Matrices for Systems of Equations

What is reduced row-echelon form?

Reduced row-echelon form of a matrix is the form of a matrix in which:

1. any row with all zeros is at the bottom of the matrix.
2. any row that has an entry other than zero has one as the first non-zero entry.
3. any row that has one as the first non-zero entry has that entry further to the right than the first non-zero entry of the row above.

How do you change a matrix into reduced row-echelon form?

There are a number of moves that can change a matrix into reduced row-echelon form. But before we learn these moves, we will first examine the connections between matrices and linear equations.

For example, a school band decides to purchase xylophones and zithers. If xylophones cost $30.00, and zithers cost $40.00, how many of each should they buy in order to purchase exactly 254 instruments and spend their entire budget of $8,500?

Two Perspectives on the Solution

Let x = the number of xylophones.

Let z = the number of zithers.

x + z = 254

30x + 40z = 8,500

Standard Elimination Method

Matrix Method

x + z = 254

30x + 40z = 8,500

The above equation can also be represented

.

In order to solve the system using matrices, we use moves (called row operations) similar to those on the left. The following row operations are acceptable:

• switching one row for another
• multiplying any row by a constant
• adding two rows together

These row operations may be used in combination.

The amount of writing is reduced by committing the variables to memory, removing the variables from the matrix, and writing the matrix equation in augmented form as shown below.

The row operations used to maneuver the matrix into reduced echelon form are described below.

• Multiply the first equation by –30 and add it to the second.

–30x – 30z = –7,620

30x + 40z = 8,500

10z = 880

• Multiply row 1 by –30 and add it to row 2. This new row becomes row 2.

• Multiply the resulting equation by 0.1.

z
= 88

• Multiply row 2 by 0.1.

This augmented matrix represents

,

meaning x + z = 254,

What is the Gauss-Jordan Elimination Method?

Reduced row-echelon form works well, but the Gauss-Jordan method is better for those who do not like to do back-substitution.

The Gauss-Jordan elimination method is similar to the reduced row-echelon method, but avoids back-substitution by continuing the row operations until the left side of the matrix has a diagonal (in a negative slope) with entries of 1 and all other entries of 0. The entries on the right side of the matrix may or may not be 0.

The matrix below has been subjected to the Gauss-Jordan elimination method. Notice how easy it is to read the solution to this system of four equations.

The first variable is equal to –2.

The second variable is equal to 0.

The third variable is equal to 6.

The fourth variable is equal to
.

How do you perform the Gauss-Jordan Elimination Method?

The approach is to start in the reduced row-echelon method form, but instead of back-substitution, continue row operations. The idea is to first focus on the column above the last initial entry of 1. When all the entries in that row are 0, then do the same on the column to the left. Continue moving to the left until the final form is achieved.

Let’s look at the former example.

For us to convert this matrix to the Gauss-Jordan elimination final form, the only thing we need to do is make the second entry in row 1 into a 0. We can achieve this by multiplying row 2 by –1 and adding it to row 1, thus replacing row 1.

This matrix can also be viewed as

.

Notice how easy it is to identify the solutions x = 166 and z = 88.

Let’s try a more complicated example.

Solve the following system of equations using the Gauss-Jordan elimination method.

2a + b + c = 5

2a + b – 3c = 9

3a + 2b + 5c = 2

The matrix equation is

.

The augmented matrix is

.

• Use row operations to create 1 in the initial position of row 1.

(Row 3 – row 1 becomes row 1.)

• Use row operations to create 0 in the first column, moving down.

(–2 times row 1 + row 2 becomes row 2.)

(–3 times row 1 + row 3 becomes row 3.)

After the first row is complete, the second entry
in row 2 needs to become 1.

(–1 times row 2 becomes row 2.)

• Work to get 0 under the 1 in row 2.

(Row 2 + row 3 becomes row 3.)

• Next, make the third entry in row 3 into a 1.

(Multiply row 3 by 0.25.)

• The above matrix is in reduced echelon form. We continue when using the Gauss-Jordan method. The task is to convert the other positions on the left side of the matrix to zeroes, beginning by moving up in column 3.

(–11 times row 3 + row 2 becomes row 2.)

(–4 times row 3 + row 1 becomes row 1.)

• By moving a column to the left, the entries above the lowest 1 become 0.

(Row 1 – row 2 becomes row 1.)

• The solutions to the system of equations can be read from this final form of the augmented matrix.

a = 5

b = –4

c = –1

Review of New Vocabulary and Concepts

• The reduced row-echelon form, also known as the row reduced echelon form, of a matrix equation requires back-substitution in order to develop the solution to the equation.
• The Gauss-Jordan elimination method, while more complex and time-consuming than reduced row-echelon method, develops a form in which the solution is easy to identify and does not require back-substitution.
• Both methods are ways to solve systems of linear equations.