### COL_8.2

```HAWKES LEARNING SYSTEMS
math courseware specialists
Hawkes Learning Systems
College Algebra
Section 8.2: Matrix Notation and Gaussian
Elimination
HAWKES LEARNING SYSTEMS
math courseware specialists
Objectives
o Linear systems, matrices, and augmented matrices.
o Gaussian elimination and row echelon form.
o Gauss-Jordan elimination and reduced row echelon
form.
HAWKES LEARNING SYSTEMS
math courseware specialists
Linear Systems and Matrices
Matrices and Matrix Notation
A matrix is a rectangular array of numbers, called elements
or entries of the matrix. They naturally form rows and
columns. We say that a matrix with m rows and n columns
is an m  n matrix (read “m by n”), or of order m  n . By
convention, the number of rows is always stated first.
5 9 7 
A


2
6
0


A is a 2x3 matrix.
HAWKES LEARNING SYSTEMS
math courseware specialists
Linear Systems and Matrices
Matrices are often labeled with capital letters. The same
letter in lower case, with a pair of subscripts attached, is
usually used to refer to its individual elements. For instance,
if A is a matrix, aij refers to the element in the i th row and
the j th column of A.
5 9 7 
A


2
6
0


a12  9
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 1: Linear Systems and Matrices
Given the matrix below, determine the following:
 5  9
 2 6 

A
 7 0
 5 3 


a. The order of A . A has 4 rows and 2 columns, so A is a 4x2 matrix.
4 2
b. The value of a32. The first subscript refers to the row and the
second subscript refers to the column, so find the
0
entry in the 3rd row and 2nd column.
c. The value of a11 . Similarly, find the entry in the 1st row, 1st column.
5
HAWKES LEARNING SYSTEMS
math courseware specialists
Linear Systems and Augmented Matrices
Augmented Matrices
The augmented matrix of a linear system of equations is a
matrix consisting of the coefficients of the variables, with
an adjoined column consisting of the constants from the
right-hand side of the system. The matrix of coefficients and
the column of constants are customarily separated by a
vertical bar.
For example, the augmented matrix for the system
3 x  7 y  4
 3  7  4
is 
.


9
 x  4 y  9
 1 4
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 2: Linear Systems and Augmented Matrices
Construct the augmented matrix for the linear system.
 2 x  5 y  1  6 z
 6 x  12 z
 2 y

3

 z  3 x  7 y  9
 2x  5 y  6z  1

 2x  y  4z  2
3x  7 y  z  9

 2 5 6 1
 2 1  4 2


 3 7 1 9 
Our first step is to
write each
equation in
standard form.
Now we can
convert the
coefficients and
constants into an
augmented matrix.
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 3: Linear Systems and Augmented Matrices
Construct the linear system for the augmented matrix.
 0 8  3 5 12  First, we need to assign each of the
 9 1 0 0  7  coefficient columns to a variable.


5 Now we can create the system of equations.
 2 0 0 2
a b c d
 0a  8b  3c  5d  12

9a  b  0c  0d  7
2a  0b  0c  2d  5

 8b  3c  5d  12

9a  b  7
2a  2d  5

HAWKES LEARNING SYSTEMS
math courseware specialists
Gaussian Elimination and Row Echelon Form
Consider the following augmented matrix.
1  2 3 10 
0 1 4  7 


0 0 1  1 
If we translate this back into system form we obtain
 x  2 y  3z  10

y  4 z  7


z  1

and can easily solve for the variables by back substitution.
z  1
y  4  1  7
x  2  3  3 1  10
x7
y  3
HAWKES LEARNING SYSTEMS
math courseware specialists
Gaussian Elimination and Row Echelon Form
The point of Gaussian elimination is that it transforms an
arbitrary augmented matrix into a form (called row echelon
form) like the one on the previous slide.
Row Echelon Form
A matrix is in row echelon form if:
1. The first non-zero entry in each row is 1.
2. Every entry below each 1 (called a leading 1) is 0, and
each leading 1 appears one digit farther to the right
than the leading 1 in the previous row.
3. All rows consisting entirely of 0’s appear at the bottom.
HAWKES LEARNING SYSTEMS
math courseware specialists
Row Echelon Form
The matrix below is in row echelon form.
1 9 1 1 
A  0 1 0 2


0 0 1 3 
However, the matrix below is not in row echelon form
1 2 3 4 


B 0 4 7 2


0 6 0 0 
because the first non-zero entries in the second and
third rows are not 1.
HAWKES LEARNING SYSTEMS
math courseware specialists
Gaussian Elimination and Row Echelon Form
Elementary Row Operations
Assume A is an augmented matrix corresponding to a given
system of equations. Each of the following operations on A
results in the augmented matrix of an equivalent system. In
the notation, Ri refers to row i of the matrix A.
1. Rows i and j can be interchanged. (Denoted Ri  R j )
2. Each entry in row i can be multiplied by a non-zero
constant c . (Denoted cRi )
3. Row j can be replaced with the sum of itself and a
constant multiple of row i . (Denoted cRi  R j )
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 4: Gaussian Elimination and Row
Echelon Form
Use Gaussian Elimination to solve the system.
7 x  y  4 z  11

 x  3y  2 z  13
6 x  2 y  3z  22

R1  R2
Augmented
matrix form
7 1  4 11 
1 3  2 13 


6  2 3  22 
1 3  2 13  7R  R
7 1  4 11  1 2

 6R1  R3
6  2 3  22 
3  2 13 
1
0  20 10  80 


0  20 15  100 
Continued on the next slide…
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 4: Gaussian Elimination and Row
Echelon Form (Cont.)
3  2 13 
1
0  20 10  80 


0  20 15  100 
20R2  R3
1 3  2 13 


1
0 1
4
2


0 0 5  20 


 1 
  R2
 20 
1
  R3
5
1

0

0

1

0

0

3  2 13 

1
1
4 
2

 20 15  100 
3  2 13

1
1
4
2

0
1  4 
The final matrix is in row echelon form.
Continued on the next slide…
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 4: Gaussian Elimination and Row
Echelon Form (Cont.)
1 3  2 13


1
0 1
4
2


0 0

1

4


Now we can solve for x, y and z.
z  4 Given by the last row of the matrix.
Plug the value found for z into the equation
1
y
4  4 given by the 2nd row of the matrix.
2
 
y2
x  3 2   2  4   13
x  1
 1,2, 4
Plug the values found for y and z into the 1st row
of the matrix.
The solution set to this system.
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 5: Gaussian Elimination and Row
Echelon Form
Use Gaussian Elimination to solve the system.
 x  y  9 z  16

 x  3y  4 z  6
2 x  6 y  38z  10

R1  R2
2R1  R3
1 1  9  16 
0  2  5  10 


0 8 20  22 
 1 1  9  16
Augmented
 1  3 4

6 matrix form


 2 6 38 10 
4R2  R3
1 1  9  16 
0  2  5  10 


0 0 0  62 
We can stop here because 0  62 is a false statement. Therefore,
the solution set  
HAWKES LEARNING SYSTEMS
math courseware specialists
Gauss-Jordan Elimination and Reduced Row
Echelon Form
The goal of Gauss-Jordan elimination is to put a given
matrix into reduced row echelon form.
Reduced Row Echelon Form
A matrix is said to be in reduced row echelon form if:
1. It is in row echelon form.
2. Each entry above a leading 1 is also 0.
For example, the following matrix is in reduced row
echelon form. 1 0 0 2 


0 1 0 17 
0 0 1 3 
HAWKES LEARNING SYSTEMS
math courseware specialists
Gauss-Jordan Elimination and Reduced Row
Echelon Form
Consider the last matrix obtained in Example 4.
1 3 2 13  1
5
R3  R2 1 3 0


2
1
0 1 0 2 
0 1 
4

2

 2R3  R1 
0 0 1 4 
0 0 1 4 


3R2  R1
1 0 0 1
0 1 0 2 


0 0 1 4 
Reduced row echelon
form.
Now we can write
the system as
 x  1

y  2
 z  4

which is equivalent to the original system, but in a form that tells us the
solution to the system.
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 6: Gauss-Jordan Elimination and
Reduced Row Echelon Form
Use Gauss-Jordan elimination to solve the system.
 x  2 y  3 z  7

3x    y   5 z  7
8 x  10 y  19 z  1

3R1  R2
8R1  R3
2R2  R1
6R2  R3
1 2 3 7 


3

1
5
7


8 10 19 1
1 2 3 7 


0

7
14
28


0 6 43 55 
1
R2
7
1 2 3 7 


0
1

2

4


0 6 43 55 
1 0 1 1 


0
1

2

4


0 0 31 31 
1
R3
31
1 0 1 1 


0 1 2 4 
0 0 1 1 
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 6: Gauss-Jordan Elimination and
Reduced Row Echelon Form (Cont.)
1 0 1 1 


0 1 2 4 
0 0 1 1 
2R3  R2
1R3  R1
1 0 0 0 


0 1 0 2 
0 0 1 1 
Thus, we can write this in system form
x  0

 y  2
z  1

and the solution set for this system is the ordered triple
 0, 2,1 .
```