Skip to content Skip to navigation Skip to collection information

OpenStax_CNX

You are here: Home » Content » Applied Finite Mathematics » Matrices

Navigation

Lenses

What is a lens?

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

This content is ...

Endorsed by Endorsed (What does "Endorsed by" mean?)

This content has been endorsed by the organizations listed. Click each link for a list of all content endorsed by the organization.
  • College Open Textbooks display tagshide tags

    This collection is included inLens: Community College Open Textbook Collaborative
    By: CC Open Textbook Collaborative

    Comments:

    "Reviewer's Comments: 'I recommend this book for undergraduates. The content is especially useful for those in finance, probability statistics, and linear programming. The course material is […]"

    Click the "College Open Textbooks" link to see all content they endorse.

    Click the tag icon tag icon to display tags associated with this content.

Affiliated with (What does "Affiliated with" mean?)

This content is either by members of the organizations listed or about topics related to the organizations listed. Click each link to see a list of all content affiliated with the organization.
  • Bookshare

    This collection is included inLens: Bookshare's Lens
    By: Bookshare - A Benetech Initiative

    Comments:

    "Accessible versions of this collection are available at Bookshare. DAISY and BRF provided."

    Click the "Bookshare" link to see all content affiliated with them.

  • Featured Content display tagshide tags

    This collection is included inLens: Connexions Featured Content
    By: Connexions

    Comments:

    "Applied Finite Mathematics covers topics including linear equations, matrices, linear programming, the mathematics of finance, sets and counting, probability, Markov chains, and game theory."

    Click the "Featured Content" link to see all content affiliated with them.

    Click the tag icon tag icon to display tags associated with this content.

Recently Viewed

This feature requires Javascript to be enabled.

Tags

(What is a tag?)

These tags come from the endorsement, affiliation, and other lenses that include this content.
 

Matrices

Module by: Rupinder Sekhon. E-mail the author

Summary: This chapter covers principles of matrices. After completing this chapter students should be able to: complete matrix operations; solve linear systems using Gauss-Jordan method; Solve linear systems using the matrix inverse method and complete application problems.

Chapter Overview

In this chapter, you will learn to:

  1. Do matrix operations.
  2. Solve linear systems using the Gauss-Jordan method.
  3. Solve linear systems using the matrix inverse method.
  4. Do application problems.

Introduction to Matrices

Section Overview

In this section you will learn to:

  1. Add and subtract matrices.
  2. Multiply a matrix by a scalar.
  3. Multiply two matrices.

A matrix is a rectangular array of numbers. Matrices are useful in organizing and manipulating large amounts of data. In order to get some idea of what matrices are all about, we will look at the following example.

Example 1

Problem 1

Fine Furniture Company makes chairs and tables at its San Jose, Hayward, and Oakland factories. The total production, in hundreds, from the three factories for the years 1994 and 1995 is listed in the table below.

Table 1
 
1994 1995
Chairs Tables Chairs Tables
San Jose 30 18 36 20
Hayward 20 12 24 18
Oakland 16 10 20 12
  1. Represent the production for the years 1994 and 1995 as the matrices A and B.
  2. Find the difference in sales between the years 1994 and 1995.
  3. The company predicts that in the year 2000 the production at these factories will double that of the year 1994. What will the production be for the year 2000?
Solution
  1. The matrices are as follows: A=301820121610A=301820121610 size 12{A= left [ matrix { "30" {} # "18" {} ## "20" {} # "12" {} ## "16" {} # "10"{} } right ]} {}B=362024182012B=362024182012 size 12{B= left [ matrix { "36" {} # "20" {} ## "24" {} # "18" {} ## "20" {} # "12"{} } right ]} {}
  2. We are looking for the matrix BABA size 12{B - A} {}. When two matrices have the same number of rows and columns, the matrices can be added or subtracted entry by entry. Therefore, we get

    BA=363020182420181220161210=624642BA=363020182420181220161210=624642 size 12{B - A= left [ matrix { "36" - "30" {} # "20" - "18" {} ## "24" - "20" {} # "18" - "12" {} ## "20" - "16" {} # "12" - "10"{} } right ]= left [ matrix { 6 {} # 2 {} ## 4 {} # 6 {} ## 4 {} # 2{} } right ]} {}
    (1)
  3. We would like a matrix that is twice the matrix of 1994, i.e., 2A2A size 12{2A} {}.

    Whenever a matrix is multiplied by a number, each entry is multiplied by the number.

    2A=2301820121610=6036402432202A=2301820121610=603640243220 size 12{2A=2 left [ matrix { "30" {} # "18" {} ## "20" {} # "12" {} ## "16" {} # "10"{} } right ]= left [ matrix { "60" {} # "36" {} ## "40" {} # "24" {} ## "32" {} # "20"{} } right ]} {}
    (2)

Before we go any further, we need to familiarize ourselves with some terms that are associated with matrices. The numbers in a matrix are called the entries or the elements of a matrix. Whenever we talk about a matrix, we need to know the size or the dimension of the matrix. The dimension of a matrix is the number of rows and columns it has. When we say a matrix is a 3 by 4 matrix, we are saying that it has 3 rows and 4 columns. The rows are always mentioned first and the columns second. This means that a 3×43×4 size 12{3 times 4} {} matrix does not have the same dimension as a 4×34×3 size 12{4 times 3} {} matrix. A matrix that has the same number of rows as columns is called a square matrix. A matrix with all entries zero is called a zero matrix. A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. A matrix with only one row is called a row matrix or a row vector, and a matrix with only one column is called a column matrix or a column vector. Two matrices are equal if they have the same size and the corresponding entries are equal.

Matrix Addition and Subtraction

If two matrices have the same size, they can be added or subtracted. The operations are performed on corresponding entries.

Example 2

Problem 1

Given the matrices AA size 12{A} {}, BB size 12{B} {}, CC size 12{C} {} and DD size 12{D} {}, below

A=124231503B=213242361C=423D=234A=124231503 size 12{A= left [ matrix { 1 {} # 2 {} # 4 {} ## 2 {} # 3 {} # 1 {} ## 5 {} # 0 {} # 3{} } right ]} {}B=213242361 size 12{B= left [ matrix { 2 {} # - 1 {} # 3 {} ## 2 {} # 4 {} # 2 {} ## 3 {} # 6 {} # 1{} } right ]} {}C=423 size 12{C= left [ matrix { 4 {} ## 2 {} ## 3 } right ]} {}D=234 size 12{D= left [ matrix { - 2 {} ## - 3 {} ## 4 } right ]} {}
(3)

Find, if possible.

  1. A+BA+B size 12{A+B} {}
  2. CDCD size 12{C - D} {}
  3. A+DA+D size 12{A+D} {}.
Solution

As we mentioned earlier, matrix addition and subtraction involves performing these operations entry by entry.

  1. We add each element of AA size 12{A} {} to the corresponding entry of BB size 12{B} {}.

    A+B=317473864A+B=317473864 size 12{A+B= left [ matrix { 3 {} # 1 {} # 7 {} ## 4 {} # 7 {} # 3 {} ## 8 {} # 6 {} # 4{} } right ]} {}
    (4)
  2. Just like the problem above, we perform the subtraction entry by entry.

    CD=651CD=651 size 12{C - D= left [ matrix { 6 {} ## 5 {} ## - 1 } right ]} {}
    (5)
  3. The sum A+DA+D size 12{A+D} {} cannot be found because the two matrices have different sizes.

Note:

Two matrices can only be added or subtracted if they have the same dimension.

Multiplying a Matrix by a Scalar

If a matrix is multiplied by a scalar (a constant number), each entry is multiplied by that scalar.

Example 3

Problem 1

Given the matrix AA size 12{A} {} and CC size 12{C} {} in Example 2, find 2A2A size 12{2A} {} and 3C3C size 12{ - 3C} {}.

Solution

To find 2A2A size 12{2A} {}, we multiply each entry of matrix AA size 12{A} {} by 22 size 12{2} {}, and to find 3C3C size 12{ - 3C} {}, we multiply each entry of CC size 12{C} {} by 33 size 12{ - 3} {}. The results are given below.

  1. We multiply each entry of AA size 12{A} {} by 22 size 12{2} {}.

    2A=24846210062A=2484621006 size 12{2A= left [ matrix { 2 {} # 4 {} # 8 {} ## 4 {} # 6 {} # 2 {} ## "10" {} # 0 {} # 6{} } right ]} {}
    (6)
  2. We multiply each entry of CC size 12{C} {} by 33 size 12{ - 3} {}.

    3C=12693C=1269 size 12{ - 3C= left [ matrix { - "12" {} ## - 6 {} ## - 9 } right ]} {}
    (7)

Multiplication of Two Matrices

To multiply a matrix by another is not as easy as the addition, subtraction, or scalar multiplication of matrices. Because of its wide use in application problems, it is important that we learn it well. Therefore, we will try to learn the process in a step by step manner. We first begin by finding a product of a row matrix and a column matrix.

Example 4

Problem 1

Given A=234A=234 size 12{A= left [ matrix { 2 {} # 3 {} # 4{} } right ]} {} and B=abcB=abc size 12{B= left [ matrix { a {} ## b {} ## c } right ]} {}, find the product ABAB size 12{ ital "AB"} {}.

Solution

The product is a 1×11×1 size 12{1 times 1} {} matrix whose entry is obtained by multiplying the corresponding entries and then forming the sum.

AB=234abc=2a+3b+4cAB=234abc=2a+3b+4c size 12{ ital "AB"= left [ matrix { 2 {} # 3 {} # 4{} } right ] left [ matrix { a {} ## b {} ## c } right ]= left [ left (2a+3b+4c right ) right ]} {}
(8)

Note that ABAB size 12{ ital "AB"} {} is a 1×11×1 size 12{1 times 1} {} matrix, and its only entry is 2a+3b+4c2a+3b+4c size 12{2a+3b+4c} {}.

Example 5

Problem 1

Given A=234A=234 size 12{A= left [ matrix { 2 {} # 3 {} # 4{} } right ]} {} and B=567B=567 size 12{B= left [ matrix { 5 {} ## 6 {} ## 7 } right ]} {} , find the product ABAB size 12{ ital "AB"} {}.

Solution

Again, we multiply the corresponding entries and add.

AB=234567=25+36+47=10+18+28=56AB=234567=25+36+47=10+18+28=56 size 12{ matrix { ital "AB" {} # ={} {} # left [ matrix { 2 {} # 3 {} # 4{} } right ] left [ matrix { 5 {} ## 6 {} ## 7 } right ] {} ## {} # ={} {} # left [2 cdot 5+3 cdot 6+4 cdot 7 right ] {} ## {} # ={} {} # left ["10"+"18"+"28" right ] {} ## {} # ={} {} # left ["56" right ]{} } } {}
(9)

Note:

In order for a product of a row matrix and a column matrix to exist, the number of entries in the row matrix must be the same as the number of entries in the column matrix.

Example 6

Problem 1

Given A=234A=234 size 12{A= left [ matrix { 2 {} # 3 {} # 4{} } right ]} {} and B=536475B=536475 size 12{B= left [ matrix { 5 {} # 3 {} ## 6 {} # 4 {} ## 7 {} # 5{} } right ]} {} , find the product ABAB size 12{ ital "AB"} {}.

Solution

We already know how to multiply a row matrix by a column matrix. To find the product ABAB size 12{ ital "AB"} {}, in this example, we will be multiplying the row matrix AA size 12{A} {} to both the first and second columns of matrix BB size 12{B} {}, resulting in a 1×21×2 size 12{1 times 2} {} matrix.

AB=25+36+4723+34+45=5638AB=25+36+4723+34+45=5638 size 12{ ital "AB"= left [ matrix { 2 cdot 5+3 cdot 6+4 cdot 7 {} # 2 cdot 3+3 cdot 4+4 cdot 5{} } right ]= left [ matrix { "56" {} # "38"{} } right ]} {}
(10)

We have just multiplied a 1×31×3 size 12{1 times 3} {} matrix by a matrix whose size is 3×23×2 size 12{3 times 2} {}. So unlike addition and subtraction, it is possible to multiply two matrices with different dimensions as long as the number of entries in the rows of the first matrix are the same as the number of entries in columns of the second matrix.

Example 7

Problem 1

Given A=234123A=234123 size 12{A= left [ matrix { 2 {} # 3 {} # 4 {} ## 1 {} # 2 {} # 3{} } right ]} {} and B=536475B=536475 size 12{B= left [ matrix { 5 {} # 3 {} ## 6 {} # 4 {} ## 7 {} # 5{} } right ]} {} , find the product ABAB size 12{ ital "AB"} {}.

Solution

This time we are multiplying two rows of the matrix AA size 12{A} {} with two columns of the matrix BB size 12{B} {}. Since the number of entries in each row of AA size 12{A} {} are the same as the number of entries in each column of BB size 12{B} {}, the product is possible. We do exactly what we did in Example 6. The only difference is that the matrix AA size 12{A} {} has one more row.

We multiply the first row of the matrix AA size 12{A} {} with the two columns of BB size 12{B} {}, one at a time, and then repeat the process with the second row of AA size 12{A} {}. We get

AB = 2 3 4 1 2 3 5 3 6 4 7 5 = 25+36+4723+34+4515+26+3713+24+35 = 56383826 AB = 2 3 4 1 2 3 5 3 6 4 7 5 = 25+36+4723+34+4515+26+3713+24+35 = 56383826 size 12{ matrix { ital "AB" {} # ={} {} # left [ matrix { 2 {} # 3 {} # 4 {} ## 1 {} # 2 {} # 3{} } right ] left [ matrix { 5 {} # 3 {} ## 6 {} # 4 {} ## 7 {} # 5{} } right ] {} ## {} # ={} {} # left [ matrix { 2 cdot 5+3 cdot 6+4 cdot 7 {} # 2 cdot 3+3 cdot 4+4 cdot 5 {} ## 1 cdot 5+2 cdot 6+3 cdot 7 {} # 1 cdot 3+2 cdot 4+3 cdot 5{} } right ] {} ## {} # ={} {} # left [ matrix { "56" {} # "38" {} ## "38" {} # "26"{} } right ]{} } } {}
(11)

Example 8

Problem 1

Given the matrices EE size 12{E} {}, FF size 12{F} {}, GG size 12{G} {} and HH size 12{H} {}, below

E=124231F=2132G=41H=31E=124231 size 12{E= left [ matrix { 1 {} # 2 {} ## 4 {} # 2 {} ## 3 {} # 1{} } right ]} {}F=2132 size 12{F= left [ matrix { 2 {} # - 1 {} ## 3 {} # 2{} } right ]} {}G=41 size 12{G= left [ matrix { 4 {} # 1{} } right ]} {}H=31 size 12{H= left [ matrix { - 3 {} ## - 1 } right ]} {}
(12)

Find, if possible.

  1. EFEF size 12{ ital "EF"} {}
  2. FEFE size 12{ ital "FE"} {}
  3. FHFH size 12{ ital "FH"} {}
  4. GHGH size 12{ ital "GH"} {}
Solution
  1. To find EFEF size 12{ ital "EF"} {}, we multiply the first row 1212 size 12{ left [ matrix { 1 {} # 2{} } right ]} {} of EE size 12{E} {} with the columns 2323 size 12{ left [ matrix { 2 {} ## 3 } right ]} {} and 1212 size 12{ left [ matrix { - 1 {} ## 2 } right ]} {} of the matrix FF size 12{F} {}, and then repeat the process by multiplying the other two rows of EE size 12{E} {} with these columns of FF size 12{F} {}. The result is as follows:

    EF = 1 2 4 2 3 1 21 32 = 12+2311+2242+2341+2232+1331+12=8314091EF = 1 2 4 2 3 1 21 32 = 12+2311+2242+2341+2232+1331+12=8314091 size 12{ matrix { ital "EF" {} # ={} {} # left [ matrix { 1 {} # 2 {} ## 4 {} # 2 {} ## 3 {} # 1{} } right ] left [ matrix { 2 {} # - 1 {} ## 3 {} # 2{} } right ] {} ## {} # ={} {} # left [ matrix { 1 cdot 2+2 cdot 3 {} # 1 cdot - 1+2 cdot 2 {} ## 4 cdot 2+2 cdot 3 {} # 4 cdot - 1+2 cdot 2 {} ## 3 cdot 2+1 cdot 3 {} # 3 cdot - 1+1 cdot 2{} } right ]= left [ matrix { 8 {} # 3 {} ## "14" {} # 0 {} ## 9 {} # - 1{} } right ]{} } } {}
    (13)
  2. The product FEFE size 12{ ital "FE"} {} is not possible because the matrix FF size 12{F} {} has two entries in each row, while the matrix EE size 12{E} {} has three entries in each column. In other words, the matrix FF size 12{F} {} has two columns, while the matrix EE size 12{E} {} has three rows.

  3. FH=213231=23+1133+21=511FH=213231=23+1133+21=511 size 12{ ital "FH"= left [ matrix { 2 {} # - 1 {} ## 3 {} # 2{} } right ] left [ matrix { - 3 {} ## - 1 } right ]= left [ matrix { 2 cdot - 3+ - 1 cdot - 1 {} ## 3 cdot - 3+2 cdot - 1 } right ]= left [ matrix { - 5 {} ## - "11" } right ]} {}

  4. GH=4131=43+11=13GH=4131=43+11=13 size 12{ ital "GH"= left [ matrix { 4 {} # 1{} } right ] left [ matrix { - 3 {} ## - 1 } right ]= left [4 cdot - 3+1 cdot - 1 right ]= left [ - "13" right ]} {}

We summarize matrix multiplication as follows:

In order for product ABAB size 12{ ital "AB"} {} to exist, the number of columns of AA size 12{A} {}, must equal the number of rows of BB size 12{B} {}. If matrix AA size 12{A} {} is of dimension m×nm×n size 12{m times n} {} and BB size 12{B} {} of dimension n×pn×p size 12{n times p} {}, the product will have the dimension m×pm×p size 12{m times p} {}. Furthermore, matrix multiplication is not commutative.

Example 9

Problem 1

Given the matrices RR size 12{R} {}, SS size 12{S} {}, and TT size 12{T} {} below.

R=102215231S=012310421T=230322110R=102215231 size 12{R= left [ matrix { 1 {} # 0 {} # 2 {} ## 2 {} # 1 {} # 5 {} ## 2 {} # 3 {} # 1{} } right ]} {}S=012310421 size 12{S= left [ matrix { 0 {} # - 1 {} # 2 {} ## 3 {} # 1 {} # 0 {} ## 4 {} # 2 {} # 1{} } right ]} {}T=230322110 size 12{T= left [ matrix { - 2 {} # 3 {} # 0 {} ## - 3 {} # 2 {} # 2 {} ## - 1 {} # 1 {} # 0{} } right ]} {}
(14)

Find 2RS3ST2RS3ST size 12{2 ital "RS" - 3 ital "ST"} {}.

Solution

We multiply the matrices RR size 12{R} {} and SS size 12{S} {}.

RS=83423991335RS=83423991335 size 12{ ital "RS"= left [ matrix { 8 {} # 3 {} # 4 {} ## "23" {} # 9 {} # 9 {} ## "13" {} # 3 {} # 5{} } right ]} {}
(15)
2RS=283423991335=1668461818266102RS=283423991335=166846181826610 size 12{2 ital "RS"=2 left [ matrix { 8 {} # 3 {} # 4 {} ## "23" {} # 9 {} # 9 {} ## "13" {} # 3 {} # 5{} } right ]= left [ matrix { "16" {} # 6 {} # 8 {} ## "46" {} # "18" {} # "18" {} ## "26" {} # 6 {} # "10"{} } right ]} {}
(16)
ST=102911215174ST=102911215174 size 12{ ital "ST"= left [ matrix { 1 {} # 0 {} # - 2 {} ## - 9 {} # "11" {} # 2 {} ## - "15" {} # "17" {} # 4{} } right ]} {}
(17)
3ST=3102911215174=306273364551123ST=3102911215174=30627336455112 size 12{3 ital "ST"=3 left [ matrix { 1 {} # 0 {} # - 2 {} ## - 9 {} # "11" {} # 2 {} ## - "15" {} # "17" {} # 4{} } right ]= left [ matrix { 3 {} # 0 {} # - 6 {} ## - "27" {} # "33" {} # 6 {} ## - "45" {} # "51" {} # "12"{} } right ]} {}
(18)
2RS3ST=16684618182661030627336455112=13614731512714522RS3ST=16684618182661030627336455112=1361473151271452 size 12{2 ital "RS" - 3 ital "ST"= left [ matrix { "16" {} # 6 {} # 8 {} ## "46" {} # "18" {} # "18" {} ## "26" {} # 6 {} # "10"{} } right ] - left [ matrix { 3 {} # 0 {} # - 6 {} ## - "27" {} # "33" {} # 6 {} ## - "45" {} # "51" {} # "12"{} } right ]= left [ matrix { "13" {} # 6 {} # "14" {} ## "73" {} # - "15" {} # "12" {} ## "71" {} # - "45" {} # - 2{} } right ]} {}
(19)

In this chapter, we will be using matrices to solve linear systems. In Section 7, we will be asked to express linear systems as the matrix equation AX=BAX=B size 12{ ital "AX"=B} {}, where AA size 12{A} {}, XX size 12{X} {}, and BB size 12{B} {} are matrices. The matrix AA size 12{A} {} is called the coefficient matrix.

Example 10

Problem 1

Verify that the system of two linear equations with two unknowns:

ax+by=hax+by=h size 12{ ital "ax"+ ital "by"=h} {}
(20)
cx+dy=kcx+dy=k size 12{ ital "cx"+ ital "dy"=k} {}
(21)

can be written as AX=BAX=B size 12{ ital "AX"=B} {}, where

A=abcdX=xyand   B=hkA=abcd size 12{A= left [ matrix { a {} # b {} ## c {} # d{} } right ]} {}X=xy size 12{X= left [ matrix { x {} ## y } right ]} {}and   B=hk size 12{B= left [ matrix { h {} ## k } right ]} {}

Solution

If we multiply the matrices AA size 12{A} {} and XX size 12{X} {}, we get

AX=abcdxy=ax+bycx+dyAX=abcdxy=ax+bycx+dy size 12{ ital "AX"= left [ matrix { a {} # b {} ## c {} # d{} } right ] left [ matrix { x {} ## y } right ]= left [ matrix { ital "ax"+ ital "by" {} ## ital "cx"+ ital "dy" } right ]} {}
(22)

If AX=BAX=B size 12{ ital "AX"=B} {} then

ax+bycx+dy=hkax+bycx+dy=hk size 12{ left [ matrix { ital "ax"+ ital "by" {} ## ital "cx"+ ital "dy" } right ]= left [ matrix { h {} ## k } right ]} {}
(23)

If two matrices are equal, then their corresponding entries are equal. Therefore, it follows that

ax+by=hax+by=h size 12{ ital "ax"+ ital "by"=h} {}
(24)
cx+dy=kcx+dy=k size 12{ ital "cx"+ ital "dy"=k} {}
(25)

Example 11

Problem 1

Express the following system as AX=BAX=B size 12{ ital "AX"=B} {}.

2x+3y4z=52x+3y4z=5 size 12{2x+"3y"–4z=5} {}
(26)
3x+4y5z=63x+4y5z=6 size 12{3x+4y - 5z=6} {}
(27)
5x6z=75x6z=7 size 12{5x - 6z=7} {}
(28)
Solution

The above system of equations can be expressed in the form AX=BAX=B size 12{ ital "AX"=B} {} as shown below.

234345506xyz=567234345506xyz=567 size 12{ left [ matrix { 2 {} # 3 {} # - 4 {} ## 3 {} # 4 {} # - 5 {} ## 5 {} # 0 {} # - 6{} } right ] left [ matrix { x {} ## y {} ## z } right ]= left [ matrix { 5 {} ## 6 {} ## 7 } right ]} {}
(29)

Systems of Linear Equations; Gauss-Jordan Method

In this section, we learn to solve systems of linear equations using a process called the Gauss-Jordan method. The process begins by first expressing the system as a matrix, and then reducing it to an equivalent system by simple row operations. The process is continued until the solution is obvious from the matrix. The matrix that represents the system is called the augmented matrix, and the arithmetic manipulation that is used to move from a system to a reduced equivalent system is called a row operation.

Example 12

Problem 1

Write the following system as an augmented matrix.

2x+3y4z=52x+3y4z=5 size 12{2x+3y - 4z=5} {}
(30)
3x+4y5z=63x+4y5z=6 size 12{3x+4y - 5z= - 6} {}
(31)
4x+5y6z=74x+5y6z=7 size 12{4x+5y - 6z=7} {}
(32)
Solution

We express the above information in matrix form. Since a system is entirely determined by its coefficient matrix and by its matrix of constant terms, the augmented matrix will include only the coefficient matrix and the constant matrix. So the augmented matrix we get is as follows:

234534564567234534564567 size 12{ left [ matrix { 2 {} # 3 {} # - 4 {} # \lline {} # 5 {} ## 3 {} # 4 {} # - 5 {} # \lline {} # - 6 {} ## 4 {} # 5 {} # - 6 {} # \lline {} # 7{} } right ]} {}
(33)

In the Section 2, we expressed the system of equations as AX=BAX=B size 12{ ital "AX"=B} {}, where AA size 12{A} {} represented the coefficient matrix, and BB size 12{B} {} the matrix of constant terms. As an augmented matrix, we write the matrix as ABAB size 12{ left [A \lline B right ]} {}. It is clear that all of the information is maintained in this matrix form, and only the letters xx size 12{x} {}, yy size 12{y} {} and zz size 12{z} {} are missing. A student may choose to write xx size 12{x} {}, yy size 12{y} {} and zz size 12{z} {} on top of the first three columns to help ease the transition.

Example 13

Problem 1

For the following augmented matrix, write the system of equations it represents.

135220353231135220353231 size 12{ left [ matrix { 1 {} # 3 {} # - 5 {} # \lline {} # 2 {} ## 2 {} # 0 {} # - 3 {} # \lline {} # - 5 {} ## 3 {} # 2 {} # - 3 {} # \lline {} # - 1{} } right ]} {}
(34)
Solution

The system is readily obtained as below.

x+3y5z=2x+3y5z=2 size 12{x+3y - 5z=2} {}
(35)
2x3z=52x3z=5 size 12{2x - 3z= - 5} {}
(36)
3x+2y3z=13x+2y3z=1 size 12{3x+2y - 3z= - 1} {}
(37)

Once a system is expressed as an augmented matrix, the Gauss-Jordan method reduces the system into a series of equivalent systems by employing the row operations. This row reduction continues until the system is expressed in what is called the reduced row echelon form. The reduced row echelon form of the coefficient matrix has 1's along the main diagonal and zeros elsewhere. The solution is readily obtained from this form.

The method is not much different form the algebraic operations we employed in the elimination method in the first chapter. The basic difference is that it is algorithmic in nature, and, therefore, can easily be programmed on a computer.

We will next solve a system of two equations with two unknowns, using the elimination method, and then show that the method is analogous to the Gauss-Jordan method.

Example 14

Problem 1

Solve the following system by the elimination method.

x+3y=7x+3y=7 size 12{x+3y=7} {}
(38)
3x+4y=113x+4y=11 size 12{3x+4y="11"} {}
(39)
Solution

We multiply the first equation by – 3, and add it to the second equation.

3x9y=213x+4y=11̲5y=103x9y=213x+4y=11̲5y=10 size 12{ matrix { - 3x - 9y= - "21" {} ## {underline {3x+4y="11"}} {} ## - 5y= - "10" } } {}
(40)

By doing this we have transformed our original system into an equivalent system as follows.

x+3y=7x+3y=7 size 12{x+3y=7} {}
(41)
5y=105y=10 size 12{ - 5y= - "10"} {}
(42)

We divide the second equation by – 5, and we get the next equivalent system.

x+3y=7x+3y=7 size 12{x+3y=7} {}
(43)
y=2y=2 size 12{y=2} {}
(44)

Now we multiply the second equation by – 3 and add to the first, we get

x=1x=1 size 12{x=1} {}
(45)
y=2y=2 size 12{y=2} {}
(46)

Example 15

Problem 1

Solve the following system from Example 14 by the Gauss-Jordan method, and show the similarities in both methods by writing the equations next to the matrices.

x+3y=7x+3y=7 size 12{x+3y=7} {}
(47)
3x+4y=113x+4y=11 size 12{3x+4y="11"} {}
(48)
Solution

The augmented matrix for the system is as follows.

1373411x+3y=73x+4y=111373411 size 12{ left [ matrix { 1 {} # 3 {} # \lline {} # 7 {} ## 3 {} # 4 {} # \lline {} # "11"{} } right ]} {}x+3y=73x+4y=11 size 12{ left [ matrix { x+3y=7 {} ## 3x+4y="11" } right ]} {}
(49)

We multiply the first row by – 3, and add to the second row.

1370510x+3y=75y=101370510 size 12{ left [ matrix { 1 {} # 3 {} # \lline {} # 7 {} ## 0 {} # - 5 {} # \lline {} # - "10"{} } right ]} {}x+3y=75y=10 size 12{ left [ matrix { x+3y=7 {} ## - 5y= - "10" } right ]} {}
(50)

We divide the second row by – 5, we get,

137012x+3y=7y=2137012 size 12{ left [ matrix { 1 {} # 3 {} # \lline {} # 7 {} ## 0 {} # 1 {} # \lline {} # 2{} } right ]} {}x+3y=7y=2 size 12{ left [ matrix { x+3y=7 {} ## y=2 } right ]} {}
(51)

Finally, we multiply the second row by – 3 and add to the first row, and we get,

101012x=1y=2101012 size 12{ left [ matrix { 1 {} # 0 {} # \lline {} # 1 {} ## 0 {} # 1 {} # \lline {} # 2{} } right ]} {}x=1y=2 size 12{ left [ matrix { x=1 {} ## y=2 } right ]} {}
(52)

Now we list the three row operations the Gauss-Jordan method employs.

Row Operations

  1. Any two rows in the augmented matrix may be interchanged.
  2. Any row may be multiplied by a non-zero constant.
  3. A constant multiple of a row may be added to another row.

One can easily see that these three row operation may make the system look different, but they do not change the solution of the system.

The first row operation states that if any two rows of a system are interchanged, the new system obtained has the same solution as the old one. Let us look at an example in two equations with two unknowns. Consider the system

x+3y=7x+3y=7 size 12{x+3y=7} {}
(53)
3x+4y=113x+4y=11 size 12{3x+4y="11"} {}
(54)

We interchange the rows, and we get,

3x+4y=113x+4y=11 size 12{3x+4y="11"} {}
(55)
x+3y=7x+3y=7 size 12{x+3y=7} {}
(56)

Clearly, this system has the same solution as the one above.

The second operation states that if a row is multiplied by any non-zero constant, the new system obtained has the same solution as the old one. Consider the above system again,

x+3y=7x+3y=7 size 12{x+3y=7} {}
(57)
3x+4y=113x+4y=11 size 12{3x+4y="11"} {}
(58)

We multiply the first row by –3, we get,

- 3 x - 9 y = - 21 -3x-9y=-21
(59)
3x+4y=113x+4y=11 size 12{3x+4y="11"} {}
(60)

Again, it is obvious that this new system has the same solution as the original.

The third row operation states that any constant multiple of one row added to another preserves the solution. Consider our system,

x+3y=7x+3y=7 size 12{x+3y=7} {}
(61)
3x+4y=113x+4y=11 size 12{3x+4y="11"} {}
(62)

If we multiply the first row by – 3, and add it to the second row, we get,

x+3y=7x+3y=7 size 12{x+3y=7} {}
(63)
5y=105y=10 size 12{ - 5y= - "10"} {}
(64)

And once again, the same solution is maintained.

Now that we understand how the three row operations work, it is time to introduce the Gauss-Jordan method to solve systems of linear equations.

As mentioned earlier, the Gauss-Jordan method starts out with an augmented matrix, and by a series of row operations ends up with a matrix that is in the reduced row echelon form. A matrix is in the reduced row echelon form if the first nonzero entry in each row is a 1, and the columns containing these 1's have all other entries as zeros. The reduced row echelon form also requires that the leading entry in each row be to the right of the leading entry in the row above it, and the rows containing all zeros be moved down to the bottom.

We state the Gauss-Jordan method as follows.

Gauss-Jordan Method

  1. Write the augmented matrix.
  2. Interchange rows if necessary to obtain a non-zero number in the first row, first column.
  3. Use a row operation to make the entry in the first row, first column, a 1.
  4. Use row operations to make all other entries as zeros in column one.
  5. Interchange rows if necessary to obtain a nonzero number in the second row, second column. Use a row operation to make this entry 1. Use row operations to make all other entries as zeros in column two.
  6. Repeat step 5 for row 3, column 3. Continue moving along the main diagonal until you reach the last row, or until the number is zero. The final matrix is called the reduced row-echelon form.

Example 18

Problem 1

Solve the following system by the Gauss-Jordan method.

2x+y+2z=102x+y+2z=10 size 12{2x+y+2z="10"} {}
(65)
x+2y+z=8x+2y+z=8 size 12{x+2y+z=8} {}
(66)
3x+yz=23x+yz=2 size 12{3x+y - z=2} {}
(67)
Solution

We write the augmented matrix.

21210121831122121012183112 size 12{ left [ matrix { 2 {} # 1 {} # 2 {} # \lline {} # "10" {} ## 1 {} # 2 {} # 1 {} # \lline {} # 8 {} ## 3 {} # 1 {} # - 1 {} # \lline {} # 2{} } right ]} {}
(68)

We want a 1 in row one, column one. This can be obtained by dividing the first row by 2, or interchanging the second row with the first. Interchanging the rows is a better choice because that way we avoid fractions.

12182121031121218212103112 size 12{ left [ matrix { 1 {} # 2 {} # 1 {} # \lline {} # 8 {} ## 2 {} # 1 {} # 2 {} # \lline {} # "10" {} ## 3 {} # 1 {} # - 1 {} # \lline {} # 2{} } right ]} {}
(69)

we interchanged row 1(R1) and row 2(R2)

We need to make all other entries zeros in column 1. To make the entry (2) a zero in row 2, column 1, we multiply row 1 by - 2 and add it to the second row. We get,

121803063112      −2R1+R2121803063112 size 12{ left [ matrix { 1 {} # 2 {} # 1 {} # \lline {} # 8 {} ## 0 {} # - 3 {} # 0 {} # \lline {} # - 6 {} ## 3 {} # 1 {} # - 1 {} # \lline {} # 2{} } right ]} {}      −2R1+R2 size 12{ -2R1+R2} {}
(70)

To make the entry (3) a zero in row 3, column 1, we multiply row 1 by –3 and add it to the third row. We get,

1218030605422      −3R1+R31218030605422 size 12{ left [ matrix { 1 {} # 2 {} # 1 {} # \lline {} # 8 {} ## 0 {} # - 3 {} # 0 {} # \lline {} # - 6 {} ## 0 {} # - 5 {} # - 4 {} # \lline {} # - "22"{} } right ]} {}      −3R1+R3 size 12{ -3R1+R3} {}
(71)

So far we have made a 1 in the left corner and all other entries zeros in that column. Now we move to the next diagonal entry, row 2, column 2. We need to make this entry(–3) a 1 and make all other entries in this column zeros. To make row 2, column 2 entry a 1, we divide the entire second row by –3.

1218010205422      R2÷31218010205422 size 12{ left [ matrix { 1 {} # 2 {} # 1 {} # \lline {} # 8 {} ## 0 {} # 1 {} # 0 {} # \lline {} # 2 {} ## 0 {} # - 5 {} # - 4 {} # \lline {} # - "22"{} } right ]} {}      R2÷3 size 12{R2 div left ( -3 right )} {}
(72)

Next, we make all other entries zeros in the second column.

1014010200412      −2R2+R1and5R2+R31014010200412 size 12{ left [ matrix { 1 {} # 0 {} # 1 {} # \lline {} # 4 {} ## 0 {} # 1 {} # 0 {} # \lline {} # 2 {} ## 0 {} # 0 {} # - 4 {} # \lline {} # - "12"{} } right ]} {}      −2R2+R1 size 12{ - 2R2+R1} {}and5R2+R3
(73)

We make the last diagonal entry a 1, by dividing row 3 by – 4.

101401020013      R3÷4101401020013 size 12{ left [ matrix { 1 {} # 0 {} # 1 {} # \lline {} # 4 {} ## 0 {} # 1 {} # 0 {} # \lline {} # 2 {} ## 0 {} # 0 {} # 1 {} # \lline {} # 3{} } right ]} {}      R3÷4 size 12{R3 div left ( - 4 right )} {}
(74)

Finally, we make all other entries zeros in column 3.

100101020013      −R3+R1100101020013 size 12{ left [ matrix { 1 {} # 0 {} # 0 {} # \lline {} # 1 {} ## 0 {} # 1 {} # 0 {} # \lline {} # 2 {} ## 0 {} # 0 {} # 1 {} # \lline {} # 3{} } right ]} {}      −R3+R1 size 12{ -R3+R1} {}
(75)

Clearly, the solution reads x=1x=1 size 12{x=1} {}, y=2y=2 size 12{y=2} {} , and z=3z=3 size 12{z=3} {}.

Before we leave this section, we mention some terms we may need in the fourth chapter. The process of obtaining a 1 in a location, and then making all other entries zeros in that column, is called pivoting. The number that is made a 1 is called the pivot element, and the row that contains the pivot element is called the pivot row. We often multiply the pivot row by a number and add it to another row to obtain a zero in the latter. The row to which a multiple of pivot row is added is called the target row.

Systems of Linear Equations – Special Cases

Section Overview

In this section you will learn to:

  1. Determine the linear systems that have no solution.
  2. Solve the linear systems that have infinitely many solutions.

If we consider the intersection of two lines in a plane, three things can happen.

  1. The lines intersect in exactly one point. This is called an independent system.
  2. The lines are parallel, so they do not intersect. This is called an inconsistent system.
  3. The lines coincide, so they intersect at infinitely many points. This is a dependent system.

The figures below shows all three cases.

Figure 1
 The graph on the right depicts an independent system. The graph in the middle depicts an inconsistent system. The graph on the left depicts a dependent system.

Every system of equations has either one solution, no solution, or infinitely many solutions.

In the Section 4, we used the Gauss-Jordan method to solve systems that had exactly one solution. In this section, we will determine the systems that have no solution, and solve the systems that have infinitely many solutions.

Example 19

Problem 1

Solve the following system of equations.

x+y=7x+y=7 size 12{x+y=7} {}
(76)
x+y=9x+y=9 size 12{x+y=9} {}
(77)
Solution

Let us use the Gauss-Jordan method to solve this system. The augmented matrix is as follows.

117119x+y=7x+y=9117119 size 12{ left [ matrix { 1 {} # 1 {} # \lline {} # 7 {} ## 1 {} # 1 {} # \lline {} # 9{} } right ]} {}x+y=7x+y=9 size 12{ left [ matrix { x+y=7 {} ## x+y=9 } right ]} {}
(78)

If we multiply the first row by – 1 and add to the second row, we get

117002x+y=70x+0y=2117002 size 12{ left [ matrix { 1 {} # 1 {} # \lline {} # 7 {} ## 0 {} # 0 {} # \lline {} # 2{} } right ]} {}x+y=70x+0y=2 size 12{ left [ matrix { x+y=7 {} ## 0x+0y=2 } right ]} {}
(79)

Since 0 cannot equal 2, the last equation cannot be true for any choices of xx size 12{x} {} and yy size 12{y} {}.

Alternatively, it is clear that the two lines are parallel; therefore, they do not intersect.

At this stage, we are going to start using a calculator to row reduce the augmented matrix.

Example 20

Problem 1

Solve the following system of equations.

2x+3y4z=72x+3y4z=7 size 12{2x+3y - 4z=7} {}
(80)
3x+4y2z=93x+4y2z=9 size 12{3x+4y - 2z=9} {}
(81)
5x+7y6z=205x+7y6z=20 size 12{5x+7y - 6z="20"} {}
(82)
Solution

We enter the following augmented matrix in the calculator.

23473429576202347342957620 size 12{ left [ matrix { 2 {} # 3 {} # - 4 {} # \lline {} # 7 {} ## 3 {} # 4 {} # - 2 {} # \lline {} # 9 {} ## 5 {} # 7 {} # - 6 {} # \lline {} # "20"{} } right ]} {}
(83)

Now by pressing the key to obtain the reduced row-echelon form, we get

10100018000011010001800001 size 12{ left [ matrix { 1 {} # 0 {} # "10" {} # \lline {} # 0 {} ## 0 {} # 1 {} # - 8 {} # \lline {} # 0 {} ## 0 {} # 0 {} # 0 {} # \lline {} # 1{} } right ]} {}
(84)

The last row indicates that the system is inconsistent; therefore, there is no solution.

Example 21

Problem 1

Solve the following system of equations.

x+y=7x+y=7 size 12{x+y=7} {}
(85)
x+y=7x+y=7 size 12{x+y=7} {}
(86)
Solution

The problem clearly asks for the intersection of two lines that are the same; that is, the lines coincide. This means the lines intersect at an infinite number of points.

A few intersection points are listed as follows: (3, 4), (5, 2), (–1, 8), (–6, 13) etc. However, when a system has an infinite number of solutions, the solution is often expressed in the parametric form. This can be accomplished by assigning an arbitrary constant, tt size 12{t} {}, to one of the variables, and then solving for the remaining variables. Therefore, if we let y=ty=t size 12{y=t} {}, then x=7tx=7t size 12{x=7 - t} {}. Or we can say all ordered pairs of the form ( 7t7t size 12{7 - t} {}, tt size 12{t} {}) satisfy the given system of equations.

Alternatively, while solving the Gauss-Jordan method, we will get the reduced row-echelon form given below.

117000117000 size 12{ left [ matrix { 1 {} # 1 {} # \lline {} # 7 {} ## 0 {} # 0 {} # \lline {} # 0{} } right ]} {}
(87)

The row of all zeros, can simply be discarded in a manner that it never existed. This leaves us with only one equation but two variables. And whenever there are more variables than the equations, the solution must be expressed in terms of an arbitrary constant, as above. That is, x=7tx=7t size 12{x=7 - t} {}, y=ty=t size 12{y=t} {}.

Example 22

Problem 1

Solve the following system of equations.

x+y+z=2x+y+z=2 size 12{x+y+z=2} {}
(88)
2x+yz=32x+yz=3 size 12{2x+y - z=3} {}
(89)
3x+2y=53x+2y=5 size 12{3x+2y=5} {}
(90)
Solution

The augmented matrix and the reduced row-echelon form are given below.

111221133205111221133205 size 12{ left [ matrix { 1 {} # 1 {} # 1 {} # \lline {} # 2 {} ## 2 {} # 1 {} # - 1 {} # \lline {} # 3 {} ## 3 {} # 2 {} # 0 {} # \lline {} # 5{} } right ]} {}
(91)
102101310000102101310000 size 12{ left [ matrix { 1 {} # 0 {} # - 2 {} # \lline {} # 1 {} ## 0 {} # 1 {} # 3 {} # \lline {} # 1 {} ## 0 {} # 0 {} # 0 {} # \lline {} # 0{} } right ]} {}
(92)

Since the last equation dropped out, we are left with two equations and three variables. This means the system has infinite number of solutions. We express those solutions in the parametric form by letting the last variable zz size 12{z} {} equal the parameter tt size 12{t} {}.

The first equation reads x2z=1x2z=1 size 12{x - 2z=1} {}, therefore, x=1+2zx=1+2z size 12{x=1+2z} {}.

The second equation reads y+3z=1y+3z=1 size 12{y+3z=1} {}, therefore, y=13zy=13z size 12{y=1 - 3z} {}.

And now if we let z=tz=t size 12{z=t} {}, the solution is expressed as follows:

x=1+2t, y=13t, z=t.x=1+2t size 12{x=1+2t} {},y=13t size 12{y=1 - 3t} {},z=t size 12{z=t} {}.
(93)

The reader should note that particular solutions to the system can be obtained by assigning values to the parameter tt size 12{t} {}. For example, if we let t=2t=2 size 12{t=2} {}, we have the solution (5, –5, 2).

Example 23

Problem 1

Solve the following system of equations.

x+2y3z=5x+2y3z=5 size 12{x+2y - 3z=5} {}
(94)
2x+4y6z=102x+4y6z=10 size 12{2x+4y - 6z="10"} {}
(95)
3x+6y9z=153x+6y9z=15 size 12{3x+6y - 9z="15"} {}
(96)
Solution

The reduced row-echelon form is given below.

123500000000123500000000 size 12{ left [ matrix { 1 {} # 2 {} # - 3 {} # \lline {} # 5 {} ## 0 {} # 0 {} # 0 {} # \lline {} # 0 {} ## 0 {} # 0 {} # 0 {} # \lline {} # 0{} } right ]} {}
(97)

This time the last two equations drop out, and we are left with one equation and three variables. Again, there are infinite number of solutions. But this time the answer must be expressed in terms of two arbitrary constants.

If we let z=tz=t size 12{z=t} {} and let y=sy=s size 12{y=s} {}, then the first equation x+2y3z=5x+2y3z=5 size 12{x+2y - 3z=5} {} results in x=52s+3tx=52s+3t size 12{x=5 - 2s+3t} {}.

We rewrite the solution as: x=52s+3tx=52s+3t size 12{x=5 - 2s+3t} {}, y=sy=s size 12{y=s} {}, z=tz=t size 12{z=t} {}.

We summarize our discussion in the following table.

  1. If any row of the reduced row-echelon form of the matrix gives a false statement such as 0 = 1, the system is inconsistent and has no solution.
  2. If the reduced row echelon form has fewer equations than the variables and the system is consistent, then the system has an infinite number of solutions. Remember the rows that contain all zeros are dropped.
    1. If a system has an infinite number of solutions, the solution must be expressed in the parametric form.
    2. The number of arbitrary parameters equals the number of variables minus the number of equations.

Inverse Matrices

Section Overview

In this section you will learn to:

  1. Find the inverse of a matrix, if it exists.
  2. Use inverses to solve linear systems.

In this section, we will learn to find the inverse of a matrix, if it exists. Later, we will use matrix inverses to solve linear systems.

Definition of an Inverse: An n×nn×n size 12{n times n} {} matrix has an inverse if there exists a matrix BB size 12{B} {} such that AB=BA=InAB=BA=In size 12{ ital "AB"= ital "BA"=I rSub { size 8{n} } } {}, where InIn size 12{I rSub { size 8{n} } } {} is an n×nn×n size 12{n times n} {} identity matrix. The inverse of a matrix AA size 12{A} {}, if it exists, is denoted by the symbol A1A1 size 12{A rSup { size 8{ - 1} } } {}.

Example 24

Problem 1

Given matrices AA size 12{A} {} and BB size 12{B} {} below, verify that they are inverses.

A=4131B=1134A=4131 size 12{A= left [ matrix { 4 {} # 1 {} ## 3 {} # 1{} } right ]} {}B=1134 size 12{B= left [ matrix { 1 {} # - 1 {} ## - 3 {} # 4{} } right ]} {}
(98)
Solution

The matrices are inverses if the product ABAB size 12{ ital "AB"} {} and BABA size 12{ ital "BA"} {} both equal I 2 I 2, the identity matrix of dimension 2×22×2 size 12{2 times 2} {}.

AB=41311134=1001and       BA=11344131=1001AB=41311134=1001 size 12{ ital "AB"= left [ matrix { 4 {} # 1 {} ## 3 {} # 1{} } right ] left [ matrix { 1 {} # - 1 {} ## - 3 {} # 4{} } right ]= left [ matrix { 1 {} # 0 {} ## 0 {} # 1{} } right ]} {}and       BA=11344131=1001 size 12{ ital "BA"= left [ matrix { 1 {} # - 1 {} ## - 3 {} # 4{} } right ] left [ matrix { 4 {} # 1 {} ## 3 {} # 1{} } right ]= left [ matrix { 1 {} # 0 {} ## 0 {} # 1{} } right ]} {}
(99)

Clearly that is the case; therefore, the matrices A and B are inverses of each other.

Example 25

Problem 1

Find the inverse of the following matrix.

A=3152A=3152 size 12{A= left [ matrix { 3 {} # 1 {} ## 5 {} # 2{} } right ]} {}
(100)
Solution

Suppose AA size 12{A} {} has an inverse, and it is

B=abcdB=abcd size 12{A= left [ matrix { 3 {} # 1 {} ## 5 {} # 2{} } right ]} {}
(101)

Then AB=IAB=I

3152abcd=10013152abcd=1001 size 12{ left [ matrix { 3 {} # 1 {} ## 5 {} # 2{} } right ] left [ matrix { a {} # b {} ## c {} # d{} } right ]= left [ matrix { 1 {} # 0 {} ## 0 {} # 1{} } right ]} {}
(102)

After multiplying the two matrices on the left side, we get

3a+c3b+65a+2c5b+2d=10013a+c3b+65a+2c5b+2d=1001 size 12{ left [ matrix { 3a+c {} # 3b+6 {} ## 5a+2c {} # 5b+2d{} } right ]= left [ matrix { 1 {} # 0 {} ## 0 {} # 1{} } right ]} {}
(103)

Equating the corresponding entries, we get four equations with four unknowns as follows:

3a+c=1     3b+d=03a+c=1 size 12{3a+c=1} {}     3b+d=0 size 12{3b+d=0} {}
(104)
5a+2c=0      5b+2d=15a+2c=0 size 12{5a+2c=0} {}      5b+2d=1 size 12{5b+2d=1} {}
(105)

Solving this system, we get

a=2     b=1     c=5     d=3a=2 size 12{a=2} {}     b=1 size 12{b= - 1} {}     c=5 size 12{c= - 5} {}     d=3 size 12{d=3} {}
(106)

Therefore, the inverse of the matrix AA size 12{A} {} is

21532153 size 12{ left [ matrix { 2 {} # - 1 {} ## - 5 {} # 3{} } right ]} {}
(107)

In this problem, finding the inverse of matrix AA size 12{A} {} amounted to solving the system of equations:

3a+c=1      3b+d=03a+c=1 size 12{3a+c=1} {}      3b+d=0 size 12{3b+d=0} {}
(108)
5a+2c=0      5b+2d=15a+2c=0 size 12{5a+2c=0} {}      5b+2d=1 size 12{5b+2d=1} {}
(109)

Actually, it can be written as two systems, one with variables aa size 12{a} {} and cc size 12{c} {}, and the other with bb and dd size 12{d} {}. The augmented matrices for both are given below.

311520and 310521311520 size 12{ left [ matrix { 3 {} # 1 {} # \lline {} # 0 {} ## 5 {} # 2 {} # \lline {} # 1{} } right ]} {}and310521 size 12{ left [ matrix { 3 {} # 1 {} # \lline {} # 0 {} ## 5 {} # 2 {} # \lline {} # 1{} } right ]} {}
(110)

As we look at the two augmented matrices, we notice that the coefficient matrix for both the matrices is the same. Which implies the row operations of the Gauss-Jordan method will also be the same. A great deal of work can be saved if the two right hand columns are grouped together to form one augmented matrix as below.

3110520131105201 size 12{ left [ matrix { 3 {} # 1 {} # \lline {} # 1 {} # 0 {} ## 5 {} # 2 {} # \lline {} # 0 {} # 1{} } right ]} {}
(111)

And solving this system, we get

1021015310210153 size 12{ left [ matrix { 1 {} # 0 {} # \lline {} # 2 {} # - 1 {} ## 0 {} # 1 {} # \lline {} # - 5 {} # 3{} } right ]} {}
(112)

The matrix on the right side of the vertical line is the A1A1 size 12{A rSup { size 8{ - 1} } } {} matrix.

What you just witnessed is no coincidence. This is the method that is often employed in finding the inverse of a matrix.

We list the steps, as follows:

The Method for Finding the Inverse of a Matrix

  1. Write the augmented matrix AInAIn size 12{ left [A \lline I rSub { size 8{n} } right ]} {}.
  2. Write the augmented matrix in step 1 in reduced row echelon form.
  3. If the reduced row echelon form in 2 is InBInB size 12{ left [ matrix { I rSub { size 8{n} } {} # \lline {} # B{} } right ]} {}, then BB size 12{B} {} is the inverse of AA size 12{A} {}.
  4. If the left side of the row reduced echelon is not an identity matrix, the inverse does not exist.

Example 27

Problem 1

Given the matrix AA size 12{A} {} below, find its inverse.

A=111230021A=111230021 size 12{A= left [ matrix { 1 {} # - 1 {} # 1 {} ## 2 {} # 3 {} # 0 {} ## 0 {} # - 2 {} # 1{} } right ]} {}
(113)
Solution

We write the augmented matrix as follows.

111100230010021001111100230010021001 size 12{ left [ matrix { 1 {} # - 1 {} # 1 {} # \lline {} # 1 {} # 0 {} # 0 {} ## 2 {} # 3 {} # 0 {} # \lline {} # 0 {} # 1 {} # 0 {} ## 0 {} # - 2 {} # 1 {} # \lline {} # 0 {} # 0 {} # 1{} } right ]} {}
(114)

We reduce this matrix using the Gauss-Jordan method.

Multiplying the first row by –2 and adding it to the second row, we get

111100052210021001111100052210021001 size 12{ left [ matrix { 1 {} # - 1 {} # 1 {} # \lline {} # 1 {} # 0 {} # 0 {} ## 0 {} # 5 {} # - 2 {} # \lline {} # - 2 {} # 1 {} # 0 {} ## 0 {} # - 2 {} # 1 {} # \lline {} # 0 {} # 0 {} # 1{} } right ]} {}
(115)

If we swap the second and third rows, we get

111100021001052210111100021001052210 size 12{ left [ matrix { 1 {} # - 1 {} # 1 {} # \lline {} # 1 {} # 0 {} # 0 {} ## 0 {} # - 2 {} # 1 {} # \lline {} # 0 {} # 0 {} # 1 {} ## 0 {} # 5 {} # - 2 {} # \lline {} # - 2 {} # 1 {} # 0{} } right ]} {}
(116)

Divide the second row by –2. The result is

111100011/2001/2052210111100011/2001/2052210 size 12{ left [ matrix { 1 {} # - 1 {} # 1 {} # \lline {} # 1 {} # 0 {} # 0 {} ## 0 {} # 1 {} # - 1/2 {} # \lline {} # 0 {} # 0 {} # - 1/2 {} ## 0 {} # 5 {} # - 2 {} # \lline {} # - 2 {} # 1 {} # 0{} } right ]} {}
(117)

Let us do two operations here. 1) Add the second row to first, 2) Add -5 times the second row to the third. And we get

101/2101/2011/2001/2001/2215/2101/2101/2011/2001/2001/2215/2 size 12{ left [ matrix { 1 {} # 0 {} # 1/2 {} # \lline {} # 1 {} # 0 {} # - 1/2 {} ## 0 {} # 1 {} # - 1/2 {} # \lline {} # 0 {} # 0 {} # - 1/2 {} ## 0 {} # 0 {} # 1/2 {} # \lline {} # - 2 {} # 1 {} # 5/2{} } right ]} {}
(118)

Multiplication of the third row by 2 results in

101/2101/2011/2001/2001425101/2101/2011/2001/2001425 size 12{ left [ matrix { 1 {} # 0 {} # 1/2 {} # \lline {} # 1 {} # 0 {} # - 1/2 {} ## 0 {} # 1 {} # - 1/2 {} # \lline {} # 0 {} # 0 {} # - 1/2 {} ## 0 {} # 0 {} # 1 {} # \lline {} # - 4 {} # 2 {} # 5{} } right ]} {}
(119)

Multiply the third row by 1/21/2 size 12{1/2} {} and add it to the second. Also, multiply the third row by 1/21/2 size 12{ - 1/2} {} and add it to the first. We get

100313010212001425100313010212001425 size 12{ left [ matrix { 1 {} # 0 {} # 0 {} # \lline {} # 3 {} # - 1 {} # - 3 {} ## 0 {} # 1 {} # 0 {} # \lline {} # - 2 {} # 1 {} # 2 {} ## 0 {} # 0 {} # 1 {} # \lline {} # - 4 {} # 2 {} # 5{} } right ]} {}
(120)

Therefore, the inverse of matrix AA size 12{A} {} is

313212425313212425 size 12{ left [ matrix { 3 {} # - 1 {} # - 3 {} ## - 2 {} # 1 {} # 2 {} ## - 4 {} # 2 {} # 5{} } right ]} {}
(121)

One should verify the result by multiplying the two matrices to see if the product does, indeed, equal the identity matrix.

Now that we know how to find the inverse of a matrix, we will use inverses to solve systems of equations. The method is analogous to solving a simple equation like the one below.

23x=423x=4 size 12{ { {2} over {3} } x=4} {}
(122)

Example 28

Problem 1

Solve the following equation .

23x=423x=4 size 12{ { {2} over {3} } x=4} {}
(123)
Solution

To solve the above equation, we multiply both sides of the equation by the multiplicative inverse of 2323 size 12{ { {2} over {3} } } {} which happens to be 3232 size 12{ { {3} over {2} } } {} . We get

3223x=4323223x=432 size 12{ { {3} over {2} } cdot { {2} over {3} } x=4 cdot { {3} over {2} } } {}
(124)
x=6x=6 size 12{x=6} {}
(125)

We use the above example as an analogy to show how linear systems of the form AX=BAX=B size 12{ ital "AX"=B} {} are solved.

To solve a linear system, we first write the system in the matrix equation AX=BAX=B size 12{ ital "AX"=B} {}, where AA size 12{A} {} is the coefficient matrix, XX size 12{X} {} the matrix of variables, and BB size 12{B} {} the matrix of constant terms. We then multiply both sides of this equation by the multiplicative inverse of the matrix AA size 12{A} {}.

Consider the following example.

Example 29

Problem 1

Solve the following system

3x+y=33x+y=3 size 12{3x+y=3} {}
(126)
5x+2y=45x+2y=4 size 12{5x+2y=4} {}
(127)
Solution

To solve the above equation, first we express the system as

AX=BAX=B size 12{ ital "AX"=B} {}
(128)

where AA size 12{A} {} is the coefficient matrix, and BB size 12{B} {} is the matrix of constant terms. We get

3152xy=343152xy=34 size 12{ left [ matrix { 3 {} # 1 {} ## 5 {} # 2{} } right ] left [ matrix { x {} ## y } right ]= left [ matrix { 3 {} ## 4 } right ]} {}
(129)

To solve this system, we multiply both sides of the matrix equation AX=BAX=B size 12{ ital "AX"=B} {} by A1A1 size 12{A rSup { size 8{ - 1} } } {}. Since the matrix AA size 12{A} {} is the same matrix AA size 12{A} {} whose inverse we found in Example 25,

A1=2153A1=2153 size 12{A rSup { size 8{ - 1} } = left [ matrix { 2 {} # - 1 {} ## - 5 {} # 3{} } right ]} {}
(130)

Multiplying both sides by A1A1 size 12{A rSup { size 8{ - 1} } } {}, we get

21533152xy=21533421533152xy=215334 size 12{ left [ matrix { 2 {} # - 1 {} ## - 5 {} # 3{} } right ] left [ matrix { 3 {} # 1 {} ## 5 {} # 2{} } right ] left [ matrix { x {} ## y } right ]= left [ matrix { 2 {} # - 1 {} ## - 5 {} # 3{} } right ] left [ matrix { 3 {} ## 4 } right ]} {}
(131)
1001xy=231001xy=23 size 12{ left [ matrix { 1 {} # 0 {} ## 0 {} # 1{} } right ] left [ matrix { x {} ## y } right ]= left [ matrix { 2 {} ## - 3 } right ]} {}
(132)
xy=23xy=23 size 12{ left [ matrix { x {} ## y } right ]= left [ matrix { 2 {} ## - 3 } right ]} {}
(133)

Therefore, x=2x=2 size 12{x=2} {} , and y=3y=3 size 12{y= - 3} {}.

Example 30

Problem 1

Solve the following system

xy+z=6xy+z=6 size 12{x - y+z=6} {}
(134)
2x+3y=12x+3y=1 size 12{2x+3y=1} {}
(135)
2y+z=52y+z=5 size 12{ - 2y+z=5} {}
(136)
Solution

To solve the above equation, we write the system in the matrix form AX=BAX=B size 12{ ital "AX"=B} {} as follows:

111230021xyz=615111230021xyz=615 size 12{ left [ matrix { 1 {} # - 1 {} # 1 {} ## 2 {} # 3 {} # 0 {} ## 0 {} # - 2 {} # 1{} } right ] left [ matrix { x {} ## y {} ## z } right ]= left [ matrix { 6 {} ## 1 {} ## 5 } right ]} {}

To solve this system, we need inverse of AA size 12{A} {}. From Example 27, we have

A1=313212425A1=313212425 size 12{A rSup { size 8{ - 1} } = left [ matrix { 3 {} # - 1 {} # - 3 {} ## - 2 {} # 1 {} # 2 {} ## - 4 {} # 2 {} # 5{} } right ]} {}

We multiply both sides of the matrix equation AX=BAX=B size 12{ ital "AX"=B} {}, by A1A1 size 12{A rSup { size 8{ - 1} } } {}, we get

313212425113230021xyz=313212425615313212425113230021xyz=313212425615 size 12{ left [ matrix { 3 {} # - 1 {} # - 3 {} ## - 2 {} # 1 {} # 2 {} ## - 4 {} # 2 {} # 5{} } right ] left [ matrix { 1 {} # - 1 {} # - 3 {} ## 2 {} # 3 {} # 0 {} ## 0 {} # - 2 {} # 1{} } right ] left [ matrix { x {} ## y {} ## z } right ]= left [ matrix { 3 {} # - 1 {} # - 3 {} ## - 2 {} # 1 {} # 2 {} ## - 4 {} # 2 {} # 5{} } right ] left [ matrix { 6 {} ## 1 {} ## 5 } right ]} {}

After multiplying the matrices, we get

100010001xyz=213100010001xyz=213 size 12{ left [ matrix { 1 {} # 0 {} # 0 {} ## 0 {} # 1 {} # 0 {} ## 0 {} # 0 {} # 1{} } right ] left [ matrix { x {} ## y {} ## z } right ]= left [ matrix { 2 {} ## - 1 {} ## 3 } right ]} {}

xyz=213xyz=213 size 12{ left [ matrix { x {} ## y {} ## z } right ]= left [ matrix { 2 {} ## - 1 {} ## 3 } right ]} {}

Once again, we remind the reader that not every system of equations can be solved by the matrix inverse method. Although the Gauss-Jordan method works for every situation, the matrix inverse method works only in cases where the inverse of the square matrix exists. In such cases the system has a unique solution.

We summarize our discussion in the following table.

The Method for Finding the Inverse of a Matrix

  1. Write the augmented matrix AInAIn size 12{ left [ matrix { A {} # \lline {} # I rSub { size 8{n} } {} } right ]} {}.
  2. Write the augmented matrix in step 1 in reduced row echelon form.
  3. If the reduced row echelon form in 2 is InBInB size 12{ left [ matrix { I rSub { size 8{n} } {} # \lline {} # B{} } right ]} {}, then BB size 12{B} {} is the inverse of AA size 12{A} {}.
  4. If the left side of the row reduced echelon is not an identity matrix, the inverse does not exist.

The Method for Solving a System of Equations When a Unique Solution Exists

  1. Express the system in the matrix equation AX=BAX=B size 12{ ital "AX"=B} {}.
  2. To solve the equation AX=BAX=B size 12{ ital "AX"=B} {}, we multiply on both sides by A1A1 size 12{A rSup { size 8{ - 1} } } {}.

Application of Matrices in Cryptography

In this section, we see a use of matrices in encoding and decoding secret messages. There are many techniques used, but we will use a method that first converts the secret message into a string of numbers by arbitrarily assigning a number to each letter of the message. Next we convert this string of numbers into a new set of numbers by multiplying the string by a square matrix of our choice that has an inverse. This new set of numbers represents the coded message. To decode the message, we take the string of coded numbers and multiply it by the inverse of the matrix to get the original string of numbers. Finally, by associating the numbers with their corresponding letters, we obtain the original message.

In this section, we will use the correspondence where the letters A to Z correspond to the numbers 1 to 26, as shown below, and a space is represented by the number 27, and all punctuation is ignored.

Table 2
A B C D E F G H I J K L M
1 2 3 4 5 6 7 8 9 10 11 12 13
N O P Q R S T U V W X Y Z
14 15 16 17 18 19 20 21 22 23 24 25 26

Example 31

Problem 1

Use the matrix A=1213A=1213 size 12{A= left [ matrix { 1 {} # 2 {} ## 1 {} # 3{} } right ]} {} to encode the message: ATTACK NOW!

Solution

We divide the letters of the message into groups of two.

AT TA CK –N OW

We assign the numbers to these letters from the above table, and convert each pair of numbers into 2×12×1 size 12{2 times 1} {} matrices. In the case where a single letter is left over on the end, a space is added to make it into a pair.

AT=120TA=201CK=311 _N=2714OW=1523AT=120 size 12{ left [ matrix { A {} ## T } right ]= left [ matrix { 1 {} ## "20" } right ]} {}TA=201 size 12{ left [ matrix { T {} ## A } right ]= left [ matrix { "20" {} ## 1 } right ]} {}CK=311 size 12{ left [ matrix { C {} ## K } right ]= left [ matrix { 3 {} ## "11" } right ]} {} _N=2714 size 12{ left [ matrix { _ {} ## N } right ]= left [ matrix { "27" {} ## "14" } right ]} {}OW=1523 size 12{ left [ matrix { O {} ## W } right ]= left [ matrix { "15" {} ## "23" } right ]} {}
(137)

So at this stage, our message expressed as 2×12×1 size 12{2 times 1} {} matrices is as follows.

1202013112714152312020131127141523 size 12{ left [ matrix { 1 {} ## "20" } right ] left [ matrix { "20" {} ## 1 } right ] left [ matrix { 3 {} ## "11" } right ] left [ matrix { "27" {} ## "14" } right ] left [ matrix { "15" {} ## "23" } right ]} {}
(138)

Now to encode, we multiply, on the left, each matrix of our message by the matrix A. For example, the product of A with our first matrix is

1213120=41611213120=4161 size 12{ left [ matrix { 1 {} # 2 {} ## 1 {} # 3{} } right ] left [ matrix { 1 {} ## "20" } right ]= left [ matrix { "41" {} ## "61" } right ]} {}
(139)

By multiplying each of the matrices in ( I ) by the matrix A, we get the desired coded message given below.

4161222325365569618441612223253655696184 size 12{ left [ matrix { "41" {} ## "61" } right ] left [ matrix { "22" {} ## "23" } right ] left [ matrix { "25" {} ## "36" } right ] left [ matrix { "55" {} ## "69" } right ] left [ matrix { "61" {} ## "84" } right ]} {}
(140)

Example 32

Problem 1

Decode the following message that was encoded using matrix A=1213A=1213 size 12{A= left [ matrix { 1 {} # 2 {} ## 1 {} # 3{} } right ]} {}.

212637534554741015369212637534554741015369 size 12{ left [ matrix { "21" {} ## "26" } right ] left [ matrix { "37" {} ## "53" } right ] left [ matrix { "45" {} ## "54" } right ] left [ matrix { "74" {} ## "101" } right ] left [ matrix { "53" {} ## "69" } right ]} {}
(141)
Solution

Since this message was encoded by multiplying by the matrix A in Example 31, we decode this message by first multiplying each matrix, on the left, by the inverse of matrix A given below.

A1=3211A1=3211 size 12{A rSup { size 8{ - 1} } = left [ matrix { 3 {} # - 2 {} ## - 1 {} # 1{} } right ]} {}
(142)

For example,

32112126=11532112126=115 size 12{ left [ matrix { 3 {} # - 2 {} ## - 1 {} # 1{} } right ] left [ matrix { "21" {} ## "26" } right ]= left [ matrix { "11" {} ## 5 } right ]} {}
(143)

By multiplying each of the matrices in ( II ) by the matrix A1A1 size 12{A rSup { size 8{ - 1} } } {}, we get the following.

1155162792027211611551627920272116 size 12{ left [ matrix { "11" {} ## 5 } right ] left [ matrix { 5 {} ## "16" } right ] left [ matrix { "27" {} ## 9 } right ] left [ matrix { "20" {} ## "27" } right ] left [ matrix { "21" {} ## "16" } right ]} {}
(144)

Finally, by associating the numbers with their corresponding letters, we obtain the following.

K E E P _ I T _UP K E E P _ I T _UP size 12{ left [ matrix { K {} ## E } right ] left [ matrix { E {} ## P } right ] left [ matrix { _ {} ## I } right ] left [ matrix { T {} ## _ } right ] left [ matrix { U {} ## P } right ]} {}
(145)

And the message reads: KEEP IT UP.

Now suppose we wanted to use a 3×33×3 size 12{3 times 3} {} matrix to encode a message, then instead of dividing the letters into groups of two, we would divide them into groups of three.

Example 33

Problem 1

Using the matrix B=111101211B=111101211 size 12{B= left [ matrix { 1 {} # 1 {} # - 1 {} ## 1 {} # 0 {} # 1 {} ## 2 {} # 1 {} # 1{} } right ]} {}, encode the message: ATTACK NOW!

Solution

We divide the letters of the message into groups of three.

ATT ACK –NO W––

Note that since the single letter "W" was left over on the end, we added two spaces to make it into a triplet.

Now we assign the numbers their corresponding letters from the table, and convert each triplet of numbers into 3×13×1 size 12{3 times 1} {} matrices. We get

ATT=12020ACK=1311 _NO=271415 W _ _=232727ATT=12020 size 12{ left [ matrix { A {} ## T {} ## T } right ]= left [ matrix { 1 {} ## "20" {} ## "20" } right ]} {}ACK=1311 size 12{ left [ matrix { A {} ## C {} ## K } right ]= left [ matrix { 1 {} ## 3 {} ## "11" } right ]} {} _NO=271415 size 12{ left [ matrix { _ {} ## N {} ## O } right ]= left [ matrix { "27" {} ## "14" {} ## "15" } right ]} {} W _ _=232727 size 12{ left [ matrix { W {} ## _ {} ## _ } right ]= left [ matrix { "23" {} ## "27" {} ## "27" } right ]} {}
(146)

So far we have,

120201311271415232727120201311271415232727 size 12{ left [ matrix { 1 {} ## "20" {} ## "20" } right ] left [ matrix { 1 {} ## 3 {} ## "11" } right ] left [ matrix { "27" {} ## "14" {} ## "15" } right ] left [ matrix { "23" {} ## "27" {} ## "27" } right ]} {}
(147)

We multiply, on the left, each matrix of our message by the matrix B. For example,

11110121112020=1214211110121112020=12142 size 12{ left [ matrix { 1 {} # 1 {} # - 1 {} ## 1 {} # 0 {} # 1 {} ## 2 {} # 1 {} # 1{} } right ] left [ matrix { 1 {} ## "20" {} ## "20" } right ]= left [ matrix { 1 {} ## "21" {} ## "42" } right ]} {}
(148)

By multiplying each of the matrices in ( III ) by the matrix B, we get the desired coded message as follows:

1214271216264283235010012142712162642832350100 size 12{ left [ matrix { 1 {} ## "21" {} ## "42" } right ] left [ matrix { - 7 {} ## "12" {} ## "16" } right ] left [ matrix { "26" {} ## "42" {} ## "83" } right ] left [ matrix { "23" {} ## "50" {} ## "100" } right ]} {}
(149)

If we need to decode this message, we simply multiply the coded message by B1B1 size 12{B rSup { size 8{ - 1} } } {}, and associate the numbers with the corresponding letters of the alphabet.

Example 34

Problem 1

Decode the following message that was encoded using matrix B=111101211B=111101211 size 12{B= left [ matrix { 1 {} # 1 {} # - 1 {} ## 1 {} # 0 {} # 1 {} ## 2 {} # 1 {} # 1{} } right ]} {}.

112043251041221414112043251041221414 size 12{ left [ matrix { "11" {} ## "20" {} ## "43" } right ] left [ matrix { "25" {} ## "10" {} ## "41" } right ] left [ matrix { "22" {} ## "14" {} ## "14" } right ]} {}
(150)
Solution

Since this message was encoded by multiplying by the matrix B. We first determine inverse of B.

B1=121132111B1=121132111 size 12{B rSup { size 8{ - 1} } = left [ matrix { 1 {} # 2 {} # - 1 {} ## - 1 {} # - 3 {} # 2 {} ## - 1 {} # - 1 {} # 1{} } right ]} {}
(151)

To decode the message, we multiply each matrix, on the left, by B1B1 size 12{B rSup { size 8{ - 1} } } {}. For example,

121132111112043=81512121132111112043=81512 size 12{ left [ matrix { 1 {} # 2 {} # - 1 {} ## - 1 {} # - 3 {} # 2 {} ## - 1 {} # - 1 {} # 1{} } right ] left [ matrix { "11" {} ## "20" {} ## "43" } right ]= left [ matrix { 8 {} ## "15" {} ## "12" } right ]} {}
(152)

By multiplying each of the matrices in ( IV ) by the matrix B1B1 size 12{B rSup { size 8{ - 1} } } {}, we get the following.

81512427691858151242769185 size 12{ left [ matrix { 8 {} ## "15" {} ## "12" } right ] left [ matrix { 4 {} ## "27" {} ## 6 } right ] left [ matrix { 9 {} ## "18" {} ## 5 } right ]} {}
(153)

Finally, by associating the numbers with their corresponding letters, we obtain the following.

HOLDFIREHOLDFIRE size 12{ left [ matrix { H {} ## O {} ## L } right ] left [ matrix { D {} ## _ {} ## F } right ] left [ matrix { I {} ## R {} ## E } right ]} {}
(154)

And the message reads: HOLD FIRE.

We summarize.

TO ENCODE A MESSAGE

  1. Divide the letters of the message into groups of two or three.
  2. Convert each group into a string of numbers by assigning a number to each letter of the message. Remember to assign letters to blank spaces.
  3. Convert each group of numbers into column matrices.
  4. Convert these column matrices into a new set of column matrices by multiplying them with a compatible square matrix of your choice that has an inverse. This new set of numbers or matrices represents the coded message.

TO DECODE A MESSAGE

  1. Take the string of coded numbers and multiply it by the inverse of the matrix that was used to encode the message.
  2. Associate the numbers with their corresponding letters.

Applications – Leontief Models

In the 1930's, Wassily Leontief used matrices to model economic systems. His models, often referred to as the input-output models, divide the economy into sectors where each sector produces goods and services not only for itself but also for other sectors. These sectors are dependent on each other and the total input always equals the total output. In 1973, he won the Nobel Prize in Economics for his work in this field. In this section we look at both the closed and the open models that he developed.

The Closed Model

As an example of the closed model, we look at a very simple economy, where there are only three sectors: food, shelter, and clothing.

Example 36

Problem 1

We assume that in a village there is a farmer, carpenter, and a tailor, who provide the three essential goods: food, shelter, and clothing. Suppose the farmer himself consumes 40% of the food he produces, and gives 40% to the carpenter, and 20% to the tailor. Thirty percent of the carpenter's production is consumed by himself, 40% by the farmer, and 30% by the carpenter. Fifty percent of the tailor's production is used by himself, 30% by the farmer, and 20% by the tailor. Write the matrix that describes this closed model.

Solution

The table below describes the above information.

Table 3
  The proportion produced by the farmer The propotion produced by the carpenter The proportion produced by the tailor
The proportion used by the farmer .40 .40 .30
The proportion used by the carpenter .40 .30 .20
The proportion used by the tailor .20 .30 .50

In a matrix form it can be written as follows.

A=.40.40.30.40.30.20.20.30.50A=.40.40.30.40.30.20.20.30.50 size 12{A= left [ matrix { "." "40" {} # "." "40" {} # "." "30" {} ## "." "40" {} # "." "30" {} # "." "20" {} ## "." "20" {} # "." "30" {} # "." "50"{} } right ]} {}
(155)

This matrix is called the input-output matrix. It is important that we read the matrix correctly. For example the entry A23A23 size 12{A rSub { size 8{"23"} } } {}, the entry in row 2 and column 3, represents the following.

A23=20%A23=20% size 12{A rSub { size 8{"23"} } ="20"%} {} of the tailor's production is used by the carpenter.

A33=50%A33=50% size 12{A rSub { size 8{"33"} } ="50"%} {}of the tailor's production is used by the tailor.

Example 37

Problem 1

In Example 36 above, how much should each person get for his efforts?

Solution

We choose the following variables.

x=Farmer's payx=Farmer's pay size 12{x="Farmer's pay"} {}
(156)
y= Carpenter's payy= Carpenter's pay size 12{y=" Carpenter's pay"} {}
(157)
z= Tailor's payz= Tailor's pay size 12{z=" Tailor's pay"} {}
(158)

As we said earlier, in this model input must equal output. That is, the amount paid by each equals the amount received by each.

Let us say the farmer gets paid xx size 12{x} {} dollars. Let us now look at the farmer's expenses. The farmer uses up 40% of his own production, that is, of the xx size 12{x} {} dollars he gets paid, he pays himself .40x.40x size 12{ "." "40"x} {} dollars, he pays .40y.40y size 12{ "." "40"y} {} dollars to the carpenter, and .30z.30z size 12{ "." "30"z} {} to the tailor. Since the expenses equal the wages, we get the following equation.

x=.40x+.40y+.30zx=.40x+.40y+.30z size 12{x= "." "40"x+ "." "40"y+ "." "30"z} {}
(159)

In the same manner, we get

y=.40x+.30y+.20zy=.40x+.30y+.20z size 12{y= "." "40"x+ "." "30"y+ "." "20"z} {}
(160)
z=.20x+.30y+.50zz=.20x+.30y+.50z size 12{z= "." "20"x+ "." "30"y+ "." "50"z} {}
(161)

The above system can be written as

xyz=.40.40.30.40.30.20.20.30.50xyzxyz=.40.40.30.40.30.20.20.30.50xyz size 12{ left [ matrix { x {} ## y {} ## z } right ]= left [ matrix { "." "40" {} # "." "40" {} # "." "30" {} ## "." "40" {} # "." "30" {} # "." "20" {} ## "." "20" {} # "." "30" {} # "." "50"{} } right ] left [ matrix { x {} ## y {} ## z } right ]} {}
(162)

This system is often referred to as

X=AXX=AX size 12{X= ital "AX"} {}
(163)

Simplification results in

.60x.40y.30z=0.60x.40y.30z=0 size 12{ "." "60"x - "." "40"y - "." "30"z=0} {}
(164)
.40x+.70y.20z=0.40x+.70y.20z=0 size 12{ - "." "40"x+ "." "70"y - "." "20"z=0} {}
(165)
.20x.30y+.50z=0.20x.30y+.50z=0 size 12{ - "." "20"x - "." "30"y+ "." "50"z=0} {}
(166)

Solving for xx size 12{x} {}, yy size 12{y} {}, and zz size 12{z} {} using the Gauss-Jordan method, we get

x=2926t      y=1213t      and       z=tx=2926t size 12{x= { {"29"} over {"26"} } t} {}      y=1213t size 12{y= { {"12"} over {"13"} } t} {}      and       z=t size 12{z=t} {}
(167)

Since we are only trying to determine the proportions of the pay, we can choose t to be any value. Suppose we let t=$2600t=$2600 size 12{t=$"2600"} {}, then we get

x=$2900      y=$2400      and       z=$2600x=$2900 size 12{x=$"2900"} {}      y=$2400 size 12{y=$"2400"} {}      and      z=$2600 size 12{z=$"2600"} {}
(168)

Note:

The use of a calculator in solving these problems is strongly recommended. Although we at De Anza College use TI-85 calculators, any calculator that handles matrices will do.

The Open Model

The open model is more realistic, as it deals with the economy where sectors of the economy not only satisfy each others needs, but they also satisfy some outside demands. In this case, the outside demands are put on by the consumer. But the basic assumption is still the same; that is, whatever is produced is consumed.

Let us again look at a very simple scenario. Suppose the economy consists of three people, the farmer FF size 12{F} {}, the carpenter CC size 12{C} {}, and the tailor TT size 12{T} {}. A part of the farmer's production is used by all three, and the rest is used by the consumer. In the same manner, a part of the carpenter's and the tailor's production is used by all three, and rest is used by the consumer.

Let us assume that whatever the farmer produces, 20% is used by him, 15% by the carpenter, 10% by the tailor, and the consumer uses the other 40 billion dollars worth of the food. Ten percent of the carpenter's production is used by him, 25% by the farmer, 5% by the tailor, and 50 billion dollars worth by the consumer. Fifteen percent of the clothing is used by the tailor, 10% by the farmer, 5% by the carpenter, and the remaining 60 billion dollars worth by the consumer. We write the internal consumption in the following table, and express the demand as the matrix D.

Table 4
  FF size 12{F} {} produces CC size 12{C} {} produces TT size 12{T} {} produces
FF size 12{F} {} uses .20 .25 .10
CC size 12{C} {} uses .15 .10 .05
TT size 12{T} {} uses .10 .05 .15

The consumer demand for each industry in billions of dollars is given below.

D=405060D=405060 size 12{D= left [ matrix { "40" {} ## "50" {} ## "60" } right ]} {}
(169)

Example 38

Problem 1

In Example 37, what should be, in billions of dollars, the required output by each industry to meet the demand given by the matrix D?

Solution

We choose the following variables.

x=Farmer's outputx=Farmer's output size 12{x="Farmer's output"} {}
(170)
y= Carpenter's outputy= Carpenter's output size 12{y=" Carpenter's output"} {}
(171)
z= Tailor's outputz= Tailor's output size 12{z=" Tailor's output"} {}
(172)

In the closed model, our equation was X=AXX=AX size 12{X = ital "AX"} {}, that is, the total input equals the total output. This time our equation is similar with the exception of the demand by the consumer.

So our equation for the open model should be X=AX+DX=AX+D size 12{X= ital "AX"+D} {}, where DD size 12{D} {} represents the demand matrix. We express it as follows:

X=AX+DX=AX+D size 12{X= ital "AX"+D} {}
(173)
xyz=.20.25.10.15.10.05.10.05.15xyz+405060xyz=.20.25.10.15.10.05.10.05.15xyz+405060 size 12{ left [ matrix { x {} ## y {} ## z } right ]= left [ matrix { "." "20" {} # "." "25" {} # "." "10" {} ## "." "15" {} # "." "10" {} # "." "05" {} ## "." "10" {} # "." "05" {} # "." "15"{} } right ] left [ matrix { x {} ## y {} ## z } right ]+ left [ matrix { "40" {} ## "50" {} ## "60" } right ]} {}
(174)

To solve this system, we write it as

X=AX+DX=AX+D size 12{X= ital "AX"+D} {}
(175)
IAX=DIAX=D size 12{ left (I - A right )X=D} {}
(176)

where I is a 3 by 3 identity matrix

X=IA1DX=IA1D size 12{X= left (I - A right ) rSup { size 8{ - 1} } D} {}
(177)
IA=.80.25.10.15.90.05.10.05.85IA=.80.25.10.15.90.05.10.05.85 size 12{I - A= left [ matrix { "." "80" {} # - "." "25" {} # - "." "10" {} ## - "." "15" {} # "." "90" {} # - "." "05" {} ## - "." "10" {} # - "." "05" {} # "." "85"{} } right ]} {}
(178)
IA1=1.3445.3835.1807.23361.1814.097.1719.11461.2034IA1=1.3445.3835.1807.23361.1814.097.1719.11461.2034 size 12{ left (I - A right ) rSup { size 8{ - 1} } = left [ matrix { 1 "." "3445" {} # "." "3835" {} # "." "1807" {} ## "." "2336" {} # 1 "." "1814" {} # "." "097" {} ## "." "1719" {} # "." "1146" {} # 1 "." "2034"{} } right ]} {}
(179)
X=1.3445.3835.1807.23361.1814.097.1719.11461.2034405060X=1.3445.3835.1807.23361.1814.097.1719.11461.2034405060 size 12{X= left [ matrix { 1 "." "3445" {} # "." "3835" {} # "." "1807" {} ## "." "2336" {} # 1 "." "1814" {} # "." "097" {} ## "." "1719" {} # "." "1146" {} # 1 "." "2034"{} } right ] left [ matrix { "40" {} ## "50" {} ## "60" } right ]} {}
(180)
X=83.799974.234184.8138X=83.799974.234184.8138 size 12{X= left [ matrix { "83" "." "7999" {} ## "74" "." "2341" {} ## "84" "." "8138" } right ]} {}
(181)

Therefore, the three industries must produce the following amount of goods in billions of dollars.

Farmer=$83.7999      Carpenter=$74.2341     Tailor=$84.813Farmer=$83.7999 size 12{"Farmer"=$"83" "." "7999"} {}      Carpenter=$74.2341 size 12{"Carpenter"=$"74" "." "2341"} {}     Tailor=$84.813 size 12{"Tailor"=$"84" "." "813"} {}
(182)

We will do one more problem like the one above, except this time we give the amount of internal and external consumption in dollars and ask for the proportion of the amounts consumed by each of the industries. In other words, we ask for the matrix A.

Example 39

Problem 1

Suppose an economy consists of three industries FF size 12{F} {}, CC size 12{C} {}, and TT size 12{T} {}. Again, each of the industries produces for internal consumption among themselves, as well as, the external demand by the consumer. The following table gives information about the use of each industry's production in dollars.

Table 5
  F F size 12{F} {} C C size 12{C} {} T T size 12{T} {} Demand Total
F F size 12{F} {} 40 50 60 100 250
C C size 12{C} {} 30 40 40 110 220
T T size 12{T} {} 20 30 30 120 200

The first row says that of the $250 dollars worth of production by the industry FF size 12{F} {}, $40 is used by FF size 12{F} {}, $50 is used by CC size 12{C} {}, $60 is used by TT size 12{T} {}, and the remainder of $100 is used by the consumer. The other rows are described in a similar manner.

Find the proportion of the amounts consumed by each of the industries. In other words, find the matrix A.

Once again, the total input equals the total output.

Solution

We are being asked to determine the following:

How much of the production of each of the three industries, FF size 12{F} {}, CC size 12{C} {}, and TT size 12{T} {} is required to produce one unit of FF size 12{F} {}? In the same way, how much of the production of each of the three industries, FF size 12{F} {}, CC size 12{C} {}, and TT size 12{T} {} is required to produce one unit of CC size 12{C} {}? And finally, how much of the production of each of the three industries, FF size 12{F} {}, CC size 12{C} {}, and TT size 12{T} {} is required to produce one unit of TT size 12{T} {}?

Since we are looking for proportions, we need to divide the production of each industry by the total production for each industry.

We analyze as follows:

To produce 250 units of FF size 12{F} {}, we need to use 40 units of FF size 12{F} {}, 30 units of CC size 12{C} {}, and 20 units of TT size 12{T} {}.

Therefore, to produce 1 unit of FF size 12{F} {}, we need to use 40/250 units of FF size 12{F} {}, 30/250 units of CC size 12{C} {}, and 20/250 units of TT size 12{T} {}.

To produce 220 units of CC size 12{C} {}, we need to use 50 units of FF size 12{F} {}, 40 units of CC size 12{C} {}, and 30 units of TT size 12{T} {}.

Therefore, to produce 1 unit of CC size 12{C} {}, we need to use 50/220 units of FF size 12{F} {}, 40/220 units of CC size 12{C} {}, and 30/220 units of TT size 12{T} {}.

To produce 200 units of TT size 12{T} {}, we need to use 60 units of FF size 12{F} {}, 40 units of CC size 12{C} {}, and 30 units of TT size 12{T} {}.

Therefore, to produce 1 unit of TT size 12{T} {}, we need to use 60/200 units of FF size 12{F} {}, 40/200 units of CC size 12{C} {}, and 30/200 units of TT size 12{T} {}.

We obtain the following matrix.

A=40/25050/22060/20030/25040/22040/22020/25030/22030/220A=40/25050/22060/20030/25040/22040/22020/25030/22030/220 size 12{A= left [ matrix { "40"/"250" {} # "50"/"220" {} # "60"/"200" {} ## "30"/"250" {} # "40"/"220" {} # "40"/"220" {} ## "20"/"250" {} # "30"/"220" {} # "30"/"220"{} } right ]} {}
(183)

or

A=.1600.2273.3000.1200.1818.2000.0800.1364.1500A=.1600.2273.3000.1200.1818.2000.0800.1364.1500 size 12{A= left [ matrix { "." "1600" {} # "." "2273" {} # "." "3000" {} ## "." "1200" {} # "." "1818" {} # "." "2000" {} ## "." "0800" {} # "." "1364" {} # "." "1500"{} } right ]} {}
(184)

Clearly

AX+D=XAX+D=X size 12{ ital "AX"+D=X} {}
(185)
40/25050/22060/20030/25040/22040/20020/25030/22030/200250220200+100110120=25022020040/25050/22060/20030/25040/22040/20020/25030/22030/200250220200+100110120=250220200 size 12{ left [ matrix { "40"/"250" {} # "50"/"220" {} # "60"/"200" {} ## "30"/"250" {} # "40"/"220" {} # "40"/"200" {} ## "20"/"250" {} # "30"/"220" {} # "30"/"200"{} } right ] left [ matrix { "250" {} ## "220" {} ## "200" } right ]+ left [ matrix { "100" {} ## "110" {} ## "120" } right ]= left [ matrix { "250" {} ## "220" {} ## "200" } right ]} {}
(186)

We summarize as follows:

LEONTIEF'S MODELS

THE CLOSED MODEL

  1. All consumption is within the industries. There is no external demand.
  2. Input = Output Input = Output size 12{"Input"=" Output"} {}
    (187)
  3. X=AXX=AX size 12{X= ital "AX"} {} or IAX=0IAX=0 size 12{ left (I - A right )X=0} {}

THE OPEN MODEL

  1. In addition to internal consumption, there is an outside demand by the consumer.
  2. Input = Output Input = Output size 12{"Input "=" Output"} {}
    (188)
  3. X=AX+DX=AX+D size 12{X= ital "AX"+D} {} or X=IA1DX=IA1D size 12{X= left (I - A right ) rSup { size 8{ - 1} } D} {}

Collection Navigation

Content actions

Download:

Collection as:

PDF | EPUB (?)

What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

Downloading to a reading device

For detailed instructions on how to download this content's EPUB to your specific device, click the "(?)" link.

| More downloads ...

Module as:

PDF | EPUB (?)

What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

Downloading to a reading device

For detailed instructions on how to download this content's EPUB to your specific device, click the "(?)" link.

| More downloads ...

Add:

Collection to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

Module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks