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COLLEGE ALGEBRA. LIAL HORNSBY SCHNEIDER. 5.2. Matrix Solution of Linear Systems. The Gauss-Jordan Method Special Systems. Matrix Solutions.

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### COLLEGE ALGEBRA

LIAL

HORNSBY

SCHNEIDER

Matrix Solution of Linear Systems

The Gauss-Jordan Method

Special Systems

Since systems of linear equations occur in so many practical situations, computer methods have been developed for efficiently solving linear systems. Computer solutions of linear systems depend on the idea of a matrix (plural matrices), a rectangular array of numbers enclosed in brackets. Each number iscalled an element of the matrix.

Matrices in general are discussed in more detail later in this chapter. In this section, we develop a method for solving linear systems using matrices. As an example, start with a system and write the coefficients of the variables and the constants as a matrix, called the augmented matrix of the system.

Linear system of equations

The vertical line, which is optional, separates the coefficients from the constants. Because this matrix has 3 rows (horizontal) and 4 columns (vertical), we say its size is 3  4 (read “three by four”).The number of rows is always given first. To refer to a number in the matrix, use its row and column numbers.

For any augmented matrix of a system of linear equations, the following row transformations will result in the matrix of an equivalent system.

1. Interchange any two rows.

2. Multiply or divide the elements of any row by a nonzero real number.

3. Replace any row of the matrix by the sum of the elements of that row and a multiple of the elements of another row.

These transformations are just restatements in matrix form of the transformations of systems discussed in the previous section. From now on, when referring to the third transformation, “a multiple of the elements of a row” will be abbreviated as “a multiple of a row.”

The Gauss-Jordan method is a systematic technique for applying matrix row transformations in an attempt to reduce a matrix to diagonal form, with 1s along the diagonal, such as

from which the solutions are easily obtained. This form is also called reduced-row echelon form.

Step 1 Obtain 1 as the first element of the first column.

Step 2 Use the first row to transform the remaining entries in the first column to 0.

Step 3 Obtain 1 as the second entry in the second column.

Step 4 Use the second row to transform the remaining entries in the second column to 0.

Step 5 Continue in this manner as far as possible.

Note FormThe Gauss-Jordan method proceeds column by column, from left

to right.

When you are working with a particular column, no row operation should undo the form of a preceding column.

Example 1

Solve the system.

Solution

Both equations are in the same form, with variable terms in the same order on the left, and constant terms on the right.

Write the augmented matrix.

Example 1

The goal is to transform the augmented matrix into one in which the value of the variables will be easy to see. That is, since each column in the matrix represents the coefficients of one variable, the augmented matrix should be transformed so that it is of the form for real numbers k and j. Once the augmented matrix is in this form, the matrix can be rewritten as a linear system to get

This form is our goal.

Example 1

It is best to work in columns beginning in each column with the element that is to become 1.

In the augmented matrix

3 is in the first row, first column position. Use transformation 2, multiplying each entry in the first row by to get 1 in this position. (This step is abbreviated as ⅓R1.)

Example 1

Introduce 0 in the second row, first column by multiplying each element of the first row by –5 and adding the result to the corresponding element in the second row, using transformation 3.

Example 1

Obtain 1 in the second row, second column by multiplying each element of the second row by using transformation 2.

Example 1

Finally, get 0 in the first row, second column by multiplying each element of the second row by and adding the result to the corresponding element in the first row.

Example 1

This last matrix corresponds to the system

that has solution set {(3, 2)}. We can read this solution directly from the third column of the final matrix. Check the solution in both equations of the original system.

Example 2

Solve the system.

Solution

Example 2

There is already a 1 in the first row, first column. Introduce 0 in the second row of the first column by multiplying each element in the first row by –3 and adding the result to the corresponding element in the second row.

Example 2

To change the third element in the first column to 0, multiply each element of the first row by –1, and add the result to the corresponding element of the

third row.

Example 2

Use the same procedure to transform the second and third columns. For both of these columns, perform the additional step of getting 1 in the appropriate position

of each column. Do this by multiplying the elements of the row by the reciprocal of the number in that position.

Example 2

The linear system associated with this final matrix is

The solution set is {(1, 2, –1)}. Check the solution in the original system.

Example 3

Use the Gauss-Jordan method to solve the system.

Solution

Example 3

The next step would be to get 1 in the second row, second column. Because of the 0 there, it is impossible to go further. Since the second row corresponds to the equation 0x + 0y = 1, which has no solution, the system is inconsistent and the solution set is ø.

Example 4

Use the Gauss-Jordan method to solve the system.

Solution

Recall from the previous section that a system with two equations in three variables usually has an infinite number of solutions. We can use the

Gauss-Jordan method to give the solution with z arbitrary.

Example 4

It is not possible to go further with the Gauss-Jordan method.

Example 4

The equations that correspond to the final matrix are

Solve these equations for x and y, respectively.

The solution set, written with z arbitrary, is

When matrix methods are used to solve a system of linear equations and the

resulting matrix is written in diagonal form:

1. If the number of rows with nonzero elements to the left of the vertical line is equal to the number of variables in the system, then the system has a single solution. See Examples 1 and 2.

2. If one of the rows has the form [0 0    0 a] with a ≠ 0, then the system has no solution. See Example 3.

3. If there are fewer rows in the matrix containing nonzero elements than the number of variables, then the system has either no solution or infinitely many solutions. If there are infinitely many solutions, give the solutions in terms of one or more arbitrary variables. See Example 4.