Linear Equations in Linear Algebra

1.1 Systems of Linear Equations (선형방정식계,연립일차방정식)

Linear Equation

A linear equation in the variables $x_1, \cdots, x_n$ is an equation that can be represented in the form

$$a_1x_1+a_2x_2+\cdots+a_nx_n = b \tag{1}$$

where $b$ and the coefficients $a_1, \cdots, a_n$ are real or complex numbers, usually known in advance.

A system of linear equations (or a linear system) is a collection of one or more linear equations involving the same variables -- say, $x_1, \cdots, x_n$.

$$\begin{align} 2x_1 -x_2 +1.5x_3 &= 8 \\ x_1 -4x_3 = -7 \tag{2} \end{align}$$

A solution of the system is a list $\left(s_1,s_2,\cdots,s_n\right)$ of numbers that makes each equation a true statement when the values $\left(s_1,\cdots,s_n\right)$ are substituted for $x_1, \cdots, x_n$ respectively. The set of all possible solutions is called the solution set of the linear system.

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

A system of linear equations has

1. no solution, or
2. exactly one solution, or
3. infinitely many solutions.

A system of linear equations is said to be consistent if it has either

• one solution or
• infinitely many solutions.

A system is inconsistent if it has

• no solution.

• linear equation,
• system of linear equations, linear system
• solution(해)
• equivalent(동치)
• no solution, (one solution, infinitely many solution)
• consistent system, inconsistent system

Matrix Notation

The essential information of a linear system can be recorded compactly in a rectangular array called a matrix. Given the system,

$$\begin{align*} x_1-2x_2+x_3 &= 0 \\ 2x_2-8x_3 &= 8 \\ -4x_1+5x_2+9x_3 &= -9 \\ \end{align*} \tag{3}$$ ,

with the coefficients of each variable aligned in columns, the matrix

$$\begin{bmatrix} 1 & -2 & 1 \\ 0 & 2 & -8 \\ -4 & 5 & 9 \end{bmatrix}$$

is called the coefficient matrix (or matrix of coefficients) of the system.

An augmented matrix of a system consists of the coefficient matrix with an added column containing the constants from the right sides of the equations.

For the given system of equations,

$$\begin{bmatrix} 1 & -2 & 1 &0\\ 0 & 2 & -8 &8\\ -4 & 5 & 9 &-9 \end{bmatrix} \tag{4}$$

is called the augmented matrix of the system.

• matrix
• coefficient matrix (계수행렬)
• augmented matrix (첨가행렬)

Matrix Size

The size of a matrix tells how many rows and columns it has. If $m$ and $n$ are positive integers, an $m \times n$ matrix is a rectangular array of numbers with $m$ rows and $n$ columns. (The number of rows always comes first.)

Solving a Linear System

The basic strategy for solving a linear system is to replace one system with an equivalent system (i.e., one with the same solution set) that is easier to solve.

Example 1

Solve the given system of equations.

$$\begin{align} x_1-2x_2+x_3 &= 0 \tag{1}\\ 2x_2-8x_3 &= 8 \tag{2}\\ -4x_1+5x_2+9x_3 &= -9 \tag{3}\\ \end{align}$$ ,

Solution of Example 1

The elimination procedure is shown here with and without matrix notation, and the results are placed side by side for comparison.

$$\begin{matrix} \begin{align*} x_1-2x_2+x_3 &= 0 \\ 2x_2-8x_3 &= 8 \\ -4x_1+5x_2+9x_3 &= -9 \end{align*} & \begin{bmatrix} 1 & -2 & 1 &0\\ 0 & 2 & -8 &8\\ -4 & 5 & 9 &-9 \end{bmatrix} \end{matrix}$$

Keep $x_1$ in the first equation and eliminate it from the other equations. To do so, add 4 times equation 1 to equation 3.

$$\begin{align*} 4x_1-8x_2+4x_3 &= 0 \\ -4x_1+5x_2+9x_3 &= -9 \\ \hline -3x_2+13x_3 &= -9 \end{align*}$$

The result of this calculation is written in place of the original third equation:

$$\begin{matrix} \begin{align*} x_1-2x_2+x_3 &= 0 \\ 2x_2-8x_3 &= 8 \\ -3x_2+13x_3 &= -9 \end{align*} & \begin{bmatrix} 1 & -2 & 1 &0\\ 0 & 2 & -8 &8\\ 0 & -3 & 13 &-9 \end{bmatrix} \end{matrix}$$

Now, multiply equation 2 by $\frac{1}{2}$ in order to obtain 1 as the coefficient for $x_2$

$$\begin{align} \begin{matrix} x_1-2x_2+x_3 &= 0 \\ x_2-4x_3 &= 4 \\ -3x_2+13x_3 &= -9 \end{matrix} & & \begin{bmatrix} 1 & -2 & 1 &0\\ 0 & 1 & -4 &4\\ 0 & -3 & 13 &-9 \end{bmatrix} \end{align}$$

Use the $x_2$ in equation 2 to eliminate the $-3x_2$ in equation 3.

$$\begin{align*} 3x_2-12x_3 &= 12 \\ -3x_2+13x_3 &= -9 \\ \hline x_3 &= 3 \end{align*}$$

The new system has a triangular form.

$$\begin{align} \begin{matrix} x_1-2x_2+x_3 &= 0 \\ x_2-4x_3 &= 4 \\ x_3 &= 3 \end{matrix} & & \begin{bmatrix} 1 & -2 & 1 & 0\\ 0 & 1 & -4 & 4\\ 0 & 0 & 1 & 3 \end{bmatrix} \end{align}$$

Eventually, you want to eliminate $-2x_2$ the term from equation 1, but it is more efficient to use the $x_3$ term in equation 3 first to eliminate the $-4x_3$ and $x_3$ terms in equations 2 and 1.

$$\begin{align} \begin{matrix} &&4x_3 &= 12 \\ &x_2&-4x_3 &= 4 \\ \hline &x_2& &= 16 \end{matrix} & & \begin{matrix} &&-x_3 &= -3 \\ x_1&-2x_2&+x_3 &= 0 \\ \hline x_1&-2x_2& &= -3 \end{matrix} \end{align}$$

Now, combine the results of these two operations.

$$\begin{align} \begin{matrix} x_1-2x_2 &= -3 \\ x_2 &= 16 \\ x_3 &= 3 \end{matrix} & & \begin{bmatrix} 1 & -2 & 0 & -3\\ 0 & 1 & 0 & 16\\ 0 & 0 & 1 & 3 \end{bmatrix} \end{align}$$

Move back to the $x_2$ in equation 2, and use it to eliminate the $-2x_2$ above it. Because of the previous work with $x_3$, there is now no arithmetic involving $x_3$ terms. Add 2 times equation 2 to equation 1 and obtain the system:

$$\begin{align} \begin{matrix} x_1 &= 29 \\ x_2 &= 16 \\ x_3 &= 3 \end{matrix} & & \begin{bmatrix} 1 & 0 & 0 & 29\\ 0 & 1 & 0 & 16\\ 0 & 0 & 1 & 3 \end{bmatrix} \end{align}$$

Thus, the only solution of the original system is (29,16,3). To verify that (29,16,3) is a solution, substitute these values into the left side of the original system, and compute.

$$\begin{align*} (29)-2(16)+(3) &= 29-32+3 &= 0 \\ 2(16)-8(3) &= 32-24 &= 8 \\ -4(29)+5(16)+9(3) &= -116+80+27 &= -9 \end{align*}$$

The results agree with the right side(우변) of the original system, so (29,16,3) is a solution of the system. $\blacksquare$

Elementary Row Operations

Three basic operations are used to simplify a linear system.

These Elementary row operations include the following:

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

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

It is important to note that row operations are reversible.

If the augmented matrices of two linear systems are row equivalent, then the two systems have the same solution set.

• elementary row operations (RIS)
• row equivalent
• row operations are reversible.

Existence and Uniqueness of System of Equations

Two fundamental questions about a linear system are as follows:

1. Is the system consistent; that is, does at least one solution exist?
2. If a solution exists, is it the only one; that is, is the solution unique?

Example 3:

Determine if the following system is consistent:

$$\begin{align} x_2-4x_3 &= 8 \\ 2x_1-3x_2+2x_3 &= 1 \\ 5x_1-8x_2+7x_3 &= 1 \end{align} \tag{5}$$

Solution:

• The augmented matrix is

$$\begin{bmatrix} 0 & 1 & -4 & 8 \\ 2 & -3 & 2 & 1 \\ 5 & -8 & 7 & 1 \end{bmatrix}$$

• To obtain an $x_1$ in the first equation, interchange rows 1 and 2:

$$\begin{bmatrix} 2 & -3 & 2 & 1 \\ 0 & 1 & -4 & 8 \\ 5 & -8 & 7 & 1 \end{bmatrix}$$

• To eliminate the $5x_1$ term in the third equation, add $\frac{-5}{2}$ times row 1 to row 3.

$$\begin{bmatrix} 2 & -3 & 2 & 1 \\ 0 & 1 & -4 & 8 \\ 0 & \frac{-1}{2} & 2 & \frac{-3}{2} \end{bmatrix} \tag{6}$$

• Next, use the $x_2$ term in the second equation to eliminate the $-\frac{1}{2}x_2$ term from the third equaton. Add $\frac{1}{2}$ times row 2 to row 3.

$$\begin{bmatrix} 2 & -3 & 2 & 1 \\ 0 & 1 & -4 & 8 \\ 0 & 0 & 0 & \frac{5}{2} \end{bmatrix} \tag{7}$$

• The augmented matrix is now in triangular form. To interpret it correctly, go back to equation notation.

$$\begin{align} 2x_1-3x_2+2x_3 &= 1 \\ x_2-4x_3 &= 8 \\ 0 &= \frac{5}{2} \end{align} \tag{8}$$

• The equation $0=\frac{5}{2}$ is a short form of $0x_1+0x_2+0x_3=\frac{5}{2}$.
• There are no values of $x_1,x_2,x_3$ that satisfy (8) because the equation $0=\frac{5}{2}$ is never true.
• Since (8) and (5) have the same solution set, the original system is inconsistent (i.e., has no solution).