Transcript 4.1 Systems of Linear Equations in two variables

Lecture 4
Rules of Matricies
Addition and Subtraction
of Matrices

To add or subtract matrices, they must be of the same order,
m x n where m is the number of rows, and n is the number of
columns. We always say the number of rows, then the
number of columns in this order. To add matrices of the same
order, add their corresponding entries. To subtract matrices
of the same order, subtract their corresponding entries. The
general rule is as follows using mathematical notation: The i
represents the row number, and j the column number
A  B   aij  bij 
A  B   aij  bij 
Example: Addition

Add the matrices
 4 3 1   1 2 3
0 5 2   6 7 9

 

5 6 0   0 4 8
Example: Addition
Solution

Add the matrices
 4 3 1   1 2 3
0 5 2   6 7 9

 

5 6 0   0 4 8

Solution: First note that
each matrix has dimensions
of 3x3, so we are able to
perform the addition. The
result is shown at right:

Adding corresponding
entries, we have
3 1 4
6 2 7 


5 10 8 
Example: Subtraction

Now, we will subtract the
same two matrices
 4 3 1   1 2 3
0 5 2   6 7 9

 

5 6 0   0 4 8
Example: Subtraction
Solution

Now, we will subtract the
same two matrices
 4 3 1   1 2 3
0 5 2   6 7 9

 

5 6 0   0 4 8

Subtract corresponding
entries as follows:
3  2
1 3 
 4  (1)
 06

5

(

7)

2

9


 5  0
6  (4) 0  8 
=
 5 5 2 
 6 12 11


 5 2 8 
Scalar Multiplication

The scalar product of a number k and a matrix A is
the matrix denoted by kA, obtained by multiplying
each entry of A by the number k. The number k is
called a scalar. In mathematical notation,
kA   kaij 
Example: Scalar Multiplication

Find (-1)A, where A =
 1 2 3
 6 7 9


 0 4 8
Example: Scalar Multiplication
Solution
 Solution:

Find (-1)A, where A =
 1 2 3
 6 7 9


 0 4 8
 1 2 3
 6 7 9

(-1)A= -1 
 0 4 8
 1 2 3  1 2 3
 (1)  6 7 9    6 7 9 
 0 4 8  0 4 8
Alternate Definition of
Subtraction of Matrices

The definition of
subtraction of two real
numbers a and b is
a – b = a + (-1)b or
“a plus the opposite of b”.
We can define subtraction
of matrices similarly:

If A and B are two matrices
of the same dimensions,
then
A – B = A + (-1)B,
where (-1) is a scalar.
Example

The example on the
right illustrates this
procedure for two
2x2 matrices.
1 2 2  1
3 4  3  1

 

1 2
2  1

 (1) 


3 4
3  1
1 2  2 1




3
4

3
1

 

 1 3


 0 5
Matrix Equations
Example: Find a, b, c, and d so that
 a b   2 1  4 3
 c d    5 6    2 4

 
 

Matrix Equations
Example: Find a, b, c, and d so that
 a b   2 1  4 3
 c d    5 6    2 4

 
 

Solution: Subtract the matrices on the left side:
a  2 b  1   4 3
 c  5 d  6    2 4 

 

Use the definition of equality to change this matrix equation
into 4 real number equations:
a-2=4
a=6
b+1=3
b=2
c + 5 = -2
c = -7
d-6=4
d = 10
Product of a Row Matrix
and a Column Matrix


In order to understand the general procedure of
matrix multiplication, we will introduce the concept
of the product of a row matrix by a column matrix.
A row matrix (or ROW VECTOR) consists of a single
row of numbers, while a column matrix (or COLUMN
VECTOR) consists of a single column of numbers. If
the number of columns of a row matrix equals the
number of rows of a column matrix, the product of a
row matrix and column matrix is defined. Otherwise,
the product is not defined.
Row by Column Multiplication
Example: A row matrix consists of 1 row of 4
numbers so this matrix has four columns. It has
dimensions 1 x 4. This matrix can be multiplied
by a column matrix consisting of 4 numbers in
a single column (this matrix has dimensions 4 x
1).
 1x4 row matrix multiplied by a 4x1 column
matrix. Notice the manner in which
corresponding entries of each matrix are
multiplied:

Matricies in Mathematica



When multiplying matricies
in Mathematica, you have
to use what is called the
“Dot” operator, see RV.CV
RV is 1x4
CV is 4x1, thus they are
conformable, and result in a
1x1 (otherwise known as a
scalar)
Pay attention to order
Be careful however,
as we will see, it
REALLY matters
what order you put
things in when doing
matrix mathematics
 Multiplying CV.RV,
gives you a 4x4.
Why?

Example:
Revenue of a Car Dealer

A car dealer sells four model types: A, B, C, D. In a
given week, this dealer sold 10 cars of model A, 5 of
model B, 8 of model C and 3 of model D. The selling
prices of each automobile are respectively $12,500,
$11,800, $15,900 and $25,300. Represent the data
using matrices and use matrix multiplication to find
the total revenue.
Solution using Matrix
Multiplication

We represent the number of each model sold using a
row matrix (4x1), and we use a 1x4 column matrix to
represent the sales price of each model. When a 4x1
matrix is multiplied by a 1x4 matrix, the result is a 1x1
matrix containing a single number.
12,500 
11,800 
  10(12,500)  5(11,800)  8(15,900)  3(25,300)  387,100
10 5 8 3 
15,900 


 25,300 
Mathematica… you need the dot
Matrix Product (Conformity)

If A is an m x p matrix and B is a p x n matrix, the
matrix product of A and B, denoted by AB, is an m x
n matrix whose element in the i th row and j th
column is the real number obtained from the
product of the i th row of A and the j th column of B.
If the number of columns of A does not equal the
number of rows of B, the matrix product AB is not
defined. Very easy to keep track of if you always
write down the matrix dimensions. Inner numbers
must be equal, outer numbers give you new matrix
size with m rows and n columns
Premultiply and Postmultiply are
not the same (noncommutative)


Premultiply A by B means
 BA
Postmultiply A by B means
 AB
Multiplying a 2x4 matrix by a
4x3 matrix to obtain a 2x3

The following is an illustration of the product of a
2x4 matrix with a 4x3. (Note 4=4, and 2x3 is new
matrix dimension. First, the number of columns of
the matrix on the left must equal the number of
rows of the matrix on the right, so matrix
multiplication is defined. A row-by column
multiplication is performed three times to obtain the
first row of the product: 70 80 90.
Final Result
Undefined Matrix Multiplication
Why is the matrix multiplication below not
defined?
4 by 3 and a 2 by 4. 3 is not 2… can’t
proceed!
Undefined Matrix Multiplication
Solution
Why is the matrix multiplication below not defined?
The answer is that the left matrix has three columns but the
matrix on the right has only two rows. To multiply the second
row [4 5 6] by the third column, 3 , there is no number to pair
with 6 to multiply.
7
Example
Given A =
3 1 1
2 0 3 


B=
1
3

 2
6
5
4 
Find AB if it is defined:
3 1 1
2 0 3 


2x3
1
3

 2
3x2
6
5
4 
3=3
2x2 is new size
Solution

Since A is a 2 x 3 matrix and
B is a 3 x 2 matrix, AB will
be a 2 x 2 matrix.
1. Multiply first row of A by
first column of B:
3(1) + 1(3) +(-1)(-2)=8
2. First row of A times
second column of B:
3(6)+1(-5)+ (-1)(4)= 9
3. Proceeding as above the
final result is
1
6

3 1 1 

3

5
2 0 3  



 2 4 
=
8 9
 4 2 4 


Is Matrix Multiplication
Commutative?
Now we will attempt to multiply the matrices in reverse
order: BA = ?
 We are multiplying a 3 x 2 matrix by a 2 x 3 matrix. This
matrix multiplication is defined, but the result will be a 3 x 3
matrix. Since AB does not equal BA, matrix multiplication is
not commutative.
15 1 17 
6  3 1 1
1
 3 5 

 = 
2
0
3

1
3

18





 2 4 

 2
2
14 
Transposes of Matricies

Now, we will find the
Transpose of the matrix.
You just make the rows into
columns.

If it is a column vector, it
becomes a row vector, and
vice versa

It is denoted by a PRIME or
what you see as an
apostrophe.
a
d

 g
a
b

 c
b
e
h
d
e
f
c
f 
i 

g
h 
i 
Rules of Transpose



(A’)’=A
(A+B)’=A’+B’
(AB)’=B’A’
 The third is the most
confusing, but most
important!!!
 We use this property
quite a bit in
economics
Practical Application

Suppose you a business owner and sell clothing. The following
represents the number of items sold and the cost for each
item. Use matrix operations to determine the total revenue
over the two days:
Monday: 3 T-shirts at $10 each, 4 hats at $15 each, and
1 pair of shorts at $20.
Tuesday: 4 T-shirts at $10 each, 2 hats at $15 each, and
3 pairs of shorts at $20.
Solution of Practical Application

Represent the information using two matrices: The product
of the two matrices gives the total revenue:
Unit price of
each item:
3 4
4 2
10
15
20

 
1 3 

Qty sold
of each
item on
Monday
Qty sold of
each item on
Tuesday
Then your total revenue for the two days is = [110 130]
Price times Quantity = Revenue
Identity Matrix for Multiplication

1 is called the multiplicative identity for real
numbers since
a(1) = (1)a = a
For example 5(1) = 5
A matrix is called square if it has the same number of rows and
columns, that is, it has size n x n.
The set of all square matrices of size n x n also has a
multiplicative identity In, with the property
AIn = InA = A
In is called the n x n identity matrix.
Identity Matrices

2 x 2 identity matrix:
1 0 
0 1 



3 x 3 identity matrix
1 0 0 
0 1 0 


0 0 1 
Identity Matrix Multiplication


AI = A (Verify the multiplication)
We can also show that IA = A and in general AI = IA =
A for all square matrices A.
 a11 a12 a13  1 0 0  a11 a12 a13 





a
a
a
0
1
0

a
a
a
21
22
23
21
22
23 


 
 a a a  0 0 1   a a a 
 31 32 33  
  31 32 33 
Inverse of a Matrix

All real numbers (excluding 0) have an inverse.
1
a 1
a
.
For example
1
5 1
5
Matrix Inverses




Matrix must be square to have an inverse
Some (not all) square matrices also have matrix inverses
If the inverse of a matrix A exists, we shall call it A-1
Then
1
1
A  A  A  A  In
Notice this is ‘commutative’ and a square multiplied by it’s
inverse is equal to the identity matrix. More specifically, the
equation to find an inverse is 1/determinant of A, multiplied
by the adjoint (or minor) of A. Determinants? Adjoints?
1
A  adjA, where adjA  C ' , where C is the adjoint of A
A
1
This formula comes in handy
1
A  adjA, where adjA  C ' , where C is the adjoint of A
A
1
Rules of Matrix Inversion

1 1
(A )  A
1
1
( AB)  B A
1
1
( A' )  ( A )'
1
The second rule here is
the most frequently
used. It’s really
important you get the
order correct.
Inverse of a 2x2 Matrix, formulaic



There is a simple procedure to find the inverse of a two by
two matrix. This procedure only works for the 2 x 2 case.
An example will be used to illustrate the procedure.
Example: Find the inverse of
2 3
1 2 


Inverse of a 2x2 matrix (memorize)




Step 1: Determine whether or not the inverse actually
exists. We define Δ = the determinant of the matrix.
2 3
1 2 


In order for the inverse of a 2 x 2 matrix to exist, Δ cannot
equal zero. (We will call this a determinant)
If Δ happens to be zero, then we conclude the inverse does
not exist, and we stop all calculations.
In our case Δ = 2(2)-1(3) = 1, so we can proceed.
Det[] finds the matrix
determinant ad-bc for
a 2x2
Inverse of a 2x2 matrix (continued)
2 3
1 2 


 2 3 
 1 2 


Step 2. Reverse the entries on
the main diagonal.
In this example, both entries
are 2, and no change is visible.
Step 3. Reverse the signs of
the other diagonal entries 3
and 1 so they become -3 and 1
Step 4. Divide each element
of the matrix by the
‘determinant’ which in this
case is 1, so no apparent
change will be noticed.
Inverse of a 2x2 matrix (continued)
 2 3 
 1 2 



The inverse of the matrix is then

To verify that this is the inverse, we will multiply the original
matrix by its inverse and hopefully obtain the 2 x 2 identity
matrix:
 2 3   2 3 
 1 2   1 2 


 
=
 4  3 6  6  1 0 
 2  2 3  4   0 1 

 

General 2x2 Matrix Inversion
Adjoint (Minors and Cofactors)

The adjoint is defined as a matrix of ‘cofactors’ (see
the C) which are determinants of minor matricies
multiplied by negative 1 if the sum of R&C is odd

| C11 |

C '  adjA  | C21 |

| C |
 n1

| C12 |  | C1n | 

| C22 |  | C2 n |, where | Cij | (1) i  j | M ij |



| Cn 2 |  | Cnn |
M ij is defined as the minor of ELEMENT ij. Or all elements of the matrix tha t
are NOT IN ROW i, and COLUMN j
The minor of element a11 below is
 a11 a12
a
 21 a22
a31 a32
a13 
a22

a23 , where M11  
a32

a33 
a23 
 a11 a13 
, and M 22  


a33 
a
a
 31 33 
First, a 2x2
 2 3

, First, the determinan t of A, is (2)(2) - (3)(1)  1. So our first term in Chiang' s
1
2


formula for finding an inverse is 1/1 or more simply 1.
Now, the cofactor matrix. The minor M11  2 , the lower right term that is not in row one or
column one, and i  j  1  1  2, thus (-1)2  1, and is positive
The determinan t of M11  2 since the determinan t of a scalar is the value of the scalar
2 
.
Thus element 1,1 in our cofactor matrix is 2. 


M12  1, elements not in row 1, column 2. Determinan t  1, Sign changes since 1  2  3 and (-1)3  1
 2  1
 2  1

, Now M 21  3, with Determinan t 3, and sign change. 
,

3




 2  1

Now M 22  2 , with Determinan t 2, and no sign change (-1)even  1 giving 

3
2


 2  3

Now, we must trans pose this ' cofactor matrix' giving 

1
2


The last step... multiply t his by 1/ | A | or 1 in this example.
Example: Inverse of a 3x3 Matrix
 1 1 3 
 2 1 2


 2 2 1 

Find the inverse of

Start by finding the determinant of the matrix (expand along
first column 1((1)(1)-(2)(-2)) + (-1)2((-1)(1)-(3)(-2)) + (-2)((-1)(2)-(3)(1)) =
1(5)+-2(5)+(-2)(-5) = 5. The first term in the formula is 1/|A| = 1/5 or 0.2.
 1 1 3


 1 2
2


2
1
2
,
where
M

,
and
i

j

1

1

2,
thus
(-1)
 1, and is positive


11
 2 1


 2  2 1



The determinan t of M11  1*1  (2 * 2)  5, thus element 1,1 in our cofactor matrix is 5
 2 2
, giving a determinan t of 6, multiplied by (-1)3  1, giving 1,2 element 6
M12  
 2 1
1 
 2
, determinan t is - 2, multiplied by 1, giving element 1,3 a value of - 2
M13  

2

2


 1 1

1
 2
 2  2

 1
M11  
 2
3

1
2 , A -1  adjA. adjA  C' , where C  matrix of minor determinan ts.
.2
1 
2
, and i  j  1  1  2, thus (-1)2  1, and is positive
1
The determinan t of M11  1(1)  (2)( 2)  5, thus element 1,1 (C11 ) in our cofactor matrix is 5
 2 2
, giving a determinan t of 6, multiplied by (-1)3  1, giving C12  -6
M12  
 2 1
1 
 2
, determinan t is - 2, multiplied by 1, giving element 1,3 a value of C13  -2.
M13  

2

2


5 - 6 - 2


C





 1 1

1
 2
 2  2

 1
M 21  
 2
3

1
2 , A -1  adjA. adjA  C' , where C  matrix of minor determinan ts.
.2
1 
3
, D  5, sign change, C 21  5
1 
 1 3
, D  7, no sign change, C 22  7
M 22  

2
1


 5 - 6 - 2


 1 1 
, D  4, sign change, C 23  4, C    5 7 4 
M 23  
  2  2




 1 3 
, D  5, no sign change, C31  5
M 31  
1
2


1 3
, D  4, sign change, C32  4
M 32  
 2 2
 5 -6

1

1


, D  3, no sign change, C33  3, C    5 7
M 33  
2 1 
5 4

-1
-1 
 5 - 5 - 5  1




1

6
7
4


6
/
5
7
/
5
4
/
5

 
. NOW it is an inverse
5
 

 2 4 3   2 / 5 4 / 5 3/ 5
- 2
 5 - 5 - 5



4 , C'    6 7 4 
 2 4 3 
3 


of our original. Check it!
Inverse of a General Square Matrix
1. Augment the matrix with the n x n identity matrix.
2. Use elementary row operations to transform the
matrix on the left side of the vertical line to the n x n
identity matrix. The row operations are used for the
entire row, so that the matrix on the right hand side
of the vertical line will also change.
3. When the matrix on the left is transformed to the n x
n identity matrix, the matrix on the right of the
vertical line is the inverse.
Example: Inverse of a 3x3 Matrix

Find the inverse of


 1 1 3 
 2 1 2


 2 2 1 
Step 1. Multiply R1 by (-2) and add the result to R2.
Step 2. Multiply R1 by 2 and add the result to R3
Example (continued)
 Step 3. Multiply row 2 by (1/3) to get a 1 in the second
row, first position.
 Step 4. Add R2 to R1.
 Step 5. Multiply R2 by 4 and add the result to R3.
 Step 6. Multiply R3 by 3/5 to get a 1 in the third row, third
position.
Example (continued)
 Step 7. Eliminate the (5/3) in the first row, third position by
multiplying R3 by (-5/3) and adding result to R1.
 Step 8. Eliminate the (-4/3) in the second row, third position
by multiplying R3 by (4/3) and adding result to R2.
 Step 9. You now have the identity matrix on the left, which
is our goal.
Example Solution

The inverse matrix appears on the right hand side of
the vertical line and is displayed below. Many
calculators as well as computers have software
programs that can calculate the inverse of a matrix
quite easily.
Determinant and Inverse of 3x3
GENERAL
Rearrangement of the Mathematica output in the second line you should see a(ei-fh)
+ b(ei-fg) + c(dh-ge). This is called ‘expanding’ along the first row, and multiplying
by the ‘sub-determinants’ or determinant of the matrix made up of those values that
are not in the same row or column as your expansion term…
Cramer’s Rule


Cramer’s Rule, extends from the previous method of
inversion using the adjoint (or transpose of the
cofactor matrix… or more thoroughly the transpose
of the matrix of minor determinants), to a system of
equations.
If we have a problem, such as Ax=d where x is a
column vector of unknowns, and d is a column
vector of answers, we can find xi by replacing the ith
column of A with the values from d, take the
determinant, and then divide it by the determinant
of A
Cramer’s Rule


Generally… x*j is the solution to the jth variable.
x*j is found by
 a11 a12  d1  a1n 


| Aj |
1  a21 a22  d 2  a2 n 





 
| A| | A| 


a

a

d

a
n2
n
nn 
 n1
Cramer’s Rule in Mathematica
Matrix Equations

Let’s review one property of solving equations involving real
numbers. Recall
1
b
b , or
If ax = b then x =
a
a

A similar property of matrices will be used to solve systems of
linear equations.

Many of the basic properties of matrices are similar to the
properties of real numbers, with the exception that matrix
multiplication is not commutative.
Basic Properties of Matrices
Assuming that all products and sums are defined for
the indicated matrices A, B, C, I, and 0, we have
 Addition Properties
 Associative: (A + B) + C = A + (B+ C)
 Commutative: A + B = B + A
 Additive Identity: A + 0 = 0 + A = A
 Additive Inverse: A + (-A) = (-A) + A = 0
Basic Properties of Matrices
(continued)


Multiplication Properties
 Associative Property: A(BC) = (AB)C
 Multiplicative identity: AI = IA = A
 Multiplicative inverse: If A is a square matrix and
A-1 exists, then AA-1 = A-1A = I
Combined Properties
 Left distributive: A(B + C) = AB + AC
 Right distributive: (B + C)A = BA + CA
Basic Properties of Matrices
(continued)

Equality
 Addition: If A = B, then A + C = B + C
 Left multiplication: If A = B, then CA = CB
 Right multiplication: If A = B, then AC = BC
The use of these properties is best illustrated by an example of
solving a matrix equation.
Example: Given an n x n matrix A and an n x p matrix B and a
third matrix denoted by X, we will solve the matrix equation
AX = B for X.
Solving a Matrix Equation
AX  B
A
1
 AX   A
 A A X  A
1
 In  X  A
1
XA B
1
1
1
B
B
B
Reasons for each step:
1. Given; since A is n x n,
X must by n x p.
2. Multiply on the left by A-1.
3. Associative property of matrices
4. Property of matrix inverses.
5. Property of the identity matrix
6. Solution. Note A-1 is on the left of
B. The order cannot be reversed
because matrix multiplication is
not commutative.
Example

Example: Use matrix inverses to
solve the system
x  y 2 z  1
2x  y
2
x 2 y 2 z  3
Example

Example: Use matrix inverses to
solve the system

Solution:
 Write out the matrix of
coefficients A, the matrix X
containing the variables x, y,
and z, and the column matrix
B containing the numbers on
the right hand side of the
equal sign.
x  y 2 z  1
2x  y
2
x 2 y 2 z  3
1 1 2 
A   2 1 0 
1 2 2 
1 
x
B   2 
X   y 
 3 
 z 
Example (continued)
 Form the matrix equation AX = B. Multiply the 3 x 3
matrix A by the 3 x 1 matrix X to verify that this
multiplication produces the 3 x 3 system at the bottom:
1 1 2   x 
2 1 0  

  y
1 2 2   z 
 
x
2x
x
y
y
2 y
2 z
1 
  2 
 3 
1
2
2 z  3
Example (continued)
 If the matrix A-1 exists, then the
solution is determined by
multiplying A-1 by the matrix B.
Since A-1 is 3 x 3 and B is 3 x 1,
the resulting product will have
dimensions 3 x 1 and will store
the values of x, y and z.
 A-1 can be determined by the
methods of a previous section
or by using a computer or
calculator. The resulting
equation is shown at the right:
1
XA B
1 1
2 2

X   1 0
 3 1

4 4
1 
2  1 
 
1  2
1   3 

4
Example Solution
The product of A-1 and B is
X  A1B
1 1
2 2

X   1 0
 3 1

4 4
1 
2  1 

1   2 
1   3 

4
 
0
 
X  2 
 1 
 
2
The solution can be read off
from the X matrix:
x = 0,
y = 2,
z = -1/2
Written as an ordered triple
of numbers, the solution is
(0, 2, -1/2)
Doing this using Mathematica
Another Example
Example: Solve the system on the right x  2 y  z  1
using the inverse matrix method.
2x  y  2z  2
You can enter this in Mathematica the 3 x  y  3 z  4
old way as well if you desire.
Notice, that this won’t work, what is
Mathematica telling us?
Another Example
Example: Solve the system on the right x  2 y  z  1
using the inverse matrix method.
2x  y  2z  2
3x  y  3z  4
Solution:
The coefficient matrix A is displayed at
the right. The inverse of A does not
exist. (We can determine this by using
a calculator.) We cannot use the
inverse matrix method. Whenever the
inverse of a matrix does not exist, we
say that the matrix is singular. We can
figure this out because the
determinant of this matrix is ZERO
1 2 1 
 2 1 2


 3 1 3
For matrix
 Determinant is
1((-1)*3-2*1)+
2(2*3-2*3)+
1(2*1-(-1)*3)

Singular Matrix



Determinant is ZERO
For a system to be
solvable… it must have
a NON-ZERO
DETERMINANT
Also called
“NONSINGULAR”
OR
-5+0+5=0… This is singular
Try expanding along the
second row, or second
column… always the
same
Cases When Matrix Techniques
Do Not Work

There are two cases when inverse methods will not work:
1. If the coefficient matrix is singular
2. If the number of variables is not the same as the number
of equations.
Application

Production scheduling: Labor and material costs for
manufacturing two guitar models are given in the table
below: Suppose that in a given week $1800 is used for labor
and $1200 used for materials. How many of each model
should be produced to use exactly each of these allocations?
Guitar model
Labor cost
Material cost
A
$30
$20
B
$40
$30
Solution
Let x be the number of model
A guitars to produce and y
represent the number of
model B guitars. Then,
multiplying the labor costs for
each guitar by the number of
guitars produced, we have
30x + 40y = 1800
Since the material costs are
$20 and $30 for models A and
B respectively, we have
20x + 30y = 1200.
This gives us the system of
linear equations:
30x + 40y = 1800
20x + 30y = 1200
We can write this as a matrix
equation:
30 40  x  1800
20 30  y   1200

  

Solution (continued)
X  A1B
30 40
A

 20 30 
The inverse of matrix A is

Solution:
Produce 60 model A guitars
and no model B guitars.
 0.3 0.4 
 0.2 0.3 


 x   0.3  0.4 1800 60
 y    0.2 0.3  1200   0 
  

  
Leontief Input-Output Analysis


In this section, we will study an important economic
application of matrix inverses and matrix multiplication.
This branch of applied mathematics is called input-output
analysis and was first proposed by Wassily Leontief, who
won the Nobel Prize in economics in 1973 for his work in
this area.
Wassily Leontief (continued)
At Harvard, he developed his theories and methods of InputOutput analysis. This work earned him the Nobel prize in
Economics in 1973 for his analysis of America's production
machinery. His analytic methods, as the Nobel committee
observed, became a permanent part of production planning
and forecasting in scores of industrialized nations and in
private corporations all over the world.
Professor Leontief passed away on Friday February 6th, 1999.
For more information on his life, visit
http://www.iioa.org/leontief/Life.html
Basic Input-Output Problem
The main problem of input-output analysis is
the following:
Consider an economy with several industries.
Each industry has a demand for products from
other instrustries (internal demand). There
are also external demands from the outside.
Find a production level for the industries that
will meet both internal and external demands.
Two-Industry Model Example
We start with an economy that has only two industries
(agriculture and energy) to illustrate the method. Later, this
method will generalized to three or more industries.
These two industries depend upon each other. For example,
each dollar’s worth of agriculture produced requires $0.40 of
agriculture and $0.20 of energy. Each dollar’s worth of energy
produced requires $0.20 of agriculture and $0.10 of energy.
So, both industries have an internal demand for each others
resources. Let us suppose there is an external demand of
$12,000,000 of agriculture and $9,000,000 dollars of energy.
Example: Matrix Equations
Let x represent the total output
from agriculture and y
represent the total output of
energy (in millions of $)
The expressions
0.4x + 0.2y
0.2x + 0.1y
can be used to represent the
internal demands for
agriculture and energy.
The external demands of 12 and
9 million must also be met, so
the revised equations are :
x = 0.4x + 0.2y + 12
y = 0.2x + 0.1y + 9
These equations can be
represented by the following
matrix equation:
 x  0.4 0.2  x  12
 y   0.2 0.1  y    9 
  
   
Example:Technology Matrix (M )
0.4 0.2 
0.2 0.1 


A
E
A
Read left to right,
then up
 input from



agriculture


 to produce



 $1 of agriculture 
  input of energy 


  to produce $1 


  of agriculture 
E





 =M
 input of energy  


to
produce
$1
of


 energy


 
 input from

 agriculture
 to produce $1

 of energy






Example: Solving the Matrix Equations
 x  0.4 0.2  x  12
 y   0.2 0.1  y    9 
  
   
We can solve this matrix equation as follows:
X = MX+D
X – MX = D
IX – MX = D
(I – M)X = D
X  ( I  M )1 D
if the inverse of (I – M) exists.
Example: Solution
1
We will now find X  ( I  M ) D
First, find (I – M):
1 0 0.4 0.2  0.6 0.2
0 1   0.2 0.1   0.2 0.9 

 
 

The inverse of (I – M) is:
1.8 .4 
 .4 1.2 


Solution (continued)
After finding the inverse of (I – M), multiply that result
by the external demand matrix D. The answer is:
Produce a total of $25.2 million of agriculture and
$15.6 million of energy to meet both the internal
demands of each resource and the external demand.
1.8 .4  12   25.2 
 .4 1.2   9   15.6 


  
Another Example
Suppose consumer demand changes from $12
million dollars of agriculture to $8 million
dollars and energy consumption changes from
$9 million to $5 million. Find the output for
each sector that is needed to satisfy this final
demand.
Example (continued)
Suppose consumer demand changes from $12
million dollars of agriculture to $8 million
dollars and energy consumption changes from
$9 million to $5 million. Find the output for
each sector that is needed to satisfy this final
demand. X  ( I  M )1 D
Solution: Recall that our general solution of the
problem is
Example
Solution
1
X  (I  M ) D
1.8 .4  8 16.4 

 .4 1.2  5  9.2 

  

More Than Two Sectors
of the Economy
This method can also be used if there are more than
two sectors of the economy. If there are three sectors,
say agriculture, building and energy, the technology
matrix M will be a 3 x 3 matrix. The solution to the
problem will still be
1
X  (I  M ) D
although in this case it is necessary to determine the
inverse of a 3 x 3 matrix.
Example: Three-Industry Model
An economy is based on three sectors, agriculture (A), energy
(E), and manufacturing (M). Production of a dollar’s worth of
agriculture requires an input of $0.20 from the agriculture
sector and $0.40 from the energy sector. Production of a
dollar’s worth of energy requires an input of $0.20 from the
energy sector and $0.40 from the manufacturing sector.
Production of a dollar’s worth of manufacturing requires an
input of $0.10 from the agriculture sector, $0.10 from the
energy sector, and $0.30 from the manufacturing sector.
Find the output from each sector that is needed to satisfy a final
demand of $20 billion for agriculture, $10 billion for energy,
and $30 billion for manufacturing.
Example
(continued)
The technology matrix is as follows:
Output
Input
A E
Ma
A 0.2 0 0.1
E 0.4 0.2 0.1
Ma  0 0.4 0.3
0
0.1
 0.8
I  M   0.4 0.8 0.1
 0
0.4 0.7 
=M
1.3 0.1 0.2
( I  M )1  0.7 1.4 0.3
0.4 0.8 1.6 
Example
(continued)
Thus, the output matrix X is given by:
X
(I - M)-1
D
 x1  1.3 0.1 0.2  20  33
 x   0.7 1.4 0.3 10   37 
 2 
   
 x3   0.4 0.8 1.6  30  64
An output of $33 billion for agriculture, $37 billion for
energy, and $64 billion for manufacturing will meet the
given final demands.