Example. Find the coordinates of the point $Q$, symmetric to the point $P(2,-1,2)$ with respect to the line

$$
\frac{x-1}{1}=\frac{y}{0}=\frac{z+1}{-2}
$$
Solution.

1. Let's find the projection of point $P$ onto the given line, i.e., point $P^{\prime}$. For this:

a) we will form the equation of the plane passing through point $P$ and perpendicular to the given line. As the normal vector $\vec{n}$ of this plane, we can take the direction vector of the given line: $\vec{n}=\vec{a}=\{1,0,-2\}$. Then

$$
1(x-2)+0(y+1)-2(z-2)=0 \Longrightarrow x-2 z+2=0
$$

b) we will find the point of intersection of the given line and the plane $x-2 z+2=0$. For this, we will write the equations of the line in parametric form:

$$
\left\{\begin{array}{l}
x=t+1 \\
y=0 \\
z=-2 t-1
\end{array}\right.
$$

Substituting these expressions for $x, y$, and $z$ into the equation of the plane, we find the value of the parameter $t$ at which the line and the plane intersect: $t_{0}=-1$;

c) substituting the found value $t_{0}=-1$ into the parametric equations of the line, we get

$$
x_{P^{\prime}}=0, y_{P^{\prime}}=0, z_{P^{\prime}}=1
$$

Thus, the point of intersection of the line and the plane, and therefore the projection of point $P$ onto the line, is $P^{\prime}(0,0,1)$.

2. The coordinates of point $Q$, symmetric to point $P$ with respect to the given line, are determined from the conditions (1):

$$
\begin{aligned}
& x_{Q}=2 x_{P^{\prime}}-x_{P}=-2 \\
& y_{Q}=2 y_{P^{\prime}}-y_{P}=1 \\
& z_{Q}=2 z_{P^{\prime}}-z_{P}=0
\end{aligned}
$$

Answer. Point $Q$ has coordinates $(-2,1,0)$.

Conditions of the Problem. Find the coordinates of the point symmetric to point $P$ with respect to the given line.

1. $P(0,-1,3), \quad \frac{x-1}{1}=\frac{y}{-1}=\frac{z}{1}$.
2. $\quad P\left((2,1,-1), \quad \frac{x-2}{2}=\frac{y}{0}=\frac{z+1}{-1}\right.$.
3. $P(-1,0,3), \quad \frac{x}{0}=\frac{y+1}{2}=\frac{z}{1}$.
4. $P(3,0,-1), \quad \frac{x}{1}=\frac{y-1}{1}=\frac{z}{-1}$.
5. $P(-1,2,1), \quad \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{0}$.
6. $P(3,-1,0), \quad \frac{x}{1}=\frac{y}{0}=\frac{z+1}{2}$.
7. $P(-1,3,0), \quad \frac{x}{1}=\frac{y}{-1}=\frac{z-1}{-1}$.
8. $P(1,-1,2), \quad \frac{x}{0}=\frac{y+1}{1}=\frac{z-2}{-2}$.
9. $P(0,3,-1), \quad \frac{x+1}{2}=\frac{y}{1}=\frac{z}{0}$.
10. $P(0,2,1), \quad \frac{x-4}{2}=\frac{y+1}{-1}=\frac{z-2}{3}$.

Answers. 1. ( $4,-1,-1)$. 2. $(2,-1,-1)$. 3. $(1,2,-1)$. 4. ( $-1,4,-1)$. 5. $(-1,2,-1)$. 6. $(-1,1,2)$. 7. $(-1,-1,4)$. 8. $(-1,-1,2)$. 9. $(2,-1,1)$. 10. $(4,-2,-3)$.

## Chapter 2

## LINEAR ALGEBRA

In studying the topic of LINEAR ALGEBRA, you will become familiar with the concepts of linear (vector) space, linear operator, its matrix, image, kernel, rank, defect, eigenvectors, and eigenvalues through examples. You will learn to perform various operations with operators and matrices, investigate and solve systems of linear equations, obtain all information about an operator (matrix, image, kernel, rank, and defect, eigenvectors, and eigenvalues) from its matrix, and transform vectors and matrices when changing bases.

Using the SOLVER.PM package, you can perform all matrix operations, reduce a matrix to a reduced (Gaussian) form, compute determinants, the inverse matrix, solve systems of equations, check the linearity of an operator, solve the characteristic equation, find the eigenvectors and eigenvalues of an operator, perform all numerical calculations, and verify the correctness of the results you have obtained.

## 2.1. Cramer's Rule

Problem Statement. Solve the system of three linear equations with three unknowns

$$
\left\{\begin{array}{l}
c_{11} x_{1}+c_{12} x_{2}+c_{13} x_{3}=d_{1} \\
c_{21} x_{1}+c_{22} x_{2}+c_{23} x_{3}=d_{2} \\
c_{31} x_{1}+c_{32} x_{2}+c_{33} x_{3}=d_{3}
\end{array}\right.
$$

using Cramer's rule.

Plan of Solution. If the determinant of the system matrix

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

is not equal to zero, then the system has a solution and only one. This solution is determined by the formulas

$$
x_{i}=\frac{\Delta_{i}}{\Delta}, \quad i=1,2,3
$$

where $\Delta_{i}$ is the determinant of the matrix obtained from the system matrix by replacing the $i$-th column with the column of free terms.

1. We compute the determinant of the system matrix

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

and ensure that it is not equal to zero. Therefore, the system of equations has a unique solution.

2. We compute the determinants

$$
\Delta_{1}=\left|\begin{array}{lll}
d_{1} & c_{12} & c_{13} \\
d_{2} & c_{22} & c_{23} \\
d_{3} & c_{32} & c_{33}
\end{array}\right|, \quad \Delta_{2}=\left|\begin{array}{lll}
c_{11} & d_{1} & c_{13} \\
c_{21} & d_{2} & c_{23} \\
c_{31} & d_{3} & c_{33}
\end{array}\right|, \Delta_{3}=\left|\begin{array}{lll}
c_{11} & c_{12} & d_{1} \\
c_{21} & c_{22} & d_{2} \\
c_{31} & c_{32} & d_{3}
\end{array}\right|
$$

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

$$
x_{1}=\frac{\Delta_{1}}{\Delta}, \quad x_{2}=\frac{\Delta_{2}}{\Delta}, \quad x_{3}=\frac{\Delta_{3}}{\Delta}
$$