Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 1

Section 2.2

Solution of Linear Systems by Gauss-Jordan Method

A matrix is an ordered rectangular array of numbers, letters, symbols or algebraic

expressions. A matrix with 𝑚 rows and 𝑛 columns has size or dimension 𝑚 × 𝑛.

The real numbers that make up the matrix are called entries or elements of the matrix. The

entry in the ith row and jth column is denoted by ija

A matrix with only one column or one row is called a column matrix (or column vector) or

row matrix (or row vector), respectively.

Example 1: Given

1131

20100

935

772

A,

a. what is the dimension of A?

b. identify 𝑎43.

c. identify 𝑎21.

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 2

Systems of Linear Equations in Matrix Form

In order to write a system of linear equations in matrix form, first make sure the like

variables occur in the same column. Then we’ll leave out the variables of the system and simply

use the coefficients and constants to write the matrix form.

Given the following system of equations:

2𝑥 + 4𝑦 + 6𝑧 = 223𝑥 + 8𝑦 + 5𝑧 = 27−𝑥 + 𝑦 + 2𝑧 = 2

The coefficient matrix is:

211

583

642

The constant matrix is: (22272)

The augmented matrix is:

2

27

22

|

|

|

211

583

642

Example 2: Give the coefficient, constant and augmented matrix for the system of equations.

2𝑥 − 4𝑦 = 15−3𝑦 + 2𝑧 = 9𝑥 + 𝑦 − 3𝑧 = −8

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 3

As you may recall from College Algebra, you can solve a system of linear equations in two

variables easily by applying the substitution or addition method. Since these methods become

tedious when solving a large system of equations, a suitable technique for solving such systems

of linear equations will consist of Row Operations. The sequence of operations on a system of

linear equations are referred to equivalent systems, which have the same solution set.

Row Operations

1. Interchange any two rows.

531

312 21 RR

312

531

2. Multiply any row by a nonzero constant

824

312 22

4

1RR

22

11

312

3. Replace any row by the sum of that row and a constant multiple of any other row.

312

531 221 RRR2

770

531

Reduced Row Echelon Form(RREF)

An 𝑚× 𝑛 augmented matrix is in row-reduced form if it satisfies the following conditions:

1. Each row consisting entirely of zeros lies below any other row having nonzero entries.

210

000

301

the correct row-reduced form

000

210

301

2. The first nonzero entry in each row is 1 (called a leading 1).

5100

3020

1001

the correct row-reduced form

51002

3010

1001

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 4

3. If a column contains a leading 1, then the other entries in that column are zeros.

1100

3221 the correct row-reduced form

1100

1021

4. In any two successive (nonzero) rows, the leading 1 in the lower row lies to the right of the

leading 1 in the upper row.

301

210 the correct row-reduced form

210

301

Example 3: Determine which of the following matrices are in row-reduced form. If a matrix is

not in row-reduced form, state which condition is violated.

a.

2100

0010

0001

d.

0000

3100

0091

b.

0000

0010

0021

e.

6200

4010

3001

c.

2010

3001

0000

f.

501

110

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 5

The Gauss-Jordan Elimination Method

1. Write the augmented matrix corresponding to the linear system.

2. Use row operations to write the augmented matrix in row reduced form. If at any point a row

in the matrix contains zeros to the left of the vertical line and a nonzero number to its right, stop

the process, as the problem has no solution.

3. Read off the solution(s).

There are three types of possibilities after doing this process.

Unique Solution

Example 4: The following augmented matrix in row-reduced form is equivalent to the

augmented matrix of a certain system of linear equations. Use this result to solve the system of

equations.

(a) (1 0 40 1 −2

) (b) (1 0 0 20 1 0 −70 0 1 3

)

Example 5: Solve the system of linear equations using the Gauss-Jordan elimination method.

1y3x2

1y2x

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 6

Example 6: Solve the system of linear equations using the Gauss-Jordan elimination method.

3𝑥 + 3𝑦 = 10 + 𝑧𝑥 + 5𝑧 = −6 + 𝑦𝑥 + 3𝑦 + 2𝑧 = 5

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 7

Infinite Number of Solutions

Example 7: The following augmented matrix in row-reduced form is equivalent to the

augmented matrix of a certain system of linear equations. Use this result to solve the system of

equations.

0000

2510

3101

Example 8: Solve the system of linear equations using the Gauss-Jordan elimination method.

3532

123

232

zyx

zyx

zyx

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 8

Example 9: Solve the system of linear equations using the Gauss-Jordan elimination method.

𝑥 + 2𝑦 + 3𝑧 − 𝑤 = 42𝑥 + 3𝑦 + 𝑤 = −33𝑥 + 5𝑦 + 3𝑧 = 1

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 9

A System of Equations That Has No Solution

In using the Gauss-Jordan elimination method the following equivalent matrix was obtained

(note this matrix is not in row-reduced form, let’s see why):

1000

1440

1111

Look at the last row. It reads: 0x + 0y + 0z = -1, in other words, 0 = -1!!! This is never true.

So the system is inconsistent and has no solution.

Systems with No Solution

If there is a row in the augmented matrix containing all zeros to the left of the vertical line and a

nonzero entry to the right of the line, then the system of equations has no solution.

Example 10: Solve the system of linear equations using the Gauss-Jordan elimination method.

2𝑥 − 3𝑦 = 13𝑥 + 𝑦 = −1𝑥 − 4𝑦 = 15

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 10

Example 11: Solve the system of linear equations using the Gauss-Jordan elimination method.

−𝑥 + 3𝑦 − 4𝑧 = 124𝑥 − 12𝑦 + 16𝑧 = −36

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 11

Example 12: A convenience store sells 23 sodas one summer afternoon in 12-, 16-, and 20-oz

cups (small, medium, and large). The total volume of soda sold was 376 oz.

(a) Suppose that the prices for a small, medium, and large soda are $1, $1.25, and $1.40,

respectively, and that the total sales were $28.45. How many of each size did the store sell?

Let 𝑥 = the number of small soda;

𝑦 = the number of medium soda;

𝑧 = the number of large soda.

Soda Sales

Small Medium Large Total

Number

Weight

Price

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 12

(b) Suppose the prices for small, medium, and large sodas are changed to $1, $2, and $3,

respectively, but all other information is kept the same. How many of each size did the store

sell?

Soda Sales

Small Medium Large Total

Number

Weight

Price

Section 2.2 Solutions of Linear Systems by the Gauss-Jordan Method Page | 13

(c) Suppose the prices are the same as in part (b), but the total revenue is $48. Now how many

of each size did the store sell?

Soda Sales

Small Medium Large Total

Number

Weight

Price

(d) Give the solutions from part (c) that have the smallest and largest numbers of large sodas.