Transcript Section 6.1 - Canton Local

Chapter 9
Matrices and
Determinants
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All rights reserved
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SECTION 9.1
Matrices and Systems of Equations
OBJECTIVES
1
2
3
4
Define a matrix.
Use matrices to solve a system.
Use Gaussian elimination to solve a system.
Use Gauss–Jordan elimination to solve a
system.
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DEFINITION OF A MATRIX
A matrix is a rectangular array of numbers
denoted by
 a11
a
21

A

a
 m1
a12
a22
am 2
... a1n 

... a2 n



... amn 
Row 1
Row 2
Row m
Column 1 Column 2 Column n
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DEFINITION OF A MATRIX
If a matrix A has m rows and n columns, then A
is said to be of order m by n (written m × n).
The entry or element in the ith row and jth
column is a real number and is denoted by the
double-subscript notation aij. We can call aij the
(i, j)th entry.
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DEFINITION OF A MATRIX
If A has n rows and n columns, then A is called
a square matrix of order n.
The entries a11, a22, …, ann form the main
diagonal of A.
A 1 × n matrix is called a row matrix, and an
n × 1 matrix is called a column matrix.
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EXAMPLE 1
Determining the Order of Matrices
Determine the order of each matrix. Identify
square, row, and column matrices. Identify
entries in the main diagonal of each square
matrix.
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EXAMPLE 1
Determining the Order of Matrices
Solution
a. Matrix A with one row and one column is a
1 × 1 matrix. A is a square matrix of order 1. In
A, a11 = 3. A is also a column and a row matrix.
b. Matrix B with one row and three columns is a
1 × 3 matrix. B is a row matrix.
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EXAMPLE 1
Determining the Order of Matrices
Solution
c. Matrix C, a 2 × 2 matrix, is a square matrix of
order 2. In C, the entries c11 = 0 and c22 = 4 form
the main diagonal.
d. Matrix D, a 3 × 3 matrix, is a square matrix of
order 3. In D, the entries d11 = 1, d22 = 5, and
d33 = 9 form the main diagonal.
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MATRIX AND LINEAR SYSTEMS
We can display the constants and coefficients of
a system in matrix called the augmented
matrix of the system.
Constants
 x y z 1

2x  3y  z  10
 x  y  2z  0

1 1 1 1 


2

3
1
1
0


1 1 2 0 
Coefficients of x
Coefficients of y
Coefficients of z
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ELEMENTARY ROW OPERATIONS
Two matrices are row equivalent if one can be
obtained from the other by a sequence of
elementary row operations.
Row Operation
In Symbols
Interchange two rows
Ri n Rj
Description
Interchange the ith and jth
rows.
Multiply a row by a
cRj
Multiply the jth row by c,
nonzero constant
c ≠ 0.
Add a multiple of one cRi + Rj g Rj
Replace the jth row by
row to another row
adding c times ith row to it.
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EXAMPLE 3
Applying Elementary Row Operations
Perform the indicated row operations (a), (b),
and (c) in order on the following matrix:
Solution
a. Interchange the 1st and 2nd rows.
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EXAMPLE 3
Applying Elementary Row Operations
Solution continued
b. Multiply the 1st row by
.
c. Add –3 times the 1st row to the 2nd row.
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ROW-ECHELON FORM AND
REDUCED ROW-ECHELON FORM
An m × n matrix is in row-echelon form if it
has the following three properties:
1. The leading entry of each nonzero row is 1.
2. The leading entry in a row is to the right of
the leading entry in the row above it.
3. All nonzero rows are above the rows
consisting entirely of zeros.
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ROW-ECHELON FORM AND
REDUCED ROW-ECHELON FORM
A matrix in row-echelon form having the
following property is in reduced row-echelon
form:
Each leading 1 is the only nonzero entry in its
column.
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SOLVING LINEAR SYSTEMS BY USING
GAUSSIAN ELIMINATION
Step 1 Write the augmented matrix.
Step 2 Use elementary row operations to
transform the augmented matrix into
row-echelon form.
Step 3 Write the system of linear equations that
corresponds to the last in Step 2.
Step 4 Use the system of equations from Step 3,
together with back-substitution, to find
the system’s solution set.
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EXAMPLE 6
Solving a System by Using Gaussian
Elimination
Solve by Gaussian elimination.
 2x  y  z  6

3 x  4 y  2 z  4
 x  y  z  2

Solution
Step 1 The augmented matrix of the system is.
 2
1 1 6


A   3 4 2 4 
 1 1 1 2 
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EXAMPLE 6
Solving a System by Using Gaussian
Elimination
Solution continued
Step 2
 1

R1  R3 3
 2
1 1 2 

4 2 4 
1 1 6 
1 1 1 2 
3R1  R2  R2 

0

1

1

2

2 R1  R3  R3 
0 1 3 10 
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EXAMPLE 6
Solving a System by Using Gaussian
Elimination
Solution continued
Step 2 continued
1 1 1 2 


 1 R2 0 1 1 2 
0 1 3 10 
1 1 1 2 


R2  R3  R3 0 1 1 2 
0 0 4 12 
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EXAMPLE 6
Solving a System by Using Gaussian
Elimination
Solution continued
Step 2 continued
1 1 1 2  The matrix is now
1 
 in row-echelon
R3 0 1 1 2 
4
form.
0 0 1 3 
Step 3 The system corresponding to the last
matrix in Step 2 is
 x  y  z  2 (1)

y  z  2 (2)


z  3 (3)

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EXAMPLE 6
Solving a System by Using Gaussian
Elimination
Solution continued
Step 4 Equation (3) in Step 3 gives the value
z = 3. Back-substitute z = 3 in equation
(2).
yz2
y3 2
y  1
Back-substitute z = 3 and y = –1 in
equation (1).
x  y  z  2
x  1  3  2
x2
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EXAMPLE 6
Solving a System by Using Gaussian
Elimination
Solution continued
The solution set for the system is {(2, –1, 3)}.
You should check the solution by substituting
these values for x, y, and z into the original
system of equations.
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EXAMPLE 7
Attempting to Solve a System with No
Solution
Solve the system of equations by Gaussian
elimination.
Solution
Step 1 The augmented matrix of the system is.
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EXAMPLE 7
Attempting to Solve a System with No
Solution
Solution continued
Step 2
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EXAMPLE 7
Attempting to Solve a System with No
Solution
Solution continued
Step 2 continued
The matrix is now
in row-echelon
form.
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EXAMPLE 7
Attempting to Solve a System with No
Solution
Solution continued
Step 3 The system corresponding to the last
matrix in Step 2 is
A false statement
Because the equation 0 = 1 is never true, we
conclude that this system is inconsistent.
Step 4 The solution set for the system is .
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EXAMPLE 8
Solving a System of Equations by GaussJordan Elimination
Solve the system given in Example 4 by GaussJordan elimination.
 x y z  1

2 x  3 y  z  10 The given system
 x  y  2z  0

Solution
The augmented matrix of the system is
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EXAMPLE 8
Solving a System of Equations by GaussJordan Elimination
Solution continued
row-echelon
form
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EXAMPLE 8
Solving a System of Equations by GaussJordan Elimination
Solution continued
reduced
row-echelon
form
The corresponding system of equations for the
last augmented matrix is
The solution set is {(5, 1, 3)}.
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EXAMPLE 9
Solving a System with Infinitely Many
Solutions
Solve the system of equations.
Solution
The augmented matrix of the system is
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EXAMPLE 9
Solving a System with Infinitely Many
Solutions
Solution continued
We need a zero at the (3, 1) position.
And we need a zero at the (1, 2) position.
reduced
row-echelon
form
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EXAMPLE 9
Solving a System with Infinitely Many
Solutions
Solution continued
The equivalent system is
Solving for x and y in terms of z, we obtain
Each real number z results in a solution with
y = –4z + 4 and x = 3z – 4, giving infinitely
many solutions of the form (3z – 4, –4z + 4, z).
The solution set is {(3z – 4, –4z + 4, z)}.
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EXAMPLE 10 Leontief Input–Output Model
Consider an economy that has steel, coal, and
transportation industries. There are two types of
demands (measured in dollars) on the production
of each industry: interindustry demand and
external consumer demand. The outputs and
requirements of the three industries are shown in
on the next slide. For example, $1.00 of
transportation output requires $0.10 from steel
and $0.01 from coal.
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EXAMPLE 10 Leontief Input–Output Model
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EXAMPLE 10 Leontief Input–Output Model
a. Write a system of equations that expresses
the outputs of the three industries. Assume
that all quantities are given in millions of
dollars.
b. Verify that s = 15, c = 10, and t = 8 will
meet both interindustry and consumer
demand.
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EXAMPLE 10 Leontief Input–Output Model
Solution
a. To satisfy both consumer and interindustry
demand, we obtain the following system of
outputs. (s denotes total steel output; c, total coal
output; t, total transportation output.)
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EXAMPLE 10 Leontief Input–Output Model
Solution continued
You can rewrite this system as:
b. To verify that s = 15, c = 10, and t = 8 satisfy
those equations, we substitute them into each
equation.
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