1.23. Find the fundamental system of solutions for the system of equations and write the general solution of this system:

$$
\left\{\begin{array}{l}
x_{1}+7 x_{2}-8 x_{3}+9 x_{4}=0 \\
2 x_{1}-3 x_{2}+3 x_{3}-2 x_{4}=0 \\
4 x_{1}+11 x_{2}-13 x_{3}+16 x_{4}=0 \\
7 x_{1}-2 x_{2}+x_{3}+3 x_{4}=0
\end{array}\right.
$$
Solution. The main matrix of the system has the form

$$
\tilde{\boldsymbol{A}}=\left(\begin{array}{cccc}
1 & 7 & -8 & 9 \\
2 & -3 & 3 & -2 \\
4 & 11 & -13 & 16 \\
7 & -2 & 1 & 3
\end{array}\right)
$$

The rank of this matrix is 2, and the minor of the second order located in the upper left corner of this matrix is not zero, while all other minors of the third order surrounding it are zero. This means that the third and fourth equations of the system are consequences of the first two. Therefore, all solutions of the original system are solutions of the following system, corresponding to the first two rows of the main matrix of the system:

$$
\left\{\begin{array}{l}
x_{1}+7 x_{2}-8 x_{3}+9 x_{4}=0 \\
2 x_{1}-3 x_{2}+3 x_{3}-2 x_{4}=0
\end{array}\right.
$$

Here, the free variables $x_{3}$ and $x_{4}$ should be moved to the right-hand sides of the equations, leaving the basic variables $x_{1}$ and $x_{2}$ in the left-hand sides:

$$
\left\{\begin{array}{l}
x_{1}+7 x_{2}=8 x_{3}-9 x_{4} \\
2 x_{1}-3 x_{2}=-3 x_{3}+2 x_{4}
\end{array}\right.
$$

We will solve the system using the matrix method. The inverse matrix for the square matrix of the second order

$$
A=\left(\begin{array}{cc}
1 & 7 \\
2 & -3
\end{array}\right)
$$

is constructed as follows: the elements on the main diagonal are swapped, and the elements on the secondary diagonals change their sign, and the resulting matrix is divided by the determinant of the original matrix:

$$
\operatorname{det} A=\left|\begin{array}{cc}
1 & 7 \\
2 & -3
\end{array}\right|=1 \cdot(-3)-7 \cdot 2=-17
$$

We have $A^{-1}=\frac{1}{-17}\left(\begin{array}{cc}-3 & -7 \\ -2 & 1\end{array}\right)=\left(\begin{array}{cc}\frac{3}{17} & \frac{7}{17} \\ \frac{2}{17} & -\frac{1}{17}\end{array}\right)$.

Next, we get:

$$
X=\binom{x_{1}}{x_{2}}=A^{-1} \cdot B=\left(\begin{array}{cc}
\frac{3}{17} & \frac{7}{17} \\
\frac{2}{17} & -\frac{1}{17}
\end{array}\right) \cdot\binom{8 x_{3}-9 x_{4}}{-3 x_{3}+2 x_{4}}=\binom{\frac{3}{17} x_{3}-\frac{13}{17} x_{4}}{\frac{19}{17} x_{3}-\frac{20}{17} x_{4}}
$$

By equating the column of free terms $\binom{x_{3}}{x_{4}}$ first to $\binom{1}{0}$, and then to $\binom{0}{1}$, we obtain the fundamental system of solutions

$$
X^{(1)}=\left(\begin{array}{c}
\frac{3}{17} \\
\frac{19}{17} \\
1 \\
0
\end{array}\right), \quad X^{(2)}=\left(\begin{array}{c}
-\frac{13}{17} \\
-\frac{20}{17} \\
0 \\
1
\end{array}\right)
$$

The general solution of the original homogeneous system is obtained

![](https://cdn.mathpix.com/cropped/2024_05_22_f5a9bb7960904462604bg-061.jpg?height=290&width=950&top_left_y=628&top_left_x=143)

Here $\alpha$ and $\beta$ are arbitrary real numbers.