### Chapter 2

```Chapter 2
Matrices
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Outline
2.1 Systems of Linear Equations with Unique Solutions
2.2 General Systems of Linear Equations
2.3 Arithmetic Operations on Matrices
2.4 The Inverse of a Matrix
2.5 The Gauss-Jordan Method for Calculating Inverses
2.6 Input-Output Analysis
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2.1 Systems of Linear Equations with Unique Solutions
1.
2.
3.
4.
5.
6.
7.
8.
Diagonal Form of a System of Equations
Elementary Row Operations
Elementary Row Operation 1
Elementary Row Operation 2
Elementary Row Operation 3
Gaussian Elimination Method
Matrix Form of an Equation
Using Spreadsheet to Solve System
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Diagonal Form of a System of Equations
A system of equations is in diagonal form if each
variable only appears in one equation and only
one variable appears in an equation.

For example:  x




 125
y
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50
z

50.
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Elementary Row Operations
Elementary row operations are operations on the
equations (rows) of a system that alters the
system but does not change the solutions.
Elementary row operations are often used to
transform a system of equations into a diagonal
system whose solution is simple to determine.
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Elementary Row Operation 1
Elementary Row Operation 1 Interchange
any two equations.
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Example Elementary Row Operations 1
Rearrange the equations of the system


3 x 
6 x

y 
z  0
y 

z  6
z  3
so that all the equations containing x are on
top.
6 x

[ R1 ]  [ R3 ]
 3x 


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
z  3
y 
y 
z  6
z  0
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Elementary Row Operation 2
Elementary Row Operation 2 Multiply an
equation by a nonzero number.
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Example Elementary Row Operation 2
Multiply the first row of the system
6 x

3 x 



z
y 
y 
z  6
z  0
so that the coefficient of x is 1.
x

1
[
R
]
 6  1  3x 




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 3

y 
y 
 16  z

z
z


1
2
6
0
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Elementary Row Operation 3
Elementary Row Operation 3 Change an
equation by adding to it a multiple of another
equation.
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Example Elementary Row Operation 3
Add a multiple of one row to another to change
x


3x 



y 
y 
 16  z

z
z


1
2
6
0
so that only the first equation has an x term.
x


[ R2 ]  3[ R1 ]

  



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
y 
y 
 16  z
 32 z
z

1
2
 9

2
0
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Gaussian Elimination Method
Gaussian Elimination Method transforms a
system of linear equations into diagonal form by
repeated applications of the three elementary row
operations.
1. Rearrange the equations in any order.
2. Multiply an equation by a nonzero number.
3. Change an equation by adding to it a multiple
of another equation.
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Example Gaussian Elimination Method
Continue Gaussian Elimination to transform into
diagonal form  x
 1 z  1






x


 1[ R2 ]

 



6

y 
 32 z
y 
z

y 
y 
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 16  z
 32 z
z
2
 9

2
0.

1
2
 9
2

0
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Example Gaussian Elimination (2)
x







y 
y 
 16  z
 32 z
z
x
2


[ R3 ]   1[ R2 ]
 9 
 
2


0



x

2  [ R ]


3
5

 



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
y 
 16  z
 32 z
z
 16  z
 32 z
 52 z

1
y 


1
2
 9
2
 9
2
1
2
 9
2
 9
5
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Example Gaussian Elimination ( 3)
x







y 
 16  z
 32 z
z

1
2
 9
2
 9
5
x

[ R2 ]  3 [ R3 ]

2

 






[ R1 ] 1 [ R3 ]
6
x







y

 16  z
y
z

1
2
 9
5
 9
2
4
5
 9
5
z  9
5
The solution is (x,y,z) = (4/5,-9/5,9/5).
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Matrix Form of an Equation
It is often easier to do row operations if the
coefficients and constants are set up in a table
(matrix, plural matrices).
Each row represents an equation.
Each column represents a variable’s coefficients
except the last which represents the constants.
Such a table is called the augmented matrix of
the system of equations.
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Example Matrix Form of an Equation
Write the augmented matrix for the system
 3x  6 y  9 z  0

 4 x  6 y  8 z  4
2 x  y  z  7.

 3 6 9 0 
 4 6 8 4 


 2 1 1 7 
Note: The vertical line
separates numbers that
are on opposite sides of
the equal sign.
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Example Gauss–Jordan Elimination
We must use elementary row operations to
transform this array into diagonal form— that is,
with ones down the diagonal, and zeros everywhere
else on the left:
 1 0 0 *
 0 1 0 *


 0 0 1 *
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Example Gauss–Jordan Elimination
We proceed one column at a time:
 3 6 9 0 
 4 6 8 4 


 2 1 1 7 
 1 2 3 0 
 0 2 4 4 


 2 1 1 7 
1 2 3 0 
0 1 2 2 


0 5 7 7 
 1 2 3 0 

  4 6 8 4 
 2 1 1 7 
1 R
3 1
R2  ( 4) R1


1 2 3 0 
R3  2 R1

 0 2 4 4 
0 5 7 7 
1 0 1 4 
0 1 2 2 
R1  2 R2



0 5 7 7 
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1 R
2 2


R3  5 R2

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Example Gauss–Jordan Elimination
Continuing:
1 0 1 4 
0 1 2 2 


0 0 3 3 
1 0 1 4 
 3   0 1 2 2 



0 0 1 1 
1 0 0 3
0 1 2 2 


0 0 1
1 
1 0 0 3
R2  2 R3
0 1 0 0 



0 0 1 1 
1
R3
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R1 1R3


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Example Gauss–Jordan Elimination
The last array is in diagonal form, so we just put
back the variables and read off the solution:
x = −3,
y = 0,
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z = 1.
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Using Spreadsheet to Solve System
Use a spreadsheet to solve


3 x 
6 x

y 
z

y 

z  6
z  3.
0
Enter the augmented matrix into your spreadsheet.
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Spreadsheet - Entering Left Side of Equations
A sample set up of the left side of the equations
in cells B1, B2 and B3 in Excel. The third
equation is shown. The three variables’ cells are
A1, A2 and A3.
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Spreadsheet - Entering Equations in Solver
A sample set up of the constraints (equations) in
Excel for Solver. The second constraint is
shown.
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Spreadsheet - Using Solver - Setup
Complete setup for
Solver.
Solution is
calculated once
Solve is clicked.
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Spreadsheet - Using Solver
A sample solution (in column A) in Excel using
Solver.
The solution
is
x = 0.8
y = -1.8
z = 1.8.
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Summary Section 2.1 - Part 1
 The three elementary row operations for a
system of linear equations (or a matrix) are as
follows:
Rearrange the equations (rows) in any order;
Multiply an equation (row) by a nonzero number;
Change an equation (row) by adding to it a
multiple of another equation (row).
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Summary Section 2.1 - Part 2
 When an elementary row operation is applied
to a system of linear equations (or an augmented
matrix) the solutions remain the same. The
Gaussian Elimination Method is a systematic
process that applies a sequence of elementary
row operations until the solutions can be easily
obtained.
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2.2 General Systems of Linear Equations
1.
2.
3.
4.
5.
Pivot a Matrix
Gaussian Elimination Method
Infinitely Many Solutions
Inconsistent System
Geometric Representation of System
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Pivot a Matrix
Method To pivot a matrix about a given
nonzero entry:
1. Transform the given entry into a one;
2. Transform all other entries in the same
column into zeros.
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Example Pivot a Matrix
Pivot the matrix about the circled element.
18 6 15
 5 2 4 


18 6 15
 5 2 4 


 3
1  R

6

1

 5
5 
2
2
4 
1
5 


3
1
R2  2 R1
2

 
 1 0 1 
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Gaussian Elimination Method
Gaussian Elimination Method to Transform a
System of Linear Equations into Diagonal
Form
1. Write down the matrix corresponding to the
linear system.
2. Make sure that the first entry in the first
column is nonzero. Do this by interchanging the
first row with one of the rows below it, if
necessary.
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Gaussian Elimination Method (2)
Gaussian Elimination Method to Transform a
System of Linear Equations into Diagonal
Form
3. Pivot the matrix about the first entry in the
first column.
4. Make sure that the second entry in the second
column is nonzero. Do this by interchanging the
second row with one of the rows below it, if
necessary.
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Gaussian Elimination Method (3)
Gaussian Elimination Method to Transform a
System of Linear Equations into Diagonal
Form
5. Pivot the matrix about the second entry in the
second column.
6. Continue in this manner until the left side of
the matrix is in diagonal form.
7. Write the system of linear equations
corresponding to the matrix.
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Infinitely Many Solutions
When a linear system cannot be completely
diagonalized,
1. Apply the Gaussian elimination method to as many
columns as possible. Proceed from left to right, but do
not disturb columns that have already been put into
proper form. As much as possible, each row should have
a 1 in its leftmost nonzero entry. (Such a 1 is called a
leading 1.) The column for each leading 1 should be to
the right of the columns for the leading 1s in the rows
above it.
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Infinitely Many Solutions (2)
2. Variables corresponding to columns not in
proper form can assume any value.
3. The other variables can be expressed in terms
of the variables of step 2.
4. This will give the general form of the solution.
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Example Infinitely Many Solutions
Find all solutions of
 2 2 4 8
 1 1 2 2 


 1 5 2 2 
2x  2 y  4z  8

 x  y  2z  2
 x  5 y  2 z  2.

 12  R
1 1 2 4 
0 2 0 2 
R2  ( 1) R1


R3  1 R1


0 6 0 6 
 12  R
1 0 2 3 
0 1 0 1 
R1  ( 1) R2


R3   6  R2


0 0 0 0 
2
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1
General Solution
z = any real number
x = 3 - 2z
y=1
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Inconsistent System
When using the Gaussian Elimination Method, if
a row of zeros occurs to the left of the vertical
line and a nonzero number is to the right of the
vertical line in the same row, then the system has
no solution and is said to be inconsistent.
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Example Inconsistent System
Find all solutions of
 1 1 1 3
 1 1 1 5


 2 4 4 1
 y  z  3
 x

 y  z  5
 x
2 x  4 y  4 z  1.

1 1 1 3 
0 2 2 2 
R2  ( 1) R1


R3   2  R1


0 2 2 7 
 12  R
1 0 0 4 
 0 1 1 1 
R1  (1) R2


R3   2  R2


 0 0 0 5 
2
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Because of the last
row, the system is
inconsistent.
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Summary Section 2.2 - Part 1
 The process of pivoting on a specific entry of
a matrix is to apply a sequence of elementary
row operations so that the specific entry becomes
1 and the other entries in its column become 0.
 To apply the Gaussian Elimination Method,
proceed from left to right and perform pivots on
as many columns to the left of the vertical line as
possible, with the specific entries for the pivots
coming from different rows.
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Summary Section 2.2 - Part 2
 After an augmented matrix has been
completely reduced with the Gaussian
Elimination Method, all the solutions to the
corresponding system of linear equations can be
obtained.
If the reduced augmented matrix has a 1 in
every column to the left of the vertical line, then
there is a unique solution.
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Summary Section 2.2 - Part 3
If one row of the reduced augmented matrix
has the form 0 0 0 … 0 | a where a ≠ 0, then
there is no solution.
 Otherwise, there are infinitely many solutions.
In this case, variables corresponding to columns
that have not been pivoted can assume any
values, and the values of the other variables can
be expressed in terms of those variables.
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2.3 Arithmetic Operations on Matrices
1.
2.
3.
4.
5.
6.
7.
Definition of Matrix
Column, Row and Square Matrix
Addition and Subtraction of Matrices
Multiplying Row Matrix to Column Matrix
Matrix Multiplication
Identity Matrix
Matrix Equation
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Definition of Matrix
A matrix is any rectangular array of numbers and
may be of any size.
The size of a matrix is nxk where n is the number
of rows and k is the number of columns.
The entry aij refers to the number in the ith row
and jth column of the matrix.
Two matrices are equal provided that they have
the same size and that all their corresponding
entries are equal.
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Example Definition of Matrix
 4 1 5  is a 2x3 matrix.
3 0 7

 The entry a1,2 = -1.
The entry a2,3 = 7.
3 9 3 9
7 0   7 0 

 

 2 2   2 2 
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3 9
3 7 2  

9 0 2    7 0 

 2 2


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Column, Row and Square Matrix
A row matrix or row vector only has one row.
A column matrix or column vector only has one
column.
A square matrix has the same number of rows as
columns.
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Example Column, Row & Square Matrix
 4 1
 3 7  is a 2x2 matrix and a square matrix.


2
8 7 1 is a 1x4 matrix and a row matrix.
 2 
1 
 2  is a 3x1 matrix and a column matrix.
 
3.4 
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Addition and Subtraction of Matrices
The sum A + B of two matrices A and B is
defined only if A and B are two matrices of the
same size. In this case A + B is the matrix
formed by adding the corresponding entries of A
and B.
Two matrices of the same size are subtracted by
subtracting corresponding entries.
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Example Addition & Subtraction
2 1 3  1 4 7   1 3 10 
4 0 5   8 3 2  12 3 3


 
 
2 1 3  1 4 7   3 5 4
4 0 5   8 3 2  4 3 7 

 
 

 1 8 
 2 1 3  

 4 0 5   4 3 is not defined.

 7 2


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Multiplying Row Matrix to Column Matrix
If A is a row matrix and B is a column matrix, then we
can form the product AB provided that the two matrices
have the same length. The product AB is a 1x1 matrix
obtained by multiplying corresponding entries of A and
B and then forming the sum.
 a1 a2
 b1 
b 
an    2    a1b1  a2b2 
 
 
bn 
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 anbn 
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Example Multiplying Row to Column
 3 
2
2

1
3


 
 5 
  2  3   1  2  3  5  7 
 3 


 4 0 2 1   2 
 5 
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is not defined.
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Matrix Multiplication
If A is an mxn matrix and B is an nxq matrix,
then we can form the product AB. The product
AB is an mxq matrix whose entries are obtained
by multiplying the rows of A by the columns of
B. The entry in the ith row and jth column of the
product AB is formed by multiplying the ith row
of A and jth column of B.
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Example Matrix Multiplication
 3 2 0 
 2 1 3  

 3 0 2    2 1 2 

  5 3 1


 7 12 -5 

2 
-19 0


 3 2 0 
 2 1 2   2 1 3  is not defined.

  3 0 2 


 5 3 1
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Identity Matrix
The identity matrix In of size n is the nxn square
matrix with all zeros except for ones down the
upper-left-to-lower-right diagonal.
Here are the identity matrix of sizes 2 and 3:
1 0 
I2  

0 1 
Finite Mathematics & Its Applications, 11/e by Goldstein/Schneider/Siegel
1 0 0 


I 3  0 1 0  .
0 0 1 
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Example Identity Matrix
1 0 2 1 3 
 2 1 3 
0 1  3 0 2   3 0 2





1 0 0 
2

1
3


 2 1 3  


0
1
0
 3 0 2  
  3 0 2

 0 0 1 


For all nxn matrices A, In A = A In = A.
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Matrix Equation
The matrix form of a system of linear equations
is
AX = B
where A is the coefficient matrix whose rows
correspond to the coefficients of the variables in
the equations. X is the column matrix
corresponding to the variables in the system. B is
the column matrix corresponding to the constants
on the right-hand side of the equations.
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Example Matrix Equation
Write the following system as a matrix equation
 4x  7 y  9

3x  2 y  5.
x
y
constants
 4 7   x   9 
 




Equation 2  3 2   y   5
Equation 1
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Summary Section 2.3 - Part 1
 A matrix of size mxn has m rows and
n columns.
 Matrices of the same size can be added (or
subtracted) by adding (or subtracting)
corresponding elements.
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Summary Section 2.3 - Part 2
The product of an mxn and an nxr matrix is the
mxr matrix whose ijth element is obtained by
multiplying the ith row of the first matrix by the
jth column of the second matrix. (The product of
each row and column is calculated as the sum of
the products of successive entries.)
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2.4 The Inverse of a Matrix
1.
2.
3.
4.
Inverse of A
Inverse of a 2x2 Matrix
Matrix With No Inverse
Solving a Matrix Equation
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Inverse of A
The inverse of a square matrix A, denoted by A-1,
is a square matrix with the property
A-1A = AA-1 = I,
where I is an identity matrix of the same size.
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Example Inverse of A
1 
4
Verify that  11 11 is the inverse of
 3
2 
 11 11
2 1
3 4  .


1 
4
2

1
1
0




11
11





 3
2   3 4  0 1 
 11 11
checks
4
1 

 2 1  11
11 1 0 

 3 4   3


0
1
2




 11 11
checks
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Inverse of a 2x2
To determine the inverse of a b  if D = ad - bc ≠ 0,
c d 


1. Interchange a and d to get  d b  .


c
a
2. Change the signs of b and c to get  d b 
 c a  .


3.
b 
d
D .
Divide all entries by D to get  D
 c
a 
D
 D
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Example Inverse of a 2x2
2 4
Find the inverse of 
.

 3 7 
D = (-2)(7) - (4)(-3) = -2 ≠ 0
7

4


7
4


2.
1.
Interchange:  3 2 Change signs:  3 2 
4   7
7

2
3.
2    2
 2
  A1
Divide:  3
2   3
1
2   2
 2

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Matrix With No Inverse
A matrix a b  has no inverse if
c d 
D = ad - bc = 0.
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Example Inverse or No Inverse
Use D to determine which matrix has an inverse.
 2 4 
 3 6 


has no inverse.
 2 4 
 3 6


has an inverse.
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Solving a Matrix Equation
Solving a Matrix Equation If the matrix A
has an inverse, then the solution of the matrix
equation
AX = B is given by X = A-1B.
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Example Solving a Matrix Equation
2 x  4 y  2
Use a matrix equation to solve 
3x  7 y  7.
The matrix form of the equation is
 2 4  x   2
 3 7   y   7  .

   
1
7


2
2
7
 x   2 4  2  2





 
 y    3 7  7    3


  
  
1  7   4
 2

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Summary Section 2.4 - Part 1
 The inverse of a square matrix A is a square
matrix A-1 with property that A-1A = I and AA-1 = I,
where I is the identity matrix.
 A 2x2 matrix
a b 
c d 


has an inverse if
 = ad - bc ≠ 0. If so, the inverse matrix is
b 
d
 .
 
 c
a 

 
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Summary Section 2.4 - Part 2
 A system of linear equations can be written in
the form AX = B, where A is a rectangular matrix
of coefficients of the variables, X is a column of
variables, and B is a column matrix of the
constants from the right side of the system. If the
matrix A has an inverse, then the solution of the
equation is given by X = A-1B.
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2.5 The Gauss-Jordan Method for
Calculating Inverses
1. Gauss-Jordan Method for Inverses
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Gauss-Jordan Method for Inverses
Step 1: Write down the matrix A, and on its right
write an identity matrix of the same size.
Step 2: Perform elementary row operations on the
left-hand matrix so as to transform it into an
identity matrix. These same operations are
performed on the right-hand matrix.
Step 3: When the matrix on the left becomes an
identity matrix, the matrix on the right is the
desired inverse.
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Example Inverses
 4 2 3 
Find the inverse of A   8 3 5  .


 7 2 4 
 4 2 3 1 0 0 
Step 1:


Step 2:
8 3 5 0 1
7 2 4 0 0
1 1
3
1
2
4
4

0 1
1 2

5 7
0 3
2
4
4

Finite Mathematics & Its Applications, 11/e by Goldstein/Schneider/Siegel
0
1 
0 0

1 0

0 1

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Example Inverses (2)
1 0

0 1

0 0

3
0
4
4
2

1 2
1
0

1 5
3
1
4 4
2 
1
1
1 0 0 2 2 1


0 1 0 3 5 4 
0 0 1 5 6 4 
Finite Mathematics & Its Applications, 11/e by Goldstein/Schneider/Siegel
Step 3:
 2 2 1


1
A   3 5 4 
 5 6 4 
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Summary Section 2.5
 To calculate the inverse of a matrix by the
Gauss-Jordan method, append an identity matrix
to the right of the original matrix and perform
pivots to reduce the original matrix to an identity
matrix. The matrix on the right will then be the
inverse of the original matrix. (If the original
matrix cannot be reduced to an identity matrix,
then the original matrix does not have an
inverse.)
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2.6 The Input-Output Analysis
1.
2.
3.
4.
Input-Output Analysis
Input-Output Matrix
Final Demand
Production Level Problem
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Input-Output Analysis
Input-output analysis is used to analyze an
economy in order to meet given consumption
and export demands.
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Input-Output Matrix
The economy is divided into a number of
industries. Each industry produces a certain
output using the outputs of other industries as
inputs. This interdependence among the
industries can be summarized in a matrix - an
input-output matrix. There is one column for
each industry’s input requirements. The entries
in the column reflect the amount of input
required from each of the industries.
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Input-Output Matrix - Form
A typical input-output matrix looks like:
Input requirements of:
Industry 1 Industry 2 Industry 3
Industry 1 
From Industry 2 

Industry 3 



.




Each column gives the dollar values of the various inputs needed
by an industry in order to produce \$1 worth of output.
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Example Input-Output Matrix
An economy is composed of three industries coal, steel, and electricity. To make \$1 of coal, it
takes no coal, but \$.02 of steel and \$.01 of
electricity; to make \$1 of steel, it takes \$.15 of
coal, \$.03 of steel, and \$.08 of electricity; and to
make \$1 of electricity, it takes \$.43 of coal, \$.20
of steel, and \$.05 of electricity. Set up the inputoutput matrix for this economy.
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Example Input-Output Matrix Answer
Coal Steel Electricity
 0

Steel .02
Electricity .01
Coal
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.15
.43
.03
.20
.08
.05

A


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Final Demand
The final demand on the economy is a column
matrix with one entry for each industry indicating
the amount of consumable output demanded from
the industry not used by the other industries:
 amount from industry 1


final demand   amount from industry 2  .


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Example Final Demand
An economy is composed of three industries coal, steel, and electricity as in the previous
example. Consumption (amount not used for
production) is projected to be \$2 billion for coal,
\$1 billion for steel and \$3 billion for electricity.
Set up the final demand matrix.
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Example Final Demand Answer
For simplicity, set up the final demand matrix in
billions of dollars.
2


Steel 1   D
Electricity  3 
Coal
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Production Level Problem
Problem:
Find the amount of production of
each industry to meet the final demand of the
economy. Our problem is to determine the output
of each industry, X, that yields the desired
amounts left over from the production process.
X = (I - A)-1D
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Example Production Level Problem
An economy is composed of three industries coal, steel, and electricity as in the previous
examples. Find the output of each industry that
will meet the final demand.
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Example Production Level Solution (1)
1 0 0   0 .15 .43  1 -.15 -.43
I  A  0 1 0   .02 .03 .20    -.02 .97 -.20 
0 0 1  .01 .08 .05   -.01 -.08 .95 
1
 1 -.15 -.43
1.01 .20 .50 




1
( I  A)  -.02 .97 -.20    .02 1.05 .23 
 -.01 -.08 .95 
 .01 .09 1.08
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Example Production Level Solution (2)
1.01 .20 .50   2  3.72 






1
X  ( I  A) D   .02 1.05 .23   1   1.78 
 .01 .09 1.08  3  3.35
The three industries should produce \$3.72
billion of coal, \$1.78 billion of steel and \$3.35
billion of electricity.
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Summary Section 2.5 - Part 1
 An input-output matrix has rows and columns
labeled with the different industries in an
economy. The ijth entry of the matrix gives the
cost of the input from the industry in row i used
in the production of \$1 worth of the output of
industry in column j.
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Summary Section 2.5 - Part 2
 If A is an input-output matrix and D is a
demand matrix giving the dollar values of the
outputs from various industries to be supplied to
outside customers, then the matrix X = (I - A)-1D
gives the amounts that must be produced by the
various industries in order to meet the demand.
Finite Mathematics & Its Applications, 11/e by Goldstein/Schneider/Siegel