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College Algebra
Section 8.2: Matrix Notation and Gaussian
Elimination
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Objectives
o Linear systems, matrices, and augmented matrices.
o Gaussian elimination and row echelon form.
o Gauss-Jordan elimination and reduced row echelon
form.
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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.
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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
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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
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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
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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.
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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

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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
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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.
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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.
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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 )
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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…
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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…
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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.
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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  
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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 
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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.
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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 
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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 .