Autar KawHumberto Isaza

http://nm.MathForCollege.comTransforming Numerical Methods Education for STEM Undergraduates

http://nm.MathForCollege.com

After reading this chapter, you should be able to:1. setup simultaneous linear equations in matrix form and vice-versa,2. understand the concept of the inverse of a matrix,3. know the difference between a consistent and inconsistent system of linear

equations, and4. learn that a system of linear equations can have a unique solution, no solution

or infinite solutions.

Matrix algebra is used for solving systems of equations. Can you illustrate thisconcept?

Matrix algebra is used to solve a system of simultaneous linear equations. In fact, formany mathematical procedures such as the solution to a set of nonlinear equations,interpolation, integration, and differential equations, the solutions reduce to a set ofsimultaneous linear equations. Let us illustrate with an example for interpolation.

The upward velocity of a rocket is given at three different times on the following table.

Table 5.1. Velocity vs. time data for a rocket

The velocity data is approximated by a polynomial as

Set up the equations in matrix form to find the coefficients of the velocity profile.

Time, t Velocity, v(s) (m/s)

5 106.88 177.212 279.2

( ) 12.t5 , 2 ≤≤++= cbtattv

cba ,,

The polynomial is going through three data where from table 5.1.

Requiring that passes through the three data points gives

( ) ( ) ( )332211 ,t and ,, ,, vvtvt

8.106,5 11 == vt

2.177,8 22 == vt

2.279,12 33 == vt

( ) cbtattv ++= 2

( ) cbtatvtv ++== 12111

( ) cbtatvtv ++== 22222

( ) cbtatvtv ++== 32333

Substituting the data gives

or

( ) ( ) ( )332211 ,and , , , , vtvtvt

( ) ( ) 8.106552 =++ cba

( ) ( ) 2.177882 =++ cba

( ) ( ) 2.27912122 =++ cba

8.106525 =++ cba

2.177864 =++ cba

2.27912144 =++ cba

This set of equations can be rewritten in the matrix form as

The above equation can be written as a linear combination as follows

and further using matrix multiplication gives

=

++++++

2.2792.1778.106

12144864525

cbacbacba

=

+

+

2.2792.1778.106

111

1285

1446425

cba

=

2.2792.1778.106

11214418641525

cba

The previous is an illustration of why matrix algebra is needed. The complete solution to the set of equations is given later in this chapter.

A general set of linear equations and unknowns,

…………………………….…………………………….

m n

11212111 cxaxaxa nn =+++

22222121 cxaxaxa nn =+++

mnmnmm cxaxaxa =+++ ........2211

can be rewritten in the matrix form as

Denoting the matrices by , , and , the system of equation is

, where is called the coefficient matrix, is called the right hand side vector and is called the solution vector.

⋅⋅=

⋅⋅

mnmnmm

n

n

c

cc

x

xx

aaa

aaaaaa

2

1

2

1

21

22221

11211

..

..

..

[ ]A [ ]X [ ]C

[ ][ ] [ ]CXA = [ ]A [ ]C[ ]X

Sometimes systems of equations are written in the augmented form. That is

[ ][ ] [ ]CXA =

[ ]

=

nmnmm

n

n

c

cc

a......aa

a......aaa......aa

CA2

1

21

22221

11211

A system of equations can be consistent or inconsistent. What does that mean?

A system of equations is consistent if there is a solution, and it isinconsistent if there is no solution. However, a consistent system of equations doesnot mean a unique solution, that is, a consistent system of equations may have aunique solution or infinite solutions (Figure 1).

Figure 5.1. Consistent and inconsistent system of equations flow chart.

[ ][ ] [ ]CXA =

Consistent System Inconsistent System

Unique Solution Infinite Solutions

[A][X]= [B]

Give examples of consistent and inconsistent system of equations.

Solution

a) The system of equations

is a consistent system of equations as it has a unique solution, that is,

=

46

3142

yx

=

11

yx

b) The system of equations

is also a consistent system of equations but it has infinite solutions as given as follows.Expanding the above set of equations,

=

36

2142

yx

32642

=+=+

yxyx

you can see that they are the same equation. Hence, any combination of that satisfies,

is a solution. For example is a solution. Other solutions include

, , and so on.

c) The system of equations

is inconsistent as no solutions exists.

642 =+ yx

( )yx,

( ) ( )1,1, =yx

( ) )25.1,5.0(, =yx ( ) )5.1 ,0(, =yx

=

46

2142

yx

How can one distinguish between a consistent and inconsistent system of equations?

A system of equations is consistent if the rank of is equal to the rank of the augmented matrix .

A system of equations is inconsistent if the rank of is less than the rank of the augmented matrix .

But, what do you mean by rank of a matrix?

The rank of a matrix is defined as the order of the largest square submatrix whose determinant is not zero.

[ ][ ] [ ]CXA = A[ ]CA

[ ][ ] [ ]CXA =[ ]CA

A

What is the rank of

?

Solution

The largest square submatrix possible is of order 3 and that is itself. Sincethe rank of

[ ]

=

321502213

A

][A,023)det( ≠−=A .3][ =A

What is the rank of

?

SolutionThe largest square submatrix of is of order 3 and that is itself. Since , the rank of is less than 3. The next largest square submatrix would be a 2 2 matrix. One of the square submatrices of is

And . Hence the rank of is 2. There is no need to look at other 2 2submatrices to establish that the rank of is 2.

[ ]

=

715502213

A

][A ][A 0)det( =A][A ×

][A

[ ]

=

0213

B

02)det( ≠−=B ][A][A

×

How do I now use the concept of rank to find if

is a consistent or inconsistent system of equations?

=

2.2792.1778.106

11214418641525

3

2

1

xxx

SolutionThe coefficient matrix is

and the right hand side vector is

[ ]

=

11214418641525

A

The augmented matrix is

Since there are no square submatrices of order 4 as is a matrix, the rank of is at most 3. So let us look at the square submatrices of of order 3; if any of these square submatrices have determinant not equal to zero, then the rank is 3. For example, a submatrix of the augmented matrix is

][B 43× ][B][B

][B

[ ]

=

2.2791121442.17718648.1061525

B

has.Hence the rank of the augmented matrix is 3. Since , the rank of is 3. Since the rank of the augmented matrix equals the rank of the coefficient matrix , the system of equations is consistent.

][B][B

=

11214418641525

][D

084)det( ≠−=D][][ DA = ][A

][A

Use the concept of rank of matrix to find if

Is consistent or inconsistent?

=

0.2842.1778.106

2138918641525

3

2

1

xxx

Solution

The coefficient matrix is given by

and the right hand side

[ ]

=

2138918641525

A

[ ]

=

0.2842.1778.106

C

Since there are no square submatrices of order 4 as is a 4 3 matrix, the rank of the augmented is at most 3. So let us look at square submatrices of the augmented matrix of order 3 and see if any of these have determinants not equal to zero. For example, a square submatrix of the augmented matrix is

Has . This means, we need to explore other square submatrices of order 3 of the augmented matrix and find their determinants.

][B][B

][B][B

×

[ ]

=

2138918641525

D

0)det( =D][B

That is,

[ ]

=

0.2842132.177188.10615

E

0)det( =E

[ ]

=

0.28413892.1778648.106525

F

0)det( =F

[ ]

=

0.2842892.1771648.106125

G

0)det( =G

All the square submatrices of order of the augmented matrix have a zero determinant. So the rank of the augmented matrix is less than 3. Is the rank of augmented matrix equal to 2? One of the submatrices of the augmented matrix is

And

So the rank of the augmented matrix is 2.

22×

33×][B

][B][B

][B

[ ]

=

864525

H

0120)det( ≠−=H

][B

][BNow we need to find the rank of the coefficient matrix

and

So the rank of the coefficient matrix is less than 3. A square submatrix of the coefficient matrix is

So the rank of the coefficient matrix is 2.

[ ]

=

2138918641525

A

0)det( =A

][A][A

[ ]

=

1815

J

03)det( ≠−=J

][A

Hence, the rank of the coefficient matrix equals the rank of the augmented matrix . So the system of equations is consistent. ][B ][][][ CXA =

][A

Use the concept of rank to find if

is consistent or inconsistent

=

0.2802.1778.106

2138918641525

3

2

1

xxx

The augmented matrix is

Since there are no square submatrices of order 4 4 as the augmented matrix is a4 3 matrix, the rank of the augmented matrix is at most 3. So let us look at squaresubmatrices of the augmented matrix (B) of order 3 and see if any of the 3 3submatrices have a determinant not equal to zero. For example, a square submatrix oforder 3 3 of

det(D) = 0

][B

[ ]

=

0.280:213892.177:18648.106:1525

B

×

× ][B×

× ][B

[ ]

=

2138918641525

D

So it means, we need to explore other square submatrices of the augmented matrix

So the rank of the augmented matrix is 3.The rank of the coefficient matrix is 2 from the previous example.Since the rank of the coefficient matrix is less than the rank of the augmentedmatrix , the system of equations is inconsistent. Hence, no solution exists for

][B

][A

[ ]

=

0.2802132.177188.10615

E

00.120det( ≠≠E

][B

][A][B

][][][ CXA =

In a system of equations that is consistent, the rank of the coefficientmatrix is the same as the augmented matrix . If in addition, the rank of thecoefficient matrix is same as the number of unknowns, then the solution is unique;if the rank of the coefficient matrix is less than the number of unknowns, theninfinite solutions exist.

][][][ CXA =

][A ][ CA][A

][A

Figure 5.2. Flow chart of conditions for consistent and inconsistent system of equations.

Unique solution ifrank (A) = number of unknowns

Infinite solutions ifrank (A) < number of unknowns

Consistent System ifrank (A) = rank (A.B)

Inconsistent System ifrank (A) < rank (A.B)

[A] [X] = [B]

We found that the following system of equations

is a consistent system of equations. Does the system of equations have a unique solution or does it have infinite solutions?

=

2.2792.1778.106

11214418641525

3

2

1

xxx

SolutionThe coefficient matrix is

and the right hand side is

While finding out whether the above equations were consistent in an earlier example, we found that the rank of the coefficient matrix (A) equals rank of augmented matrix equals 3.The solution is unique as the number of unknowns = 3 = rank of (A).

[ ]

=

11214418641525

A

[ ]

=

2.2792.1778.106

C

[ ]CA

We found that the following system of equations

is a consistent system of equations. Is the solution unique or does it have infinite solutions.

=

0.2842.1778.106

2138918641525

3

2

1

xxx

SolutionWhile finding out whether the above equations were consistent, we found that the rank of the coefficient matrix equals the rank of augmented matrix equals 2Since the rank of < number of unknowns = 3, infinite solutions exist.

If we have more equations than unknowns in [A] [X] = [C], does it mean the system is inconsistent?No, it depends on the rank of the augmented matrix and the rank of

][A ( )CA2][ =A

[ ]CA ][A

a) For example

is consistent, sincerank of augmented matrix = 3rank of coefficient matrix = 3

Now since the rank of (A) = 3 = number of unknowns, the solution is not only consistent but also unique.

=

0.2842.2792.1778.106

2138911214418641525

3

2

1

xxx

b) For example

is inconsistent, sincerank of augmented matrix = 4rank of coefficient matrix = 3

=

0.2802.2792.1778.106

2138911214418641525

3

2

1

xxx

c) For example

is consistent, sincerank of augmented matrix = 2rank of coefficient matrix = 2

But since the rank of (A) = 2 < the number of unknowns = 3, infinite solutions exist.

=

0.2806.2132.1778.106

213892105018641525

3

2

1

xxx

Consistent systems of equations can only have a unique solution or infinite solutions. Can a system of equations have more than one but not infinite number of solutions?

No, you can only have either a unique solution or infinite solutions. Let us suppose

has two solutions and so that ][][ ][ CXA = ][Y ][Z

][][ ][ CYA =

][][ ][ CZA =

If r is a constant, then from the two equations

Adding the above two equations gives

[ ][ ] [ ]CrYAr =

( )[ ][ ] ( )[ ]CrZAr −=− 1 1

[ ][ ] ( )[ ][ ] [ ] ( )[ ]CrCrZArYAr −+=−+ 1 1

[ ] [ ] ( )[ ]( ) [ ]CZrYrA =−+ 1

Hence

is a solution to

Since r is any scalar, there are infinite solutions for of the form

[ ] ( )[ ]ZrYr −+ 1

[ ][ ] [ ]CXA =

][][][ CXA =

[ ] ( )[ ]ZrYr −+ 1

If is defined, it might seem intuitive that ,

but matrix division is not defined like that. However an inverse of a matrix can be defined for certain types of square matrices. The inverse of a square matrix , if existing, is denoted by such that

Where is the identity matrix.

][][][ CBA =[ ][ ]BCA =][

][A1][ −A

][][][][][ 11 AAIAA −− ==

][I

In other words, let [A] be a square matrix. If [B] is another square matrix of the same size such that , then [B] is the inverse of [A]. [A] is then called to be invertible or nonsingular. If does not exist, is called noninvertible or singular. If [A] and [B] are two matrices such that , then these statements are also true

• [B] is the inverse of [A]• [A] is the inverse of [B]• [A] and [B] are both invertible • [A] [B]=[I].• [A] and [B] are both nonsingular• all columns of [A] and [B]are linearly independent• all rows of [A] and [B] are linearly independent.

][][][ IAB =1][ −A ][A

nn× ][][][ IAB =

Determine if

is the inverse of

=

3523

][B

−

−=

3523

[A]

Solution

Since,

[B] is the inverse of [A] and [A] is the inverse of [B].

]][[ AB

−

−

=

3523

3523

1001

=

][I=

][][][ IAB =

Can I use the concept of the inverse of a matrix to find the solution of a set of equations [A] [X] = [C]?

Yes, if the number of equations is the same as the number of unknowns, the coefficient matrix is a square matrix. Given

Then, if exists, multiplying both sides by .

This implies that if we are able to find , the solution vector of is simply a multiplication of and the right hand side vector,

][A

1][ −A

][][][ CXA =

1][ −A

][][]][[][ 11 CAXAA −− =

][][][][ 1 CAXI −=

][][][ 1 CAX −=

][][][ CXA =1][ −A

1][ −A

][C

If is a matrix, then is a matrix and according to the definition of inverse of a matrix

Denoting

][A nn×

nn×nn×

1][ −A nn×

][][][ 1 IAA =−

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

⋅⋅⋅⋅

=

nnnn

n

n

aaa

aaaaaa

A

21

22221

11211

][

nn×nn×

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

⋅⋅⋅⋅

=−

''2

'1

'2

'22

'21

'1

'12

'11

1][

nnnn

n

n

aaa

aaaaaa

A

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

⋅⋅⋅

=

10

10

010001

][I

Using the definition of matrix multiplication, the first column of the matrix canthen be found by solving

Similarly, one can find the other columns of the matrix by changing the righthand side accordingly.

nn×nn×

1][ −A

⋅⋅=

⋅⋅

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

⋅⋅⋅⋅

0

01

'1

'21

'11

21

22221

11211

nnnnn

n

n

a

aa

aaa

aaaaaa

1][ −A

The upward velocity of the rocket is given by

Table 5.2. Velocity vs time data for a rocket

In an earlier example, we wanted to approximate the velocity profile by

Time, t (s) Velocity, v (m/s)

5 106.8

8 177.2

12 279.2

( ) 125 ,2 ≤≤++= tcbtattv

We found that the coefficients in are given by

First, find the inverse of

and then use the definition of inverse to find the coefficients

cba and,, ( )tv

=

2.2792.1778.106

cba

11214418641525

[ ]

=

11214418641525

A

cba and,,

SolutionIf

is the inverse of , then

[ ]

=−

'33

'32

'31

'23

'22

'21

'13

'12

'11

1

aaaaaaaaa

A

][A

=

100010001

11214418641525

'33

'32

'31

'23

'22

'21

'13

'12

'11

aaaaaaaaa

Gives three sets of equations,

=

001

11214418641525

'31

'21

'11

aaa

=

010

11214418641525

'32

'22

'12

aaa

=

100

11214418641525

'33

'23

'13

aaa

Solving the above three sets of equations separately gives

−=

571.49524.0

04762.0

'31

'21

'11

aaa

−

−=

000.5417.108333.0

'32

'22

'12

aaa

−=

429.14643.0

03571.0

'33

'23

'13

aaa

=

Hence

Now

where

−−−

−=−

429.1000.5571.44643.0417.19524.0

03571.008333.004762.0][ 1A

[ ][ ] [ ]CXA =

[ ]

=

cba

X

[ ]

=

2.2792.1778.106

C

Using the definition of [ ] ,1−A

[ ] [ ][ ] [ ] [ ] 11 CAXAA −− =

[ ] [ ] [ ]CA X 1−=

−−−

−

2.2792.1778.106

429.1000.5571.44643.0417.19524.0

03571.008333.004762.0

Hence

So

=

086.169.19

2905.0

cba

( ) 125 ,086.169.192905.0 2 ≤≤++= ttttv

For finding the inverse of small matrices, the inverse of an invertible matrix can be found by

Where

[ ] ( ) ( )AadjA

Adet

11 =−

( )

T

21

22221

11211

=

nnnn

n

n

CCC

CCCCCC

Aadj

where are the cofactors of . The matrix

itself is called the matrix of cofactors from [A]. Cofactors are defined in Chapter 4.

ijC ija

nnn

n

n

CC

CCCCCC

1

22221

11211

Find the inverse of

SolutionFrom Example 4.6 in Chapter 04.06, we found

Next we need to find the adjoint of . The cofactors of are found as follows.

[ ]

=

11214418641525

A

( ) 84det −=A

][A

The minor of entry is

The cofactors of entry is

11a

11214418641525

11 =M

11218

=

4−=

11a

( ) 1111

11 1 MC +−=

11M=

4−=

The minor of entry is

The cofactors of entry is

12a

11214418641525

12 =M

1144164

=

80−=

12a

( ) 1221

12 1 MC +−=

12M−=

)80(−−=

80=

Similarly

38413 −=C

721 =C

11922 −=C

42023 =C

331 −=C

3932 =C

12033 −=C

Hence the matrix of cofactors of is

The adjoint of matrix is ,

][A

[ ]

−−−

−−=

1203934201197384804

C

][A T][C

( ) [ ]TCAadj =

−−−

−−=

1204203843911980

374

Hence

[ ] ( ) ( )AadjA

Adet

11 =−

−−−

−−

−=

1204203843911980

374

841

−−−

−=

429.1000.5571.44643.0417.19524.0

03571.008333.004762.0

Yes, the inverse of a square matrix is unique, if it exists. The proof is as follows. Assume that the inverse of is and if this inverse is not unique, then let another inverse of exist called

If is the inverse of , then

Multiply both sides by

][A ][B][A ][C

][B ][A

][][][ IAB =

][C

][][][][][ CICAB =

][][][][ CCAB =

Since is inverse of

Multiply both sides by

This shows that and are the same. So the inverse of is unique.

][A

][B

][C

][][][ ICA =

][][][ CIB =

][][ CB =

][B ][C ][A

Consistent systemInconsistent systemInfinite solutionsUnique solutionRankInverse