slide 9.1- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Matrices and Systems of Equations

Learn the definition of a matrix.Learn to use matrices to solve a system of linear equations.Learn to use the Gaussian elimination procedure.Learn to use the Gauss-Jordan elimination procedure.

SECTION 9.1

1

2

3

4


DEFINITION OF A MATRIX

A

a11 a12 ... a1n

a21 a22 ... a2n

M M M

am1 am2 ... amn

A matrix is a rectangular array of numbers denoted by

Row 1

Row 2

Row m

Column 1 Column 2 Column n



A aij .

If a matrix A has m rows and n columns, then A is said to be of order m by n (written m n).

The entry or element in the ith row and jth column is a real number and is denoted by the double-subscript notation aij. The entry aij is sometimes referred as the (i, j)th entry or the entry in the (i, j) position, and we often write



We also write Amn to indicate that the matrix A has m rows and n columns.

If m = n then A is called a square matrix of order n and is denoted by An.

The entries a11, a22, …, ann form the main diagonal of An.A 1 n matrix is called a row matrix, and an n 1 matrix is called a column matrix.


MATRIX AND LINEAR SYSTEMS

We can display all the numerical information contained in a linear system in an augmented matrix of the system.

x y z 1

2x 3y z 10

x y 2z 0

1 1 1

2 3 1

1 1 2

1

10

0

Coefficients of zCoefficients of y

Coefficients of x

Constants


ELEMENTARY ROW OPERATIONS

Two matrices are row equivalent if one can be obtained from the other by a sequence of elementary row operations.

Row Operation In Symbols Description

Interchange two rows Ri Rj Interchange the ith and jth rows

Multiply a row by a nonzero constant

cRj Multiply the jth row by c.

Add a multiple of one row to another row

cRi + Rj Rj Replace the jth row by adding c times jth row to it.


ROW-ECHELON FORM ANDREDUCED ROW-ECHELON FORM

An m n matrix is in row-echelon form if it has the following three properties:

1. All nonzero rows are above the rows consisting entirely of zeros.

2. The leading entry of each nonzero row is 1.3. For two successive rows, the leading 1 in the

higher row is farther to the left of the leading 1 in the lower row.


ROW-ECHELON FORM ANDREDUCED ROW-ECHELON FORM

If a matrix in row-echelon form has the following additional property, then it is in reduced row-echelon form:

4. Each leading 1 is the only nonzero entry in its column.


PROCEDURE FOR SOLVING LINEAR SYSTEMS BY USING GAUSSIAN ELIMINATIONStep 1. Write the augmented matrix.Step 2. Use elementary row operations to

transform the augmented matrix into row-echelon form.

Step 3. Write the system of linear equations that correspond to the matrix in row-echelon form that was obtained in Step 2.

Step 4. Use the system of equations obtained in Step 3, together with back-substitution, to find the solution set of the system.


EXAMPLE 5Solving a System by Using Gaussian Elimination

Solve the system of equations by using Gaussian elimination.

2x y z 6

3x 4y 2z 4

x y z 2

SolutionStep 1 The augmented matrix of the system is.

A 2 1 1

3 4 2

1 1 1

6

4

2



Solution continuedStep 2

R1 R3u ruuuuuuu

1 1 1

3 4 2

2 1 1

2

4

6

3R1 R2 R2u ruuuuuuuuuuuuu 2R1 R3 R3u ruuuuuuuuuuuuuu

1 1 1

0 1 1

0 1 3

2

2

10




R2u ruu

1 1 1

0 1 1

0 1 3

2

2

10

R2 R3 R3u ruuuuuuuuuuuu

1 1 1

0 1 1

0 0 4

2

2

12




1

4R3

u ruuu

1 1 1

0 1 1

0 0 1

2

2

3

x y z 2 (1)

y z 2 (2)

z 3 (3)

This is in row-echelon form.

Step 3 The system corresponding to the last matrix in Step 2 is



Solution continued

y z 2

y 3 2

y 1

Back-substitute z = 3 and y = –1 in Equation 1.

Step 4 Equation (3) in Step 3 gives the value z = 3. Back-substitute z = 3 in Equation (2).

x y z 2

x 1 3 2

x 2



Solution continued

The solution set for the system is {(2, –1, 3)}.

You should check the solution by substituting these values for x, y, and z into the original system of equations.


EXAMPLE 7Solving a System of Equations by Gauss-Jordan Elimination

Solve the system given in Example 4 by Gauss-Jordan elimination. Recall that the given system is x y z 1

2x 3y z 10

x y 2z 0

SolutionThe augmented matrix of the system is.

A 1 1 1

2 3 1

1 1 2

1

10

0



Solution continued

2R1 R2 R2

1 R1 R3 R3u ruuuuuuuuuuuuuuuu

1 1 1

0 1 3

0 2 1

1

8

1

1 R2u ruuuuu

1 1 1

0 1 3

0 2 1

1

8

1



Solution continued

2R2 R3 R3u ruuuuuuuuuuuuuu

1 1 1

0 1 3

0 0 5

1

8

15

1

5R3

u ruuu

1 1 1

0 1 3

0 0 1

1

8

3

B



Solution continued

R2 R1 R1u ruuuuuuuuuuu

1 0 4

0 1 3

0 0 1

7

8

3

4R3 R1 R1

3R3 R2 R2u ruuuuuuuuuuuuu

1 0 0

0 1 0

0 0 1

5

1

3



Solution continued

We now have an equivalent matrix in reduced row-echelon form. The corresponding system of equations for the last augmented matrix is:

x 5

y 1

z 3

Hence, the solution set is {(5, 1, 3)}, as in Example 4.

slide 9.1- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley

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