Linear Equation (线性方程)

Definition

For $n \geq 1$, a linear equation in the variables $x_1, ... , x_n$ is an equation that can be written in the form

$$a_1x_1 + a_2x_2 + ... + a_nx_n = b$$

where $b$ and the coefficients $a_1, ... , a_n$ are real or complex numbers.

Examples

• $\sqrt{2}x_1 + \pi x_2 = 0$ is a linear equation.

• $x_1x_2 + x_3 = 0$ is NOT a linear equation because of the term $x_1x_2$.

• $x_1 + \sqrt{x_2} = 0$ is NOT a linear equation because of the term $\sqrt{x_2}$.

• $0x_1 = 0$ is a linear equation.

• $0x_1 = 1$ is a linear equation, although it has no solution.

Linear System

Linear System (线性方程组)

A linear system is a collection of one or more linear equations involving the same variables (e.g., $x_1, ... , x_n$).

Solution of Linear Systems (线性方程组的解)

• A solution of a linear system is a (ordered) list $(s_1, s_2, ... ,s_n)$ of numbers that makes each equation a true statement when the values $s_1, s_2, ... ,s_n$ are substituted for $x_1, ... , x_n$, respectively.

• The set of all possible solutions is called the solution set of the linear system (线性方程组的解集).

• A linear system may have:

  • no solution
  • exactly one solution
  • infinitely many solutions

• A linear system that has at least one solution is called consistent (相容的); otherwise, it is called inconsistent (不相容的).

Equivalent Linear Systems (等价线性方程组)

Two linear systems are called equivalent if they have the same solution set.

Matrix Notation of Linear Systems (线性方程组的矩阵表示法)

Take the following linear system as an example:

$$\begin{cases} \begin{aligned} x_1 - 2x_2 + x_3 = 0 \\ 2x_2 - 8x_3 = 8 \\ 5x_1 - 5x_3 = 10 \end{aligned} \end{cases}$$

the matrix

$$\begin{bmatrix} 1 & -2 & 1 \\ 0 & 2 & -8 \\ 5 & 0 & -5 \end{bmatrix}$$

is called the coefficient matrix (系数矩阵), and the matrix

$$\begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 2 & -8 & 8\\ 5 & 0 & -5 & 10 \end{bmatrix}$$

is called the augmented matrix of the linear system (线性方程组的增广矩阵).

Elementary Row Operations (初等行变换)

1. Interchange: Interchange two rows.
2. Scaling: Multiply all entries in a row by a nonzero constant.
3. Replacement: Replace one row by the sum of itself and a multiple of another row.

Two matrices are called row equivalent if there is a sequence of elementary row operations that transforms one matrix into the other.

Soving a Linear System with Augmented Matrix (用增广矩阵解线性方程组)

A Simple Example and Row Operations

Solve the following linear system:

$$\begin{cases} \begin{aligned} x_1 - 2x_2 + x_3 = 0 \\ 2x_2 - 8x_3 = 8 \\ 5x_1 - 5x_3 = 10 \end{aligned} \end{cases}$$

The augmented matrix is

$$\begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 2 & -8 & 8\\ 5 & 0 & -5 & 10 \end{bmatrix}$$

We can use row operations to transform the augmented matrix into a simpler form. The row operations are:

$$\begin{aligned} \begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 2 & -8 & 8\\ 5 & 0 & -5 & 10 \end{bmatrix} &\xrightarrow{\color{red}{- 5R_1 + R_3}} \begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 2 & -8 & 8\\ 0 & 10 & -10 & 10 \end{bmatrix} \\[8pt] &\xrightarrow{- 5R_2 + R_3} \begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 2 & -8 & 8\\ 0 & 0 & 30 & -30 \end{bmatrix} \\[8pt] &\xrightarrow{\frac{1}{30}R_3} \begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 2 & -8 & 8\\ 0 & 0 & 1 & -1 \end{bmatrix} \\[8pt] &\xrightarrow{\frac{1}{2}(R_2 + 8R_3)} \begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & -1 \end{bmatrix} \\[8pt] &\xrightarrow{R_1 + 2R_2 - R_3} \begin{bmatrix} 1 & 0 & 0 & 2\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & -1 \end{bmatrix} \end{aligned}$$

Therefore, the solution of the linear system is $(2, 0, -1)$.

Here, I’ll use the step marked in red as an example to explain the row operation. The operation $- 5R_1 + R_3$ means to multiply the first row by $-5$ and then add it to the third row, replacing the third row with the result.

A More Complex Example and Common Method

Given the following augmented matrix:

$$\begin{bmatrix} 2 & 1 & -1 & 2 & 5\\ 4 & 5 & -3 & 6 & 9\\ -2 & 5 & -2 & 6 & 4\\ 4 & 11 & -4 & 8 & 2 \end{bmatrix}$$

Let's solve the corresponding linear system using row operations:

$$\begin{aligned} \begin{bmatrix} 2 & 1 & -1 & 2 & 5\\ 4 & 5 & -3 & 6 & 9\\ -2 & 5 & -2 & 6 & 4\\ 4 & 11 & -4 & 8 & 2 \end{bmatrix} &\xrightarrow{\substack{R_2 - 2R_1 \\[4pt] R_3 + R_1 \\[4pt] R_4 - 2R_1}} \begin{bmatrix} 2 & 1 & -1 & 2 & 5\\ 0 & 3 & -1 & 2 & -1\\ 0 & 6 & -3 & 8 & 9\\ 0 & 9 & -2 & 4 & -8 \end{bmatrix} \\[8pt] &\xrightarrow{\substack{R_3 - 2R_2 \\[4pt] R_4 - 3R_2}} \begin{bmatrix} 2 & 1 & -1 & 2 & 5\\ 0 & 3 & -1 & 2 & -1\\ 0 & 0 & -1 & 4 & 11\\ 0 & 0 & 1 & -2 & -5 \end{bmatrix} \\[8pt] &\xrightarrow{R_4 + R_3} \begin{bmatrix} 2 & 1 & -1 & 2 & 5\\ 0 & 3 & -1 & 2 & -1\\ 0 & 0 & -1 & 4 & 11\\ 0 & 0 & 0 & 2 & 6 \end{bmatrix} \end{aligned}$$

We firstly use the first row to eliminate the first entries of the other rows. Then, we use the second row to eliminate the second entries of the third and fourth rows. And so on. Finally, we get a matrix like this:

$$\begin{bmatrix} a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} & b_1\\ 0 & a_{2,2} & a_{2,3} & a_{2,4} & b_2\\ 0 & 0 & a_{3,3} & a_{3,4} & b_3\\ 0 & 0 & 0 & a_{4,4} & b_4 \end{bmatrix}$$

Now, we can use the last row to eliminate the fourth entries of the other rows, then use the third row to eliminate the third entries of the first two rows, and so on:

$$\begin{aligned} \begin{bmatrix} 2 & 1 & -1 & 2 & 5\\ 0 & 3 & -1 & 2 & -1\\ 0 & 0 & -1 & 4 & 11\\ 0 & 0 & 0 & 2 & 6 \end{bmatrix} &\xrightarrow{\substack{R_1 - R_4 \\[4pt] R_2 - R_4 \\[4pt] R_3 - 2R_4}} \begin{bmatrix} 2 & 1 & -1 & 0 & -1\\ 0 & 3 & -1 & 0 & -7\\ 0 & 0 & -1 & 0 & -1\\ 0 & 0 & 0 & 2 & 6 \end{bmatrix} \\[8pt] &\xrightarrow{\substack{R_1 - R_3 \\[4pt] R_2 - R_3}} \begin{bmatrix} 2 & 1 & 0 & 0 & 0\\ 0 & 3 & 0 & 0 & -6\\ 0 & 0 & -1 & 0 & -1\\ 0 & 0 & 0 & 2 & 6 \end{bmatrix} \\[8pt] &\xrightarrow{R_1 - \frac{1}{3}R_2} \begin{bmatrix} 2 & 0 & 0 & 0 & 2\\ 0 & 3 & 0 & 0 & -6\\ 0 & 0 & -1 & 0 & -1\\ 0 & 0 & 0 & 2 & 6 \end{bmatrix} \\[8pt] &\xrightarrow{\substack{\frac{1}{2}R_1, \frac{1}{3}R_2 \\[4pt] -R_3, \frac{1}{2}R_4}} \begin{bmatrix} 1 & 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0 & -2\\ 0 & 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 1 & 3 \end{bmatrix} \end{aligned}$$

So the solution of the linear system is $(1, -2, 1, 3)$.

Copyright

License under:Attribution-NonCommercial-ShareAlike 4.0 International (CC-BY-NC-SA-4.0)