Download Chapter 4-4

Chapter 4
Systems of
Equations and
Inequalities
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 4-1
1
Chapter Sections
4.1 – Solving Systems of Linear Equations in Two
Variables
4.2 – Solving Systems of Linear Equations in
Three Variables
4.3 – Systems of Linear Equations: Applications
and Problem Solving
4.4 – Solving Systems of Equations Using
Matrices
4.5 – Solving Systems of Equations Using
Determinants and Cramer’s Rule
4.6 – Solving Systems of Linear Inequalities
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 4-2
2
§ 4.4
Solving Systems
of Equations
Using Matrices
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 4-3
3
Definitions
A matrix is a rectangular array of numbers within
brackets. The plural of matrix is matrices.
rows
 4  6
2 0 


 4 8  9
 2 3 1 


2 x 3 matrix
columns
# of rows # of columns
The dimensions of a matrix are the number of
rows by the number of columns. The numbers
inside the brackets are the elements of the matrix.
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 4-4
4
Solving Systems
An augmented matrix is a matrix made up of
two smaller matrices separated by a vertical
line.
Equations
2 x  3 y  10
4x  5 y  9
Augmented Matrix
2 - 3 10
4 - 5 9 


Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 4-5
5
Solving Systems
To solve a system of equations using
matrices, rewrite the augmented matrix in
triangular form,
Augmented Matrix
 4 8  9
 2 3 1 


Triangular Form
1 a
0 1

p
q 
where a, p, and q are any constants.
From this matrix we can write an equivalent
system of equations and use substitution to
find the solution.
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 4-6
6
Solving Systems
Example:
Triangular Form
System of Equations
1 2 4
0 1 5 


x  2y  4
y5
The system above can be easily solved by
substitution.
The solution is (-6, 5).
Row transformations are used to rewrite an
augmented matrix into triangular form.
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 4-7
7
Row Transformations
Procedures for Row Transformations
1. All the numbers in a row may be multiplied (or divided) by any
nonzero real number. (This is the same as multiplying both sides of
an equation by any nonzero real number.)
2. All the numbers in a row may be multiplied by any nonzero real
number. These products may then be added to the corresponding in
any other row. (This is equivalent to eliminating a variable from a
system of equations using the addition method.)
3. The order of the rows may be switched. (This is equivalent to
switching the order of the equations in a system of equations.)
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 4-8
8
Solving Systems
Example:
Solve the system by using matrices.
2 x  4 y  2
3x  2 y  5
Rewrite the system as an augmented matrix.
 2 4  2
3  2

5


Remember that we are trying to
get the final matrix in triangular
form:
1 a
0 1

p
q 
We begin by using row transformation procedure 1 to replace
the 2 in the first column, first row, with 1. To do so, we
multiply the first row of numbers by ½ (1/2 R1).
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Chapter 4-9
9
Solving Systems
 1   1 
 2 2  2 4

-2
 3
 1  2 
 
1

2
 
R1
2
5

or
1 2  1
3 - 2 5 


We next use the row transformations to produce a 0 in the first
column, second row where currently a 3 is in this position. We
will multiply the elements in row 1 by -3 and add the products
to row
The elements in the first row multiplied by -3 are:
-3(1)
-3
-3(2)
or
-6
-3(-1)
3
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Chapter 4-10
10
Solving Systems
Now add these products to their respective elements in the
second row to get
2
1 
 1
 3  3 - 6  (-2) 3  5  3R1 R 2


or
1 2  1
0 - 8 8 


We next use row transformation to produce a 1 in the second
row, second column where currently a -8 is in this position.
We do this by multiplying the second row by -1/8.
2
 1
 1
1



   0  - (8)
 8
 8 
1

1

  1   R2 or
  8
 8 
1
0

Copyright © 2015, 2011, 2007 Pearson Education, Inc.
2 1 
1 - 1 
Chapter 4-11
11
Solving Systems
The matrix is now in row echelon form and the equivalent
system of equations is
x + 2y = -1
y = -1
Now we can solve for x using substitution
x  2 y  1
x  2(1)  1
x  2  1
x 1
A check will show that (1, -1)
is the solution to the original
system.
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 4-12
12
Three Variables
To solve a system of three linear equations in three variables,
we use the same row transformation procedures used when
solving a system of two linear equations. Our goal is to
produce an augmented matrix in the row echelon form
1 a b p 
0 1 c q 


0 0 1 r 
where a, b, c, p, q, and r represent numbers. This matrix
represents the following system of equations.
1x  ay  bz  p
x  ay  bz  p
0 x  1 y  cz  q or
y  cz  q
0 x  0 y  1z  r
zr
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 4-13
13
Recognize Inconsistent and Dependent Systems
When solving a system of two equations, if you obtain an
augmented matrix in which one row of numbers on the
lift side of the vertical line is all zeros but a zero does not
appear in the same row on the right side of the vertical
line, the system is inconsistent and has no solution
1 2 5
 0 0 3


Inconsistent
System
1 - 3  4
0 0 0 


Dependent
System
Copyright © 2015, 2011, 2007 Pearson Education, Inc.
Chapter 4-14
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