x_{1}-5 x_{2}+2 x_{3}=-11, \\ 5 x_{1}+2 x_{2}-2 x_{3}=-3 .\end{array}\right.$
4. $\left\{\begin{array}{r}x_{1}+4 x_{2}+3 x_{3}=5, \\ 3 x_{1}-2 x_{2}+3 x_{3}=9, \\ 2 x_{1}+4 x_{2}-3 x_{3}=1 .\end{array}\right.$
5. $\left\{\begin{array}{r}x_{1}+5 x_{2}+x_{3}=-8, \\ 2 x_{1}-3 x_{2}+5 x_{3}=16, \\ 5 x_{1}+2 x_{2}-x_{3}=-6 .\end{array}\right.$
6. $\left\{\begin{aligned} x_{1}+3 x_{2}+2 x_{3} & =-5, \\ 2 x_{1}-2 x_{2}+3 x_{3} & =-8, \\ 3 x_{1}+4 x_{2}-4 x_{3} & =5 .\end{aligned}\right.$
7. $\left\{\begin{aligned} x_{1}+2 x_{2}+x_{3} & =2, \\ 3 x_{1}+2 x_{2}+3 x_{3} & =6, \\ 2 x_{1}-2 x_{2}-x_{3} & =7 .\end{aligned}\right.$
8. $\left\{\begin{array}{r}x_{1}+5 x_{2}+x_{3}=3, \\ 2 x_{1}-3 x_{2}+3 x_{3}=8, \\ 2 x_{1}+4 x_{2}-x_{3}=0 .\end{array}\right.$
9. $\left\{\begin{array}{r}x_{1}+3 x_{2}+x_{3}=-5, \\ 3 x_{1}-4 x_{2}+3 x_{3}=11, \\ 2 x_{1}+4 x_{2}-x_{3}=-9 .\end{array}\right.$
10. $\left\{\begin{array}{c}x_{1}+2 x_{2}+3 x_{3}=5, \\ 3 x_{1}-2 x_{2}+3 x_{3}=-1, \\ 2 x_{1}+3 x_{2}-2 x_{3}=8 .\end{array}\right.$

Answers. 1. $x_{1}=1, x_{2}=0, x_{3}=-1$. 2. $x_{1}=0, x_{2}=2, x_{3}=1$. 3. $x_{1}=-1, x_{2}=1, x_{3}=0 . \quad 4 . x_{1}=2, x_{2}=0, x_{3}=1 . \quad 5 . x_{1}=0$, $x_{2}=-2, x_{3}=2 . \quad 6 . x_{1}=-1, x_{2}=0, x_{3}=-2 . \quad 7 . x_{1}=3, x_{2}=0$, $x_{3}=-1$. 8. $x_{1}=1, x_{2}=0, x_{3}=2 . \quad 9 . x_{1}=0, x_{2}=-2, x_{3}=1$. 10. $x_{1}=1, x_{2}=2, x_{3}=0$.

## 2.2. Inverse Matrix

PROBLEM STATEMENT. A square matrix of the third order is given

$$
C=\left(\begin{array}{lll}
c_{11} & c_{12} & c_{13} \\
c_{21} & c_{22} & c_{23} \\
c_{31} & c_{32} & c_{33}
\end{array}\right)
$$

Determine the existence and find the inverse matrix $C^{-1}$.

SOLUTION PLAN. The matrix $C^{-1}$ is called the inverse of the square matrix $C$ if

$$
C \cdot C^{-1}=C^{-1} \cdot C=E
$$

where $E$ is the identity matrix.

If $\operatorname{det} C \neq 0$ (matrix $C$ is non-singular), then matrix $C$ has an inverse. If $\operatorname{det} C=0$, then matrix $C$ does not have an inverse.

1. We calculate the determinant of the matrix $\operatorname{det} C$. If $\operatorname{det} C \neq 0$, then matrix $C$ has an inverse.
2. We form the matrix of cofactors

$$
\tilde{C}=\left(\begin{array}{lll}
C_{11} & C_{12} & C_{13} \\
C_{21} & C_{22} & C_{23} \\
C_{31} & C_{32} & C_{33}
\end{array}\right)
$$

3. We transpose the matrix $\widetilde{C}$

$$
\tilde{C}^{\mathrm{T}}=\left(\begin{array}{lll}
C_{11} & C_{21} & C_{31} \\
C_{12} & C_{22} & C_{32} \\
C_{13} & C_{23} & C_{33}
\end{array}\right)
$$

4. By dividing the matrix $\widetilde{C}^{\mathrm{T}}$ by the determinant, we obtain the desired inverse matrix

$$
C^{-1}=\frac{1}{\operatorname{det} C} \cdot\left(\begin{array}{lll}
C_{11} & C_{21} & C_{31} \\
C_{12} & C_{22} & C_{32} \\
C_{13} & C_{23} & C_{33}
\end{array}\right)
$$

5. We check that $C \cdot C^{-1}=E$ and write the answer.