Download Section 9.3

CHAPTER 9:
Systems of Equations
and Matrices
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
Systems of Equations in Two Variables
Systems of Equations in Three Variables
Matrices and Systems of Equations
Matrix Operations
Inverses of Matrices
Determinants and Cramer’s Rule
Systems of Inequalities and Linear Programming
Partial Fractions
Copyright © 2009 Pearson Education, Inc.
9.3
Matrices and Systems of Equations

Solve systems of equations using matrices.
Copyright © 2009 Pearson Education, Inc.
Matrices
The system
4 x  2 y  3
x  5 y  4
can be expressed as
 4 2 3
1 5 4 


where we have omitted the variables and replaced
the equals signs with a vertical line.
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-4
Matrices
A rectangular array of numbers such as
 4 2 3
1 5 4 


is called a matrix (plural, matrices).
The matrix is an augmented matrix because it
contains not only the coefficients but also the
constant terms.
4 2
The matrix 
is called the coefficient matrix.

1 5 
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-5
Matrices continued
The rows of a matrix are horizontal.
The columns of a matrix are vertical.
The matrix shown has 2 rows and 3 columns.
1 2 3
4 5 6


A matrix with m rows and n columns is said to be of
order m  n.
When m = n the matrix is said to be square.
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-6
Gaussian Elimination with Matrices
Row-Equivalent Operations
1. Interchange any two rows.
2. Multiply each entry in a row by the same
nonzero constant.
3. Add a nonzero multiple of one row to another
row.
We can use the operations above on an augmented
matrix to solve the system.
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-7
Example
Solve the following system:
2x  y  z  8
x  3 y  2 z  1
 5 z  23
4x
First, we write the augmented matrix, writing 0 for the missing
y-term in the last equation.
 2 1 1 8 
1 3 2 1


 4 0 5 23
Our goal is to find a row-equivalent matrix of the form
1 a b
0 1 d

0 0 1
Copyright © 2009 Pearson Education, Inc.
c
e

f 
.
Slide 9.3-8
Example continued
 2 1 1 8 
1 3 2 1


 4 0 5 23
1 3 2 1 New row 1 = row 2
 2 1 1 8  New row 2 = row 1


 4 0 5 23
We multiply the first row by 2 and add it to the second row.
We also multiply the first row by 4 and add it to the third
row.
Row 1 is unchanged
1 3 2 1
0 5 5 10  New row 2=  2(row 1) + row 2


0 12 13 27  New row 3=  4(row 1) + row3
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-9
Example continued
We multiply the second row by 1/5 to get a 1 in the second row,
second column.
1 3 2 1
0 1 1 2  New row 2= 1 (row 2)
5


0 12 13 27 
We multiply the second row by 12 and add it to the third row.
1 3 2 1
0 1 1 2 


0 0 1 3  New row 3=  12(row 2) + row 3
Now, we can write the system of equations that corresponds to
the last matrix above: x  3 y  2 z  1
yz2
z 3
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-10
Example continued
We back-substitute 3 for z in equation (2) and solve for y.
yz2
y3 2
y  1
Next, we back-substitute 1 for y and 3 for z in equation (1) and
x  3 y  2 z  1
solve for x.
x  3(1)  2(3)  1
x  3  6  1
x  3  1
x2
The triple (2, 1, 3) checks in the original system of equations,
so it is the solution.
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-11
Row-Echelon Form
1. If a row does not consist entirely of 0’s, then the first
nonzero element in the row is a 1 (called a leading 1).
2. For any two successive nonzero rows, the leading 1 in the
lower row is farther to the right than the leading 1 in the
higher row.
3. All the rows consisting entirely of 0’s are at the bottom of
the matrix.
If a fourth property is also satisfied, a matrix is said to be in
reduced row-echelon form:
4. Each column that contains a leading 1 has 0’s everywhere
else.
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-12
Example
Which of the following matrices are in row-echelon form?
0 2 4 1 
0 0 0 0 


a)
1 6 7 5 
0 1 3 4 


0 0 1 8 
b)
c)
1 2 7 6 
0 0 0 0 


0 1 4 2 
d) 1 0 0 3.5
0 1 0

0 0 1
0.7 

4.5 
Matrices (a) and (d) satisfy the row-echelon criteria. In (b) the
first nonzero element is not 1. In (c), the row consisting
entirely of 0’s is not at the bottom of the matrix.
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-13
Gauss-Jordan Elimination
We perform row-equivalent operations on a matrix
to obtain a row-equivalent matrix in row-echelon
form. We continue to apply these operations until we
have a matrix in reduced row-echelon form.
Example: Use Gauss-Jordan elimination to solve
the system of equations from the previous example;
we had obtained the matrix
1 3 2 1
0 1 1 2 


0 0 1 3 
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-14
Gauss-Jordan Elimination continued
We continue to perform row-equivalent operations
until we have a matrix in reduced row-echelon form.
1 3 0 5 
0 1 0 1


0 0 1 3 
New row 1 = 2(row 3) + row 1
New row 2 =  1(row 3) + row 2
Next, we multiply the second row by 3 and add it to
the first row.
1 0 0 2 
0 1 0 1


0 0 1 3 
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New row 1 = 3(row 2) + row 1
Slide 9.3-15
Gauss-Jordan Elimination continued
Writing the system of equations that corresponds to
this matrix, we have
2
 1
x
y
z
3
We can actually read the solution, (2, 1, 3), directly
from the last column of the reduced row-echelon
matrix.
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-16
Special Systems
When a row consists entirely of 0’s, the equations
are dependent. For example, in the matrix
1 0 4 6 
0 1 4 8 


0 0 0 0 
the system is equivalent to
x  4 z  6,
y  4 z  8.
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-17
Special Systems
When we obtain a row whose only nonzero entry
occurs in the last column, we have an inconsistent
system of equations. For example, in the matrix
1 0 4 6 
0 1 4 8 


0 0 0 9 
the last row corresponds to the false equation 0 = 9,
so we know the original system has no solution.
Copyright © 2009 Pearson Education, Inc.
Slide 9.3-18