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MATRICES AND SYSTEMS OF EQUATIONS

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• Write matrices and identify their orders.

• Perform elementary row operations on matrices.

• Use matrices and Gaussian elimination to solve

systems of linear equations.

• Use matrices and Gauss-Jordan elimination to

solve systems of linear equations.

What You Should Learn

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Matrices

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Matrices

A matrix having m rows and n columns is said to be of order m n.

If m = n, the matrix is square of order m m (or n n). For a square

matrix, the entries a11, a22, a33, . . . are the main diagonal entries.

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Example 1 – Order of Matrices

Determine the order of each matrix.

a. b.

Solution:

a. This matrix has one row and one column. The order of

the matrix is 1 1.

b. This matrix has one row and four columns. The order of

the matrix is 1 4.

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Example 1 – Order of Matrices

Determine the order of each matrix.

c. d.

Solution:

c. This matrix has two rows and two columns. The order of

the matrix is 2 2.

d. This matrix has three rows and two columns. The order

of the matrix is 3 2.

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Matrices

Row matrix: A matrix that has only one row

Column matrix: Amatrix that has only one column

Augmented matrix of the system: A matrix derived from a

system of linear equations (each written in standard form

with the constant term on the right)

Coefficient matrix of the system: the matrix derived from

the coefficients of the system (but not including the

constant terms)

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Matrices

System:

x – 4y + 3z = 5

–x + 3y – z = –3

2x – 4z = 6

Augmented Matrix: Coefficient Matrix:

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Elementary Row Operations

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Elementary Row Operations

1. Interchange two equations.

Interchange two rows

2. Multiply an equation by a nonzero constant.

Multiply a row by a nonzero constant

3. Add a multiple of an equation to another equation.

Add a multiple of a row to another row

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Elementary Row Operations

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Gaussian Elimination with

Back-Substitution

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Example 6 – Gaussian Elimination with Back-Substitution

Solve the system

y + z – 2w = –3

x + 2y – z = 2

2x + 4y + z – 3w = – 2

x – 4y – 7z – w = –19

Solution:

Write augmented matrix.

.

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Example 6 – Solution

cont’d

Interchange R1 and R2 so

first column has leading 1

in upper left corner.

Perform operations on R3

and R4 so first column has

zeros below its leading 1.

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Example 6 – Solution

cont’d

Perform operations on R4

so second column has

zeros below its leading 1.

Perform operations on R3

and R4 so third and fourth

columns have leading 1’s.

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Example 6 – Solution

The matrix is now in row-echelon form, and the

corresponding system is

x + 2y – z = 2

y + z – 2w = –3

z – w = –2

w = 3

Using back-substitution, you can determine that the

solution is x = –1, y = 2, z = 1, and w = 3.

cont’d

.

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Gaussian Elimination with Back-Substitution

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Gauss-Jordan Elimination

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Gauss-Jordan Elimination

With Gaussian elimination, elementary row operations are

applied to a matrix to obtain a (row-equivalent) row-echelon

form of the matrix.

A second method of elimination, called Gauss-Jordan

elimination, continues the reduction process until a

reduced row-echelon form is obtained.

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Example 8 – Gauss-Jordan Elimination

Use Gauss-Jordan elimination to solve the system

x – 2y + 3z = 9

–x + 3y = –4

2x – 5y + 5z = 17

Solution:

The row-echelon form of the linear system can be obtained

using the Gaussian elimination.

.

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Example 8 – Solution

Now, apply elementary row operations until you obtain

zeros above each of the leading 1’s, as follows.

Perform operations on R1

so second column has a

zero above its leading 1.

Perform operations on R1

and R2 so third column has

zeros above its leading 1.

cont’d

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Example 8 – Solution

The matrix is now in reduced row-echelon form. Converting

back to a system of linear equations, you have

x = 1

y = –1.

z = 2

Now you can simply read the solution, x = 1, y = –1, and

z = 2 which can be written as the ordered triple(1, –1, 2).

cont’d

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10000

1000

100

10

1

10000

01000

00100

00010

00001

Gaussian Elimination with

Back substitution

Gaussian Jordon