Transcript Systems of Linear Equations and Matrices

Systems of Linear Equations
and Matrices
Chapter 2
ECO 3401 - B. Potter
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Systems of Linear Equations
and Matrices





2.2 Solution of Linear Systems by the
Gauss-Jordan Method
2.3 Addition and Subtraction of Matrices
2.4 Multiplication of Matrices
2.5 Matrix Inverses
2.6 Input-Output Models
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Systems of Linear Equations
and Matrices
2.2 Solution of Linear Systems by the
Gauss-Jordan Method
Working with some basic Matrix Algebra
 2.3 Addition and Subtraction of Matrices
 2.4 Multiplication of Matrices
 2.5 Matrix Inverses


Solving a system of linear equations using matrix
inverses (with Microsoft Excel).
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Matrix Algebra

Matrix – A rectangular array of numbers.
 a11

a21

A


 am1
a12
a22
am 2
a1n 

a2 n 


amn 
is a m  n matrix (m rows, n columns), where the entry in
the ith row and jth column is aij.
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Matrix Algebra





Matrices are often named with capital letters (M).
Matrices are classified by size (# of rows  # of
columns).
10 12 5 
M 

 15 20 8 
M is a 2 x 3 matrix
Row matrix (row vector) – a matrix containing
only 1 row.
Column matrix (column vector) – only 1 column.
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Matrix Algebra

Matrix Equality

Two matrices are equal if they are the same size
and if each pair of corresponding elements is
equal.
 8 2   3 9 



 3 1   2 5 
 1 3  x y 



r
s

1
0

 

could be true
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2.3 Addition and Subtraction
of Matrices

Adding Matrices

The sum of two m  n matrices X and Y is the
m  n matrix X + Y in which each element is the
sum of the corresponding elements of X and Y.
10 12 5 
 45 35 20 
X 
, Y  

15
20
8
65
40
35




 55 47 25 
X Y  

80
60
43


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Adding Matrices - example

A toy company has plants in Boston, Chicago, and Seattle
that manufacture toy phones and calculators. The
following matrices show the per-item production costs for
the three plants:
Seattle
Boston
Chicago
Phones
Calculators
Calculators
7.01


3.27
3.51


Phones
Material  4.27
Material  4.05
Material  4.40
Labor
Labor
Labor

6.94 


3.45
3.65


Phones
Calculators
6.90 


3.54
3.76


Use matrix addition to determine the firm’s total per-item
costs.
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Adding Matrices - example

A toy company has plants in Boston, Chicago, and Seattle
that manufacture toy phones and calculators. The
following matrices show the per-item production costs for
the three plants:
 4.27
T 
 3.45
12.72

10.26
6.94   4.05 7.01  4.40 6.90 



3.65   3.27 3.51  3.54 3.76 
20.85 

10.92 
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Adding Matrices - example

A toy company has plants in Boston, Chicago, and Seattle
that manufacture toy phones and calculators. The
following matrices show the per-item production costs for
the three plants:
Phones
Calculators
Material  12.72
Labor
20.85 


10.26
10.92


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2.3 Addition and Subtraction
of Matrices

Additive Inverse

The additive inverse (or negative) of a matrix X
is the matrix –X in which each element is the
additive inverse of the corresponding element
of X.
 1 2 3
A

 0 1 5 
 1 2 3 
A  

 0 1 5 
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2.3 Addition and Subtraction
of Matrices

Zero Matrix


The sum of matrices X and –X is a zero matrix
A matrix whose elements are all zeros.
 0 0 0
A    A  

0
0
0



If O is an m x n zero matrix, and A is any m  n
matrix, then A  O  O  A  A
Zero matrix – Additive identity matrix
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2.3 Addition and Subtraction
of Matrices

Subtracting Matrices

The difference between two m  n matrices X
and Y is the m  n matrix X  Y (or Y  X) in
which each element is found by subtracting the
corresponding elements of X and Y.
10 12 5 
 45 35 20 
X 
, Y  

15
20
8
65
40
35




 35 23 15 
X Y  


50

20

27


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2.3 Addition and Subtraction
of Matrices

Motorcycle Helmets The following table shows the
percentage of motorcyclists in various regions of the
country who used helmets compliant with federal safety
regulations and the percentage who used helmets that
were not compliant in two recent years.
2008
Compliant
Noncompliant
2009
Compliant
Noncompliant
Northeast
45
8
Northeast
61
15
Midwest
67
16
Midwest
67
8
South
61
14
South
65
6
West
71
5
West
83
4
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2.3 Addition and Subtraction
of Matrices
a.
b.
Write two matrices for the 2008 and 2009 helmet usage.
Use the two matrices to form a matrix showing the
change in helmet usage from 2008 to 2009.
2008
Compliant
Noncompliant
2009
Compliant
Noncompliant
Northeast
45
8
Northeast
61
15
Midwest
67
16
Midwest
67
8
South
61
14
South
65
6
West
71
5
West
83
4
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2.3 Addition and Subtraction
of Matrices
a.
b.
Write two matrices for the 2008 and 2009 helmet usage.
Use the two matrices to form a matrix showing the
change in helmet usage from 2008 to 2009.
 45 8 
 61 15 




67
16
67
8
 , Y (2009)  

X (2008)  
 61 14 
 65 6 




 71 5 
 83 4 
ECO 3401 - B. Potter
16 7 


0

8

Y X 
 4 8 


12 1 
16
2.4 Multiplication of Matrices

Product of a Matrix and a Scaler (real
number)

The product of a scaler k and a matrix X is the
matrix kX, each of whose elements is k times
the corresponding element of X.
 1 3   2 6 
 2 



5
6
10

12

 

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2.4 Multiplication of Matrices

The product AB of an m  n matrix A and an n  k matrix B
is found as follows.
 Multiply each element of the first row of A by the
corresponding element of the first column of B. The
sum of these n products is the first row, first column of
AB.
 Multiply each element of the first row of A by the
corresponding element of the second column of B. The
sum of these n products is the first row, second column
of AB.

Example...
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Multiplication of Matrices
A
a b

d e
g h

AB
B
c  j k   aj  bl  cn ak  bm  co 

 

f  l m    dj  el  fn dk  em  fo 
  gj  hl  in gk  hm  io 
i 
n
o

 

The product AB of two matrices A and B can be found
only if the number of columns of A is the same as the
number of rows of B.
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Now You Try
Find the matrix product.
 2 3  2 0 



 3 2  1 2 
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Try another; pg. 105, #43
Quantity
Paper
Tape
Binders
Memo
Pads
Department 1
10
4
3
5
6
Department 2
7
2
2
3
8
Department 3
4
5
1
0
10
Department 4
0
3
4
5
5
Pens
Price (in dollars)
Supplier A Supplier B
Paper
2
3
Tape
1
1
Binders
4
3
Memo Pads
3
3
Pens
1
2
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Try another; pg. 105, #43

Write the information in the “Quantity” table
as a 4  5 matrix Q.
10

7
Q
4

0
4
2
5
3
3
2
1
4
5 6

3 8
0 10 

5 5
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Try another; pg. 105, #43

Write the information in the “Price” table as a
5  2 matrix P.
2

1

P  4

3
1

3

1
3

3
2 
ECO 3401 - B. Potter
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Try another; pg. 105, #43

Find the product QP (QP will be a 4  2 matrix)
10

7

4

0
4 3 5
2 2 3
5 1 0
3 4 5
2
6 
 1
8
4
10  
 3
5 
1
3

1
3

3
2 
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Using Microsoft Excel
Array 1 is a 4 x 5 (m x n) matrix
Array 2 is a 5 x 2 (n x k) matrix
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Using Microsoft Excel
Highlight m  k cells.
1.


2.
3.
m = rows in array 1
k = columns in array 2
Enter the function,
=MMULT(array 1,array 2)
Select Ctrl + Shift + Enter
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Microsoft Excel
1. Highlight m x k cells.
(m = 4, k = 2)
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Microsoft Excel
Highlight
k cells.
1. 1.Highlight
m xmk xcells.
(m (m
= 4,=k4,=k2)= 2)
2. Enter the function,
=MMULT(array 1,array 2)
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Microsoft Excel
1. Highlight m x k cells.
(m = 4, k = 2)
2. Enter the function,
=MMULT(array 1,array 2)
3. Select Ctrl + Shift + Enter
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Microsoft Excel
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Microsoft Excel
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Microsoft Excel
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2.5 Matrix Inverses



Comparable to the reciprocal of a real number.
If A is an n x n matrix, A-1 is the multiplicative inverse of
matrix A. (A-1 does not mean 1/A)
Just as the real number 1 is the multiplicative identity for
real numbers, I is defined as the multiplicative identity
matrix.
Therefore,
AI  IA  A
and,
1
1
AA  A A  I
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Multiplicative Identity Matrix I

2 x 2 identity matrix
1 0
I 

0 1

3 x 3 identity matrix
1 0 0


I  0 1 0
0 0 1



4 x 4 identity matrix
1

0
I 
0

0
ECO 3401 - B. Potter
0
1
0
0
0
0
1
0
0

0
0

1
34
Multiplicative Identity Matrix I

AI = A
a
A
c
a
AI  
c
b

d
b  1 0 


d  0 1 
 a(1)  b(0) a(0)  b(1) 
AI  

c
(1)

d
(0)
c
(0)

d
(1)


a b 
AI  
 A
c d 
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Microsoft Excel
1.
2.
3.
Highlight n x n cells.
Enter the function,
=MINVERSE(array)
Select Ctrl + Shift + Enter
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Microsoft Excel

Example:
Find A – 1
1 0 1 


A   2 2 1
3 0 0 


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Microsoft Excel
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Microsoft Excel
1.
Highlight n  n cells
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Microsoft Excel
1.
2.
Highlight n  n cells
Enter the function,
=MINVERSE(array)
ECO 3401 - B. Potter
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Microsoft Excel
1.
2.
3.
Highlight n  n cells
Enter the function,
=MINVERSE(array)
Select Ctrl + Shift + Enter
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Systems of Linear Equations



A system of linear equations is a set of n
linear equations in k variables (or
unknowns) that are solved together.
The simplest linear system is one with 2
equations in 2 variables.
A solution of a system is a solution that
satisfies all the equations in the system.
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Solving a 22 System of Linear
Equations


Example: 2 x  3 y  12
3x  4 y  1
Three methods



Graph the lines and identify the intersection (if
any)
Substitution
Elimination
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Graphing Method
12
10
8
6
4
2x+3y=12
3x-4y=1
(x, y)
2
0
-9
-6
-3 -2 0
3
6
9
12
15
18
-4
-6
-8
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Substitution

2 x  3 y  12
3x  4 y  1
Solve the first equation for y
3 y  12  2 x
2
y  4 x
3
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Substitution


2 x  3 y  12
3x  4 y  1
2 

Solve the first equation for y 3 x  4  4  x   1
3 

3 y  12  2 x
8
2
3 x  16  x  1
y  4 x
3
3
17
Substitute this expression
x  17
for y in the second equation. 3
Solve for x
x3
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Substitution


2 x  3 y  12
3x  4 y  1
2 

Solve the first equation for y 3 x  4  4  x   1
3 

3 y  12  2 x
8
2
3 x  16  x  1
y  4 x
3
3
17
Substitute this expression
x  17
for y in the second equation. 3
Solve for x
x3
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Substitution


2 x  3 y  12
3x  4 y  1
2 

Solve the first equation for y 3 x  4  4  x   1
3 

3 y  12  2 x
8
2
3 x  16  x  1
y  4 x
3
3
17
Substitute this expression
x  17
for y in the second equation. 3
Solve for x
x3
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Substitution

2 x  3 y  12
3x  4 y  1
Substitute x = 3 in either equation to solve for y.
2  3  3 y  12
3y  6
y2
Solution: (3, 2)
2  3  3  2   12
3  3  4  2   1
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Elimination

2 x  3 y  12
3x  4 y  1
In systems of equations where the coefficients of
terms containing the same variable are opposites,
the elimination method can be applied by adding
the equations. If the coefficients of those terms are
the same, the elimination method can be applied
by subtracting the equations.
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Elimination

2 x  3 y  12
3x  4 y  1
Multiply the first equation by 4 and the second
equation by 3, so the coefficients of y are
negatives of each other.
4  equation 1  8 x  12 y  48
3  equation 2  9 x  12 y  3
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Elimination

2 x  3 y  12
3x  4 y  1
Any solution of this system must also be the
solution of the sum of the two equations
8 x  12 y  48
9 x  12 y  3
17 x
 51
x 3
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Elimination

Any solution of this system must also be the
solution of the sum of the two equations
8 x  12 y  48
2  3  3 y  12
9 x  12 y  3
3y  6
17 x

2 x  3 y  12
3x  4 y  1
 51
x 3
y2
Solution: (3, 2)
Substitute x = 3 in either equation to solve for y.
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Solving Systems of Equations With
Matrix Inverses


Most practical in solving several systems that
have the same variable matrix but different
constants.
Write the system as a matrix equation AX = B,
where
 A is the matrix of the coefficients of the
variables,
 X is the matrix of the variables,
 B is the matrix of the constants.
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Solving Systems of Equations With
Matrix Inverses

Example: Consider a system of 3 equations in 3
variables (x, y, z)
a x  b y  cz  j
dx  ey  fx  k
gx  h y  i z  l
a b

A  d e
g h

c

f
i 
 x
 
X   y
z
 
ECO 3401 - B. Potter
 j
 
B  k 
l
 
55
Solving Systems of Equations With
Matrix Inverses

Example: Consider a system of 3 equations in 3
variables (x, y, z)
a x  b y  cz  j
dx  ey  fx  k
gx  h y  i z  l
a b

AX  B or  d e
g h

c  x   j 
   
f  y    k 
 l
i 
z
   
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Solving Systems of Equations With
Matrix Inverses

Solve the matrix equation AX = B
-1
 Given: A A = I, and IX = X
AX  B
A1 ( AX )  A1 B
Multiply both sides by A-1
( A1 A) X  A1 B
Associative property
IX  A1 B
X  A1 B
Multiplication inverse property
Identity property
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Solving Systems of Equations With
Matrix Inverses

Solve the matrix equation AX = B
-1
 Given: A A = I, and IX = X
AX  B
A1 ( AX )  A1 B
Multiply both sides by A-1
( A1 A) X  A1 B
Associative property
IX  A1 B
Multiplication inverse property
X  A1 B
Identity property
ECO 3401 - B. Potter
58
Solving Systems of Equations With
Matrix Inverses

To solve a system of equations AX = B, where A is the
matrix of coefficients, X is the matrix of variables, and B is
the matrix of constants, first find A-1. Then X = A-1B.
ECO 3401 - B. Potter
59
Example

Solve the following system using matrix
notation:
x  3y  6  2z
2 x  3z  8  4 y
3 x  6 y  8 z   5
ECO 3401 - B. Potter
60
Example
1.
Write each equation in proper form.
ECO 3401 - B. Potter
61
Example

Proper form of the system:



Terms with variables on the left;
constants on the right;
variables in the same order in each equation.
ECO 3401 - B. Potter
62
Example
x  3y  6  2z
2 x  3z  8  4 y
3 x  6 y  8 z   5
becomes
x  3y  2z  6
2 x  4 y  3z  8
3 x  6 y  8 z   5
ECO 3401 - B. Potter
63
Example
1.
2.
Write each equation in proper form.
Write the corresponding matrix equation, AX=B
x  3y  2z  6
2 x  4 y  3z  8
3 x  6 y  8 z   5
 1 3 2 


A   2 4 3 
 3 6 8 


 x
 
X   y
z
 
ECO 3401 - B. Potter
6
 
B 8 
 5 
 
64
Example
1.
2.
3.
Write each equation in proper form.
Write the corresponding matrix equation, AX=B
Use Microsoft Excel to:
a.
b.
calculate the inverse of the coefficient matrix (A1)
multiply A1 by the constant matrix (B) to find the variable
matrix (X).
X  A1 B
ECO 3401 - B. Potter
65
Excel
ECO 3401 - B. Potter
66
Excel
Highlight 3  3 cells
ECO 3401 - B. Potter
67
Excel
=MINVERSE(array)
ECO 3401 - B. Potter
68
Excel
Ctrl + Shift + Enter
ECO 3401 - B. Potter
69
Excel
ECO 3401 - B. Potter
70
Excel
Highlight 3  1 cells
ECO 3401 - B. Potter
71
Excel
=MMULT(array1,array2)
ECO 3401 - B. Potter
72
Excel
Ctrl + Shift + Enter
ECO 3401 - B. Potter
73
Excel
=X
ECO 3401 - B. Potter
74
Example
x = 1, y = -3, z = 2
x  3y  2z  6
2 x  4 y  3z  8
1  3  3  2  2   6
3 x  6 y  8 z   5
3 1  6  3  8  2   5
2 1  4  3  3  2   8
ECO 3401 - B. Potter
75
Example

An electronics company produces transistors,
resistors, and computer chips. Each transistor
requires 3 units of copper, 1 unit of zinc, and 2
units of glass. Each resistor requires 3, 2, and 1
units of the three materials, and each computer
chip requires 2, 1, and 2. How many of each
product can be made with 810 units of copper,
410 units of zinc, and 490 units of glass?
ECO 3401 - B. Potter
76
Transistor (x) Resistor (y) Comp. Chip (z)
Copper
3
3
2
Zinc
1
2
1
Glass
2
1
2
3x  3 y  2 z  810 units of copper
x  2 y  z  410 units of zinc
2 x  y  2 z  490 units of glass
ECO 3401 - B. Potter
77
Transistor (x) Resistor (y) Comp. Chip (z)
Copper
3
3
2
Zinc
1
2
1
Glass
2
1
2
3x  3 y  2 z  810 units of copper
x  2 y  z  410 units of zinc
2 x  y  2 z  490 units of glass
ECO 3401 - B. Potter
78
Transistor (x) Resistor (y) Comp. Chip (z)
Copper
3
3
2
Zinc
1
2
1
Glass
2
1
2
3x  3 y  2 z  810 units of copper
x  2 y  z  410 units of zinc
2 x  y  2 z  490 units of glass
ECO 3401 - B. Potter
79
3x  3 y  2 z  810
x  2 y  z  410
2 x  y  2 z  490
 3 3 2


A  1 2 1
 2 1 2


 x
 
X   y
z
 
 810 


B   410 
 490 


AX  B
1
XA B
ECO 3401 - B. Potter
80
Microsoft Excel
ECO 3401 - B. Potter
81
Microsoft Excel
Use MINVERSE(b1:d3)
ECO 3401 - B. Potter
82
Microsoft Excel
Use MMULT(b5:d7,g1:g3)
ECO 3401 - B. Potter
83
1
XA B
4

 1 3

2

X 0

3

 1 1



 4
 1
1
 
 1 810    3  410    3  490 



3  810   100  100 transistors,
110 


1 
2
 190


 
resistors,
and
  410    110
0  810    410     490 



3 
computer
chips
3
 3  can be




  490   90 
1 
  1 810made
 1 410  1 490 






ECO 3401 - B. Potter
84
Now You Try

Pretzels cost $3 per pound, dried fruit $4
per pound, and nuts $8 per pound. How
many pounds of each should be used to
produce 140 pounds of trail mix costing $6
per pound in which there is twice as much
pretzels (by weight) than dried fruit?
ECO 3401 - B. Potter
85
Now You Try

Let:



x = the number of pounds of pretzels
y = the number of pounds of dried fruit
z = the number of pounds of nuts
The system of equations is:
ECO 3401 - B. Potter
86
2.6 Input-Output Models


Matrix models used for studying the interdependencies in
an economy.
Developed by Wassily Leontief (1906 – 1909)

1973 Nobel prize in economics
ECO 3401 - B. Potter
87
Input-Output Models



Deal with the production and flow of goods in an
economy.
In practice, very complicated with many variables.
In an economy with n basic commodities, the production
of each relies on inputs of the other commodities.


Example: Oil to run machinery to plant and harvest wheat.
Input-Output Matrix – shows the amounts of each
commodity used in the production of one unit of each
commodity.
ECO 3401 - B. Potter
88
Input-Output Models

Input-Output matrix - simplified economy with three
commodity categories (Agriculture, Manufacturing,
Transportation).
Agriculture
Manufacturing
Transportation
Agriculture
0
¼
1/3
Manufacturing
½
0
¼
Transportation
¼
¼
0
ECO 3401 - B. Potter
89
Input-Output Models


Input-Output matrix - simplified economy with three
commodity categories (Agriculture, Manufacturing,
Transportation).
Agriculture
Manufacturing
Transportation
Agriculture
0
¼
1/3
Manufacturing
½
0
¼
Transportation
¼
¼
0
The amount of each commodity needed to produce one
unit of agriculture.
ECO 3401 - B. Potter
90
Input-Output Models


Input-Output matrix - simplified economy with three
commodity categories (Agriculture, Manufacturing,
Transportation).
Agriculture
Manufacturing
Transportation
Agriculture
0
¼
1/3
Manufacturing
½
0
¼
Transportation
¼
¼
0
The amount of each commodity needed to produce one
unit of manufacturing.
ECO 3401 - B. Potter
91
Input-Output Models


Input-Output matrix - simplified economy with three
commodity categories (Agriculture, Manufacturing,
Transportation).
Agriculture
Manufacturing
Transportation
Agriculture
0
¼
1/3
Manufacturing
½
0
¼
Transportation
¼
¼
0
=A
The amount of each commodity needed to produce one
unit of transportation.
ECO 3401 - B. Potter
92
Input-Output Models

Input-Output matrix (A)

0

1

A
2

1
4

1
4
0
1
4
1
3

1
4

0 

ECO 3401 - B. Potter
93
Input-Output Models


Production matrix (X) – Column matrix that gives the
amount of each commodity produced by the economy:
If the economy produces:
 60 
 60 units of agriculture
 
X

 52 units of manufacturing
 52 
 48 
 48 units of transportation
 
ECO 3401 - B. Potter
94
Input-Output Models

Input-Output matrix (A), Production matrix (X)
 ¼ unit of agriculture is used
1 1

to produce each unit of
0 4 3
 60 


manufacturing.
1
1
 

X   52 
A
0
 52 units of manufacturing are
2
4
 48 
produced.




1 1 0
 ¼  52 = 13 units of
4 4



agriculture are used in the
production of manufacturing
ECO 3401 - B. Potter
95
Input-Output Models

Input-Output matrix (A), Production matrix (X)
 1/3 unit of agriculture is
1 1

used to produce each unit of
0 4 3
 60 


transportation.
1
1
 

X   52 
A
0
 48 units of transportation are
2
4
 48 
produced.




1 1 0
 1/3  48 = 16 units of
4 4



agriculture are used in the
production of transportation.
ECO 3401 - B. Potter
96
Input-Output Models

Input-Output matrix (A), Production matrix (X)
 Therefore, 13 + 16 = 29 units
1 1

of agriculture are used for
0 4 3
 60 


“production” in the economy
1
1
 

X   52 
A
0
2
4
 48 


 
1
1


4 4 0


ECO 3401 - B. Potter
97
Input-Output Models

Input-Output matrix (A), Production matrix (X)
 Since A gives the amount of
1 1

each commodity needed to
0 4 3
 60 


produce 1 unit of each,
1
1
 

X   52 
A
0
 and X gives the number of
2
4
 48 
units of each commodity




1 1 0
produced,
4 4



 AX gives the amount of each
commodity used in the
production process.
ECO 3401 - B. Potter
98
Input-Output Models

Input-Output matrix (A), Production matrix (X)

0

1

AX 
2

1
4

1
4
0
1
4
1
3   60 

1  
52 


4  
  48 
0 

ECO 3401 - B. Potter
99
Input-Output Models

Using Excel:
Use =MMULT(B1:D3,G1:G3)
ECO 3401 - B. Potter
100
Input-Output Models

Input-Output matrix (A), Production matrix (X)

0

1

AX 
2

1
4

1
4
0
1
4
1
3   60   29 

1    
52    42 


4    
  48   28 
0 



29 units of agriculture, 42
units of manufacturing, 28
units of transportation are
used to produce 60, 52,
and 48 units of each.
Remainder is used to
satisfy the demand outside
the production system.
ECO 3401 - B. Potter
101
Input-Output Models


Demand Matrix (D) – Represents the demand for the
various commodities outside the production process.
X  AX  D
 29 
 60   29   31 
 60 
 
     
 
X   52  AX   42  D   52    42    10 
 28 
 48   28   20 
 48 
 
     
 
Therefore, production of 60 units of agriculture, 52 units
of manufacturing, and 48 units of transportation would
satisfy a demand of 31, 10, and 20 units of the
commodities.
ECO 3401 - B. Potter
102
Input-Output Models


In practice, A and D are known and X must be found.
 Where, X = amounts of production needed to satisfy
the demands (D)
X  AX  D
Identity Property
IX  AX  D
 I  A X  D Distributive Property
If matrix I – A has an inverse, then
 I  A D  X
1
ECO 3401 - B. Potter
103
Input-Output Models

What should production of each commodity be to satisfy
demands for 516 units of agriculture, 258 units of
manufacturing, and 129 units of transportation?
 I  A D  X
1
 516 


D   258 
 129 



0
1 0 0 

 1
 I  A   0 1 0   
0 0 1  2

 1

4

1
4
0
1
4
ECO 3401 - B. Potter
1 
1


3
 
1  1
 

 2
4
 
1
0   
  4

1
4
1

1
4
1
 
3

1

4

1 

104
Input-Output Models

What should production of each commodity be to satisfy
demands for 516 units of agriculture, 258 units of
manufacturing, and 129 units of transportation?
1
I

A

 DX
 516 


D   258 
 129 


 I  A
1

ECO 3401 - B. Potter
105
Input-Output Models

Excel
 I  A
1
DX
Use =MINVERSE(B5:D7)
ECO 3401 - B. Potter
106
Input-Output Models

What should production of each commodity be to satisfy
demands for 516 units of agriculture, 258 units of
manufacturing, and 129 units of transportation?
1
I

A

 DX
 516 


D   258 
 129 


 1.393 0.494 0.583 
1


I

A

0.836
1.363
0.617

 

 0.557 0.464 1.300 


ECO 3401 - B. Potter
107
Input-Output Models

What should production of each commodity be to satisfy
demands for 516 units of agriculture, 258 units of
manufacturing, and 129 units of transportation?
1
I

A

 DX
 1.393 0.494 0.583  516   921 


 

0.836
1.363
0.617
258

862


 

 0.557 0.464 1.300  129   575 


 

ECO 3401 - B. Potter
108
Input-Output Models

Excel
 I  A
1
DX
Use =MMULT(B5:D7,G1:G3)
ECO 3401 - B. Potter
109
Input-Output Models

What should production of each commodity be to satisfy
demands for 516 units of agriculture, 258 units of
manufacturing, and 129 units of transportation?
1
I

A

 DX
 1.393 0.494 0.583  516   921 


 

0.836
1.363
0.617
258

862


 

 0.557 0.464 1.300  129   575 


 


Production of 921 units of agriculture, 862 units of
manufacturing, and 575 units if transportation is needed
to satisfy demands of 516, 258, and 129.
ECO 3401 - B. Potter
110
Chapter 2
End
ECO 3401 - B. Potter
111