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INTRODUCTORY MATHEMATICAL ANALYSIS
For Business, Economics, and the Life and Social Sciences
Chapter 6
Matrix Algebra
2007 Pearson Education Asia
INTRODUCTORY MATHEMATICAL ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and Statistics
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INTRODUCTORY MATHEMATICAL ANALYSIS
9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
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Chapter 6: Matrix Algebra
Chapter Objectives
• Concept of a matrix.
• Special types of matrices.
• Matrix addition and scalar multiplication operations.
• Express a system as a single matrix equation using
matrix multiplication.
• Matrix reduction to solve a linear system.
• Theory of homogeneous systems.
• Inverse matrix.
• Use a matrix to analyze the production of sectors of an
economy.
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Chapter 6: Matrix Algebra
Chapter Outline
6.1) Matrices
6.2) Matrix Addition and Scalar Multiplication
6.3) Matrix Multiplication
6.4) Solving Systems by Reducing Matrices
6.5) Solving Systems by Reducing Matrices
(continued)
6.6) Inverses
6.7) Leontief’s Input—Output Analysis
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Chapter 6: Matrix Algebra
6.1 Matrices
• A matrix consisting of m horizontal rows and n
vertical columns is called an m×n matrix or a
matrix of size m×n.
 a11 a12
a
 21 a12
 .
.

.
 .
 .
.

am1 am 2
... a1n 
... a2n 
...
. 

...
. 
...
. 

... amn 
• For the entry aij, we call i the row subscript and j
the column subscript.
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Chapter 6: Matrix Algebra
6.1 Matrices
Example 1 – Size of a Matrix
a. The matrix 1 2 0has size 1 3 .
1  6
b. The matrix 5 1  has size 3 2 .
9 4 
c. The matrix 7 has size 1 1 .
7  2 4
1 3
d. The matrix 9 11 5 6 8 has size3 5 .


6  2  1 1 1
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Chapter 6: Matrix Algebra
6.1 Matrices
Equality of Matrices
• Matrices A = [aij ] and B = [bij] are equal if they
have the same size and aij = bij for each i and j.
Transpose of a Matrix
• A transpose matrix is denoted by AT.
Example 3 – Constructing Matrices
If A   1 2 3 , find AT .
4 5 6
Solution:
 1 4
AT  2 5 
3 6
Observe that AT 
T
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 A.
Chapter 6: Matrix Algebra
6.2 Matrix Addition and Scalar Multiplication
Matrix Addition
• Sum A + B is the m × n matrix obtained by adding
corresponding entries of A and B.
Example 1 – Matrix Addition
a. 1
4  7  2 1  7 2  2   8 0
2 5   6 4   3  6 4  4   3 8

 
 
 

3 6  3
0  5  3 6  0   8 6
b.
1 2 2
3 4  1 is impossible as matrices are not of the same

  
size.
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Chapter 6: Matrix Algebra
6.2 Matrix Addition and Scalar Multiplication
Example 3 – Demand Vectors for an Economy
Demand for the consumers is
D1  3 2 5
D2  0 17 1
D3  4 6 12
For the industries is
DC  0 1 4
DE  20 0 8
DS  30 5 0
What is the total demand for consumers and the
industries?
Solution:
D1  D2  D3  3 2 5  0 17 1  4 6 12  7 25 18
DC  DE  DS  0 1 4  20 0 8  30 5 0  50 6 12
Total: 7 25 18  50 6 12  57 31 30
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Chapter 6: Matrix Algebra
6.2 Matrix Addition and Scalar Multiplication
Scalar Multiplication
• Properties of Scalar Multiplication:
Subtraction of Matrices
• Property of subtraction is  A  1A
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Chapter 6: Matrix Algebra
6.2 Matrix Addition and Scalar Multiplication
Example 5 – Matrix Subtraction
6 6  2  2  6 6  2  4 8 
 4 1  4 1    4  4 1  1    8 0 

 
 
 

 3 2 0 3   3  0 2  3  3  1
a. 2
6 2  6  6  0
8





0

1
2
4

2

5

 
 

b. AT  2B  
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Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
• AB is the m× p matrix C whose entry cij is given by
n
c ij   aik bkj  ai 1b1 j ai 2b2 j  ...  ain bnj
k 1
Example 1 – Sizes of Matrices and Their Product
A = 3 × 5 matrix
B = 5 × 3 matrix
AB = 3 × 3 matrix but BA = 5 × 5 matrix.
C = 3 × 5 matrix
D = 7 × 3 matrix
CD = undefined but DC = 7 × 5 matrix.
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Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 3 – Matrix Products
a.
 4
1 2 35  32
6
b.
1
1 6 
21 6  2 12
 


3
3 18
c.
0
2   16  3 11
 1 3 0  1
 2 2 1   2  1 3    10  1 0 


 

 1 0  4  2
1  2  7  4 10
d. a11 a12  b11 b12   a11b11  a12 b21 a11b12  a12 b22 
a
 b
 a b  a b

a
b
a
b

a
b
22 21
21 12
22 22 
 21 22   21 22   21 11
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Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 5 – Cost Vector
Given the price and the quantities, calculate the total
cost.
 7  units of A
P  2 3 4
Q   5  units of B
11 units of C
Solution:
The cost vector is
7
PQ  2 3 4 5   73
11
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Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 7 – Associative Property
If
 1  2
A

 3  4
 1 0
C  0 2
1 1
3 0  1
B

1 1 2 
compute ABC in two ways.
Solution 1:

 1 0 


1

2
3
0
1

 


ABC   
 1 1 2 0 2 

3

4

 
 1 1 



 1  2 2  1  4  9

 3 4    6 19 

3

4


 

Solution 2:
1 0
 1  2 3 0 1  
 0 2
AB C   





  3  4  1 1 2  1 1


 1 0
1

2

5


 4  9


 0 2   6 19 

5
4
11

 1 1 



Note that A(BC) = (AB)C.
2007 Pearson Education Asia
Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 9 – Raw Materials and Cost
Find QRC when
Q  5 7 9
5 20 16 7 17
R  7 18 12 9 21
6 25 8 5 13
Solution:
2500 
1200 


C   800 


150


1500 
2500


5 20 16 7 17 1200  75850
RC  7 18 12 9 21  800   81550


6 25 8 5 13  150  71650
1500 
75850
QRC  QRC   5 7 1281550  1,809,900
71650
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Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 11 – Matrix Operations Involving I and O
If
3 2
A

 1 4
 52
B 1
 10
 51 
3 
10 
 1 0
I

0 1
0 0
O

0 0
compute each of the following.
Solution:
1 0 3 2  2  2
a. I  A  





0 1 1 4   1  3
 3 2
 3 2 2 0  3 6
 1 0 



b. 3A  2I   3 

2

3








 3 6
1
4
0
1
1
4
0
2



 
 



3 2 0 0
c. AO  
O



1 4 0 0
3 2  52
d. AB  
  1
1
4

  10
 51  1 0

 I
3 
0
1


10 
2007 Pearson Education Asia
Chapter 6: Matrix Algebra
6.3 Matrix Multiplication
Example 13 – Matrix Form of a System Using Matrix Multiplication
2x  5 x2  4
Write the system  1
8 x1  3 x2  7
in matrix form by using matrix multiplication.
Solution:
2 5
If A  

8 3
 x1 
X  
 x2 
 4
B 
7
then the single matrix equation is
AX  B
2 5  x1  4
8 3  x   7 

 2   
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Chapter 6: Matrix Algebra
6.4 Solving Systems by Reducing Matrices
Elementary Row Operations
1. Interchanging two rows of a matrix
2. Multiplying a row of a matrix by a nonzero number
3. Adding a multiple of one row of a matrix to a
different row of that matrix
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Chapter 6: Matrix Algebra
6.4 Solving Systems by Reducing Matrices
Properties of a Reduced Matrix
•
All zero-rows at the bottom.
•
For each nonzero-row, leading entry is 1 and the
rest zeros.
•
Leading entry in each row is to the right of the
leading entry in any row above it.
2007 Pearson Education Asia
Chapter 6: Matrix Algebra
6.4 Solving Systems by Reducing Matrices
Example 1 – Reduced Matrices
For each of the following matrices, determine whether
it is reduced or not reduced.
 1 0
a. 

0
3


0 0 0
d.

0
0
0


 1 0 0
b.

0
1
0


 1 0 0
e.0 0 0
0 1 0
Solution:
a. Not reduced
c. Not reduced
e. Not reduced
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0 1
c.

1
0


0 1 0 3
f.0 0 1 2
0 0 0 0
b. Reduced
d. Reduced
f. Reduced
Chapter 6: Matrix Algebra
6.4 Solving Systems by Reducing Matrices
Example 3 – Solving a System by Reduction
By using matrix reduction, solve the system
2x  3y  1

 2x  y  5
 x  y 1

Solution:
Reducing the augmented coefficient matrix of the
system,
2 3  1


2
1
5


1 1 1 


x4
We have 
y  3
2007 Pearson Education Asia
1 0 4 


0
1

3


0 0 0 


Chapter 6: Matrix Algebra
6.4 Solving Systems by Reducing Matrices
Example 5 – Parametric Form of a Solution
Using matrix reduction, solve
2x1  3 x2  2x3  6 x 4  0

x 2  2 x3  x 4  2


3 x1  3 x3  6 x 4  9

Solution:
Reducing the matrix of the system,
2 3 2 6 10 


0
1
2
1
2


3 0  3 6 9 


We have
 x1  4  52 x 4

x2  0
x  1  1 x
2 4
 3
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 1 0 0 52 4 


0 1 0 0 0 
0 0 1 21 1


and x4 takes on any real value.
Chapter 6: Matrix Algebra
6.5 Solving Systems by Reducing Matrices
(continued)
Example 1 – Two-Parameter Family of Solutions
Using matrix reduction, solve
x1  2x2  5 x3  5 x 4  3

x1  x2  3 x3  4 x 4  1
x  x  x  2x  3
2
3
4
 1
Solution:
1
The matrix is reduced to 0
0 1 3 1

1 2 1  2
0 0 0 0 0 


The solution is
2007 Pearson Education Asia
 x1  1  r  3s
 x  2  2r  s
 2

 x3  r
x 4  s
Chapter 6: Matrix Algebra
6.5 Solving Systems by Reducing Matrices (Continue)
•
•
The system
 a11x1  a12 x 2  ...  a1n x n  c1

.


.


.


.


am1x1  am 2 x 2  ...  amn xn  c m
is called a homogeneous system if c1 = c2 = …
= cm = 0.
The system is non-homogeneous if at least one
of the c’s is not equal to 0.
Concept for number of solutions:
1. k < n  infinite solutions
2. k = n  unique solution
2007 Pearson Education Asia
Chapter 6: Matrix Algebra
6.5 Solving Systems by Reducing Matrices (Continue)
Example 3 – Number of Solutions of a Homogeneous System
Determine whether the system has a unique solution
or infinitely many solutions.
 x  y  2z  0

2x  2y  4z  0
Solution:
2 equations (k), homogeneous system, 3 unknowns
(n).
The system has infinitely many solutions.
2007 Pearson Education Asia
Chapter 6: Matrix Algebra
6.6 Inverses
•
When matrix CA = I, C is an inverse of A and A is
invertible.
Example 1 – Inverse of a Matrix
Let
 1 2
A

3
7


 2

 3 1 
and C   7
. Determine whether C is
an inverse of A.
 7  2 1 2 1 0
Solution: CA  

I




 3 1  3 7 0 1
Thus, matrix C is an inverse of A.
2007 Pearson Education Asia
Chapter 6: Matrix Algebra
6.6 Inverses
Method to Find the Inverse of a Matrix
•
When matrix is reduced, A I    R B ,
-
If R = I, A is invertible and A−1 = B.
-
If R  I, A is not invertible.
Example 3 – Determining the Invertibility of a Matrix
 1 0
A

2 2
Determine if
Solution: We have
is invertible.
 1 0 1 0
A I   2 2 0 1


Matrix A is invertible where
2007 Pearson Education Asia
 1 0 1 0
 I B

1
0 1  1 2 
 1 0
A 
1

1
2

1
Chapter 6: Matrix Algebra
6.6 Inverses
Example 5 – Using the Inverse to Solve a System
Solve the system by finding the inverse of the
coefficient matrix. x  2x
1
 1
3

4 x1  2x2  x3  2
 x  2x  10 x  1
2
3
 1
Solution: 1
We have A  4
0
2 
2
1 
 1 2  10
For
  9 2 2
inverse, A1   412 4 92 
 5 1 1
The solution is given by X =
2007 Pearson Education Asia
 x1    9 2 2  1    7 
A−1B:x2    412 4 92   2    17
 x3    5 1 1  1   4 
Chapter 6: Matrix Algebra
6.7 Leontief’s Input-Output Analysis
•
•
Entries are called input–output coefficients.
Use matrices to show inputs and outputs.
Example 1 – Input-Output Analysis
Given the input–output matrix,
suppose final demand changes to be 77 for A, 154 for
B, and 231 for C. Find the output matrix for the
economy. (The entries are in millions of dollars.)
2007 Pearson Education Asia
Chapter 6: Matrix Algebra
6.7 Leontief’s Input-Output Analysis
Example 1 – Input-Output Analysis
Solution:
Divide entries by the total value of output to get A:
Final-demand matrix:
Output matrix is
 77 
D  154
231
692.5
1
X  I  A D   380 
 495 
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