(I) Introduction

Definition: Linear Equation

A linear equation in the variables \(x_1, \dots, x_n\) is an equation that can be written in the form:

\[ c_1 x_1 + c_2 x_2 + \dots + c_n x_n = b \]

where \(b\) and the coefficients \(c_1, \dots, c_n\) are real or complex numbers, usually known in advance. The subscript \(n\) may be any positive integer.

Examples of linear equations:

These are linear because they can be rearranged algebraically into the standard form. For example:

Examples that are NOT linear equations:

Definition: System of Linear Equations

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

Example System:

\[ \begin{cases} 3x_1 - 5x_2 = -2 \\ 2x_1 + x_2 - x_3 = 2\sqrt{6} \end{cases} \]

The solution set is the set of all possible solutions. Two linear systems are called equivalent if they have the same solution set.

Possible Solutions for a Linear System

Consider two linear equations in two variables. Geometrically, this represents finding the intersection of two lines. A linear system has either:

  1. Exactly one solution
  2. No solution
  3. Infinitely many solutions

A system is called consistent if it has one or infinitely many solutions. A system is called inconsistent if it has no solution.

(II) Matrix Notation

Example 1

Given the system (System 1):

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

With the coefficients of each variable aligned in columns, we can form the coefficient matrix (size \(3 \times 3\)):

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

By including the constants from the right side, we form the augmented matrix (size \(3 \times 4\)):

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

The size of a matrix describes its dimensions. An \(m \times n\) matrix is a rectangular array of numbers with \(m\) rows and \(n\) columns.

(III) Solving a Linear System

Strategy:

  1. Use the \(x_1\) term in the first equation to eliminate \(x_1\) terms in other equations.
  2. Use the \(x_2\) term in the second equation to eliminate \(x_2\) terms in other equations.
  3. Repeat until we finally obtain a very simple equivalent system.
Example 2

Solve System (1) from Example 1.

Step 1: Keep \(x_1\) in the 1st equation and eliminate it from other equations. We apply the row operation \(R_3 = -5R_1 + R_3\):

\[ \begin{bmatrix} 1 & -2 & 1 & 0 \\ 0 & 2 & -8 & 8 \\ 5 & 0 & -5 & 10 \end{bmatrix} \xrightarrow{R_3 = -5R_1 + R_3} \begin{bmatrix} 1 & -2 & 1 & 0 \\ 0 & 2 & -8 & 8 \\ 0 & 10 & -10 & 10 \end{bmatrix} \]

Step 2: Multiply Equation 2 by \(\frac{1}{2}\) and Equation 3 by \(\frac{1}{10}\) to obtain 1 as the coefficient of \(x_2\) in the 2nd row.

\[ \begin{bmatrix} 1 & -2 & 1 & 0 \\ 0 & 2 & -8 & 8 \\ 0 & 10 & -10 & 10 \end{bmatrix} \xrightarrow[\frac{1}{10}R_3]{\frac{1}{2}R_2} \begin{bmatrix} 1 & -2 & 1 & 0 \\ 0 & 1 & -4 & 4 \\ 0 & 1 & -1 & 1 \end{bmatrix} \]

Step 3: Keep \(x_2\) in the 2nd equation and eliminate it from the 3rd equation. We apply \(R_3 = -R_2 + R_3\):

\[ \begin{bmatrix} 1 & -2 & 1 & 0 \\ 0 & 1 & -4 & 4 \\ 0 & 1 & -1 & 1 \end{bmatrix} \xrightarrow{R_3 = -R_2 + R_3} \begin{bmatrix} 1 & -2 & 1 & 0 \\ 0 & 1 & -4 & 4 \\ 0 & 0 & 3 & -3 \end{bmatrix} \]

Step 4: Multiply Row 3 by \(\frac{1}{3}\):

\[ \begin{bmatrix} 1 & -2 & 1 & 0 \\ 0 & 1 & -4 & 4 \\ 0 & 0 & 3 & -3 \end{bmatrix} \xrightarrow{\frac{1}{3}R_3} \begin{bmatrix} 1 & -2 & 1 & 0 \\ 0 & 1 & -4 & 4 \\ 0 & 0 & 1 & -1 \end{bmatrix} \]

Step 5: Work backward to eliminate the terms above the leading 1s. Apply \(R_1 = -R_3 + R_1\) and \(R_2 = 4R_3 + R_2\):

\[ \begin{bmatrix} 1 & -2 & 1 & 0 \\ 0 & 1 & -4 & 4 \\ 0 & 0 & 1 & -1 \end{bmatrix} \xrightarrow[R_2 = 4R_3 + R_2]{R_1 = -R_3 + R_1} \begin{bmatrix} 1 & -2 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1 \end{bmatrix} \]

Step 6: Eliminate \(x_2\) from the 1st row. Apply \(R_1 = 2R_2 + R_1\):

\[ \begin{bmatrix} 1 & -2 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1 \end{bmatrix} \xrightarrow{R_1 = 2R_2 + R_1} \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1 \end{bmatrix} \]

The solution is \(x_1 = 1, x_2 = 0, x_3 = -1\). This system is consistent and unique.

Elementary Row Operations

  1. Replacement: Replace one row by the sum of itself and a multiple of another row. (\(R_2' = cR_1 + R_2\))
  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. If the augmented matrices of two linear systems are row equivalent, then the two systems have the same solution set. Note that these operations are reversible.

(IV) Existence and Uniqueness Questions

There are two fundamental questions about a linear system:

  1. Existence: Is the system consistent? Does a solution exist?
  2. Uniqueness: If a solution exists, is it the only one (exactly one solution), or are there infinitely many solutions?
Example 3

Determine if the following system is consistent:

\[ \begin{cases} 0x_1 + x_2 - 4x_3 = 8 \\ 2x_1 - 3x_2 + 2x_3 = 1 \\ 4x_1 - 8x_2 + 12x_3 = 1 \end{cases} \]

Solution: Form the augmented matrix and perform row operations.

Step 1: Interchange Row 1 and Row 2.

\[ \begin{bmatrix} 0 & 1 & -4 & 8 \\ 2 & -3 & 2 & 1 \\ 4 & -8 & 12 & 1 \end{bmatrix} \xrightarrow{R_1 \leftrightarrow R_2} \begin{bmatrix} 2 & -3 & 2 & 1 \\ 0 & 1 & -4 & 8 \\ 4 & -8 & 12 & 1 \end{bmatrix} \]

Step 2: Apply \(R_3 = -2R_1 + R_3\).

\[ \begin{bmatrix} 2 & -3 & 2 & 1 \\ 0 & 1 & -4 & 8 \\ 4 & -8 & 12 & 1 \end{bmatrix} \xrightarrow{R_3 = -2R_1 + R_3} \begin{bmatrix} 2 & -3 & 2 & 1 \\ 0 & 1 & -4 & 8 \\ 0 & -2 & 8 & -1 \end{bmatrix} \]

Step 3: Apply \(R_3 = 2R_2 + R_3\) to reach triangular form.

\[ \begin{bmatrix} 2 & -3 & 2 & 1 \\ 0 & 1 & -4 & 8 \\ 0 & -2 & 8 & -1 \end{bmatrix} \xrightarrow{R_3 = 2R_2 + R_3} \begin{bmatrix} 2 & -3 & 2 & 1 \\ 0 & 1 & -4 & 8 \\ 0 & 0 & 0 & 15 \end{bmatrix} \]

The last row corresponds to the equation \(0x_1 + 0x_2 + 0x_3 = 15\), which is never true. Therefore, the system is inconsistent.

Example 4

Determine the value(s) of \(h\) such that the matrix is the augmented matrix of a consistent linear system.

Problem 1:

\[ \begin{bmatrix} 1 & 3 & -2 \\ -4 & h & 8 \end{bmatrix} \xrightarrow{R_2 = 4R_1 + R_2} \begin{bmatrix} 1 & 3 & -2 \\ 0 & 12+h & 0 \end{bmatrix} \]

The second row represents the equation \((12+h)x_2 = 0\). This equation is consistent for all values of \(h\).

Problem 2:

\[ \begin{bmatrix} 2 & -3 & h \\ -6 & 9 & 5 \end{bmatrix} \xrightarrow{R_2 = 3R_1 + R_2} \begin{bmatrix} 2 & -3 & h \\ 0 & 0 & 3h+5 \end{bmatrix} \]

The second row represents the equation \(0x_1 + 0x_2 = 3h+5\). For the system to be consistent, we must have \(3h+5 = 0\), which means \(h = -5/3\).