1

Mat 211 Dr. Firoz

2-3: System of equations and matrices

Systems, Matrices, and Applications Systems of Linear Equations

System of equation

(Has solution) Consistent Inconsistent (has no solution)

Dependent Independent

For Example: Consider the system

3 2 1

5 3 11

x y

x y

Solve it and see that it has a unique solution. The system is consistent and independent.

The system

4

2 2 3

x y

x y

Is inconsistent, it has no solution. Check the solution.

Now the system

4

2 2 8

x y

x y

has infinitely many solutions, it is consistent and dependent.

We can solve it by the following methods:

1. Algebraically Using Elimination

2. Algebraically Using Substitution

3. Using Graphical method

4. Augmented Matrices and Row Operations.

5. Matrix algebra (the 1A B form)

6. Cramer’s Rule (Determinants).

Definition: The following representation of two linear equations is called the system of

linear equations in two variables:

32 21 xx

12 21 xx

2

Solution: A solution of the given system is an ordered set or list of numbers that satisfies

all the equations simultaneously when we put ax1 and bx2 . This solution is written

as ),(),( 21 baxx . We have our solution in this case as )1,1(),( 21 xx .

If the given system has at least one solution, it is said to be consistent. When the system

has no solution, it is said to be inconsistent.

Examples

1. The system 32 21 xx , 12 21 xx has unique solution. Find the solution.

2. The system 1 2 32 2x x x , 3 23 3x x has infinitely many solutions.. Write

the solution in terms of the third variable.

3. The system 1 22 2 2x x , 1 2 3x x is inconsistent.

4. Solve the system 1 24 3 10x x , 1 23 2 18x x

5. If the system 1 22 5x kx , 1 23 7x x is inconsistent find k.

6. Solve the system 1 2 32 4x x x , 1 2 33 2 4x x x , 1 2 32 4 1x x x

7. Consider this 2 by 2 system: use all the methods discussed above to solve the

system:

3 2 1

5 3 11

x y

x y

Solution:

a) Algebraically Using Elimination

3(3 2 1) 9 6 3

2(5 3 11) 10 6 22

x y x y

x y x y19 19 1x x

By substituting 1x we get 2y

b) Algebraically Using Substitution

3 2 1 3/ 2 1/ 2

3/ 2 1/ 2

5 3 3/ 2 1/ 2 11 1

x y y x

So we can substitute y x in other equation

x x x

c) Using Graphical method

3 13 2 1

2 3

5 115 3 11

3 3

x y y x

x y y x

3

d) Augmented Matrices and Row Operations.

3 2 1

5 3 11

3 2 1 1 2 / 3 1/ 3 1 2 / 3 1/ 3 1 2 / 3 1/ 3

5 3 11 5 3 11 0 19 / 3 38/ 3 0 1 2

x y

x y

The last row gives 2y , and by backward substitute we get 1x

Answers: 1. (1, 1) 2. ( 1 3 2 3 3 33 5 / 3, 1 / 3,x x x x x x ) 3. No solution 4. (-2, 6)

5. k = 6 6. (1, 1,1)

Example 2. Let’s see another example by using Augmented Matrices and Row

Operations. Solve for x, y and z

2 6 4

13 6

2 6 1

x y z

x z y

x y z

or

2 4 6

13 6

2 6 1

x y z

x y z

x y z

Augmented Matrix:

1 2 4 6

1 1 13 6

2 6 1 1

Multiply row 1 by –1 and add with row 2 and get your new row 2

1 2 4 6

0 3 9 0

2 6 1 1

; 2 2 11R R R

Multiply 2 with row 1 and add with row 3 and get your new row 3

1 2 4 6

0 3 9 0

0 2 7 11

; 3 3 12R R R

Multiply row 2 by 1/3 gives you following

1 2 4 6

0 1 3 0

0 2 7 11

; 2 2

1

3R R

Multiply by -2 with row 2 and add with row 3 and get your row 3

1 2 4 6

0 1 3 0

0 0 1 11

; 3 3 22R R R

4

Now by backward substitution to solve for variables:

11, 3(11) 0 2( 6) 4(11) 6z y x

33 104y x

Now practice:

2 1 1 2

1 3 2 1

1 1 1 2

Write the system of equations and solve. Ans (2, -1, 1)

Reduced Row Echelon Form (rref):

The Elimination Method for solving large systems of linear equations

1. Make the leading coefficient 1 either by interchanging row or by multiplying or

dividing the first by a suitable constant.

2. Eliminate the leading coefficient each later equation by replacing the later

equation by the sum itself and a suitable multiple of the first equation

3. Repeat step 1 and 2for the 2nd

row to eliminate leading coefficient 0 and to make

2nd

element 1

4. Continue the process until all diagonal elements are 1 then do back substitution to

solve variables.

Example 3. Below are three row-reduced echelon forms for matrices of certain linear

systems. For each matrix, tell how many solutions the system has. Explain. If there are

any, find the solutions. If there are infinitely many solutions, find the general formula

and two particular solutions.

a. [1 0 0 3

0 1 0 − 2

0 0 1 5 ] The system has unique solution with x=3, y=-2, z=5

b. [1 0 − 2 0

0 1 4 0

0 0 0 1] The system does not have solution

c. [1 0 − 2 0

0 1 3 0

0 0 0 0] The system has many solutions, it is dependent and consistent.

The general solutions are 3 0

3

y z

y z and

2 0

2

x z

x z

5

Now let’s discuss system of equations again:

1. Matrix algebra (the 1A B form)

Example: let

193 9 3 (9) 3

3x x

For matrix algebra we also use this inverse concept to solve systems of equation.

Singular Matrix: Singular matrices do not have an inverse. A matrix is singular

matrix if determinant of the matrix is equal to zero, let A is a matrix then 1A exist if

0A .

So let’s discuss how to find determinant of a matrix

Finding determinant

If

1 1a b d bA then A ad bc and A

c d c aad bc

Example: 3 2

( 9 10) 19 05 3

A then A

Therefore A is not singular matrix. It has inverse. Find the inverse. And verify that 1 1A A AA I . What is I ? Answer: I is an identity matrix.

Finding Determinant of a 3 by 3 matrix

1 2 4

1 1 13

2 6 1

A

Note: The determinant of a matrix will be zero if

1. An entire row is zero.

2. Two rows or columns are equal.

3. A row or column is a constant multiple of another row or column.

Remember, that a matrix is invertible, non-singular, if and only if the determinant is not

zero. So, if the determinant is zero, the matrix is singular and does not have an inverse.

1 13 1 13 1 1| | 1 ( 2) 4

6 1 2 1 2 6

1( 1 78) 2( 1 ( 26)) 4(6 ( 2)) 3

A

6

Sarrus Rule. In order to compute

Note: Sarrus rule is only applicable if the determinant is of order 3 by 3.

Example 1. Use Sarrus rule to find the value of

2 3 4

4 3 1

1 2 4

A

12 + 4 + 48 = 64

2 3 4 2 3

4 3 1 4 3 59 64 5

1 2 4 1 2

A

24 + 3 + 32 = 59

Exercise 1. Use Sarrus rule to evaluate the determinant

8 3 4

4 6 1

5 2 4

A and verify your

answer using calculator.

Finding inverse of a matrix:

Now let’s learn how to find inverse of a matrix. There are different methods to find

inverse matrix.

Method 1. Use Shortcut for 2 by 2 matrix

Let a b

Ac d

then 1 1 d bA

c aad bc

Example:

3 2

5 3A then

1

3 2

3 21 19 19

5 3 5 33( 3) 5(2)

19 19

A

7

Method 2 (Optional). Use Gauss-Jordan elimination to transform [ A | I ] into [ I | A-1

].

Example: Consider a matrix 1 2

3 4A and write the following (use rref using

calculator)

Method 3 (Optional). Adjoint method

A-1

= (adjoint of A) or A-1

= (cofactor matrix of A)T

Let

2 3 4

4 3 1

1 2 4

A , 5then A and 1

10 4 91

15 4 145

5 1 6

A

Now we know how to find inverse, let’s go back to solution of system of equations:

Example 1. (continued) Given system is 3 2 1

5 3 11

x y

x y

If we write in matrix form then we get the following,

3 2

5 3A

xX

y

1

11B

If we do AX B , we get the given system and we can rewrite

1

AX B

X A B

8

We have seen that 3 2

5 3A then 1

3 2

3 21 19 19

5 3 5 33( 3) 5(2)

19 19

A

Now

3 2

1 119 19

5 3 11 2

19 19

x

y so 1x and 2y

Cramer’s Rule (Determinants)

Given the system

This system has the unique solution

Example 1. (Continued) Solve (using Cramer’s Rule)

3 2 1

5 3 11

x y

x y

3 2( 9 10) 19

5 3D

1 2(3 22) 19

11 3xD

3 1(33 5) 38

5 11yD

Now, 19

119

xDx

D and

382

19

yDy

D

9

Example 2. Use Cramer’s Rule to solve the system:

4x - y + z = -5

2x + 2y + 3z = 10

5x – 2y + 6z = 1

Solution. We begin by setting up four determinants:

:

D consists of the coefficients of x, y, and z from the three equations

is obtained by replacing the x-coefficients in the first column of D with the constants

from the right sides of the equations.

is obtained by replacing the y-coefficients in the second column of D with the

constants from the right sides of the equations.

is obtained by replacing the z-coefficients in the third column of D with the constants

from the right sides of the equations.

Next, we evaluate the four determinants:

= 4(12 – (-6)) + 1(12 – 15) + 1(-4 – 10)

= 4(18) + 1(-3) + 1(-14)

= 72 – 3 – 14

= 55

10

= -5(12 – (-6)) + 1(60 – 3) + 1(-20 – 2)

= -5(18)+1(57) + 1(-22)

= -90 + 57 – 22

= -55

= 4(60 – 3) + 5(12 – 15) + 1(2 – 50)

= 4(57) + 5(-3) + 1(-48)

= 228 - 15 – 48

= 165

= 4(2 – (-20)) + 1(2 – 50) – 5(-4 – 10)

= 4(22) + 1(-48) – 5(-14)

= 88 – 48 + 70

= 110

Substitute these four values into the formula from Cramer’s Rule:

So, the solution is (-1, 3, 2).

11

Matrix Algebra on the Calculator TI:

1. Hit 2nd-MATRIX, then EDIT, and select [A], [B], etc.

2. Type in the size(order) of the matrix and the entries. Remember to hit ENTER after

each entry.

3.Repeat these steps until you have all your matrices entered.

4. To calculate hit 2nd-QUIT, then hit 2nd-MATRIX, then simply select

[A]. Then type the x-1 key, then hit ENTER. To convert this to fractions, hit

MATH, then select “Frac”, and hit ENTER.

5. To calculate A-1B, do as in step 4, but select [B] also, so that you have

[A]-1[B] on your screen. Hit ENTER to get the answer.

CASIO:

1. Hit MENU, and select 3:MAT. Select the matrix, type in its size and the

entries until you have all your matrices finished.

2. To calculate A-1, hit MENU, and select 1:RUN. Hit OPTN, then select

F2:MAT, then F1:MAT. This will bring up a “Mat” on the screen. Then you

hit ALPHA (Red key) and A (just below) to bring up A. Then hit the x-1 key

above the “)”.

3. Repeat these steps for 1A B .

Section: Gaussian Elimination

In this section we will learn a general method for finding possible solutions to a linear

system of equations. The method involves systematic elimination of the unknown from

each equation in turn. We will explain the method with examples.

Example 1. Solve the system

1223

12

53

321

32

321

xxx

xx

xxx

Solution: Now applying the operation 133 3rrR we have the following

16115

12

53

32

32

321

xx

xx

xxx

Applying 22 2/1 rR we have

16115

5.5.

53

32

32

321

xx

xx

xxx

And by 233 5rrR

5.135.13

5.5.

53

3

32

321

x

xx

xxx

12

Finally we the following by applying 5.13/23 rR

1

5.5.

53

3

32

321

x

xx

xxx

We now have that 13x , and other unknowns can easily be found by backward

substitution into second and first equations. We have the solution )1,1,1(),,( 321 xxx .

This method is called the Gaussian Elimination method.

Augmented matrix: Let us consider the system of equations

1223

12

53

321

32

321

xxx

xx

xxx

The augmented matrix of the above system is

1

1

5

223

120

311

Row Echelon Form: A matrix is in row echelon form when

1. The entry in row 1, column 1 is a 1 and 0 appears below it

2. The first nonzero entry in each row after the first row is a 1, zeros appear

below it and it appears to the right of the first nonzero entry in any row above.

3. Any row that contains all zeros to the left of the vertical bar appear at the

bottom.

The row echelon form of the above augmented matrix is

1

5.

5

100

5.10

311

which exactly our Gaussian Elimination.

Reduced Row Echelon Form (rref): Sometimes it is advantageous to write a matrix in

reduced row echelon form. In this form, row operations are used to obtain entries that

are zero above as well as below the leading 1 in a row. From reduced row echelon form

one can find solution of a system directly without backward substitution. The reduced

row echelon form of the above row echelon form will look like the one given below:

1

1

1

100

010

001

. The left hand side of the vertical bar is the identity matrix.

13

Finding reduced row echelon form by calculator:

TI:

1. Hit 2nd

1x , then edit your matrix, hit 2nd

MODE to go to the normal screen.

2. Hit 2nd

1x then go to MATH, hit 8 for rref (and input your matrix A using

again 2nd

1x , hit enter.

3. You will get your reduced row echelon form.

CASIO:

1. Go to MENU, choose MAT, use right arrow to select size and edit your matrix

2. Go to OPTION and Press F1 for row operation

Example 2: Find the value of K so that the given system has no solution:

02

32

6

321

321

321

Kxxx

xxx

xxx

Solution: Find the determinant value of the coefficient matrix and set equal to zero.

0

21

112

111

K

Using Sarrus’s rule see that 20)32(3 KKK . You may also use

row/column operation to find two zeros and find your result from reduced system.

Example: Use Cramer’s Rule to explain why the system 382

14

yx

yx has no solution.

Section: Matrices and Matrix Operations

Matrix: A matrix is simply a rectangular array of numbers considered as an entity. A

matrix with m rows and n columns in the array is written as follows:

mnmm

n

n

aaa

aaa

aaa

A

21

22221

11211

and a column vector looks like

1

21

11

mb

b

b

b

which is of order 1m

14

Example 1. Given125

73A ,

43

25B and

105

65C . Find the following

results:

a) CBA 2 b) CBA 5.0

Example 2. Solve for x and y

Example 3.

4 10 1 3 2 5 0

2 3 2 3 0 1 3

6 9 0 1 5 4 7

A is a matrix 3 by 2, B is 3 by 3, C is 3 by 2

4 5 10 0 9 10

2 ( 1) 3 3 3 0

6 4 9 7 10 16

4 5 10 0

2 ( 1)

A B C

A B Not Possible

A C

A C

1 10

3 3 1 6

6 4 9 7 2 2

Example 4.

4 10 1 3 2 5 0

2 3 2 3 0 1 3

6 9 0 1 5 4 7

61 2 2 5

1 1 3 48 7 1 6

0

A B C

D E F G

Order of above matrices:

A is a matrix 3 by 2, B is 3 by 3, C is 3 by 2, D is 3 by 1, E is 1 by 3, F and G are 2 by 2

matrices.

3 2 2 2 2 6 0 0

1 0 1 4 6 5 2 5

x y

x x y

15

Find the following if possible

1. -B, 2. 3A-2C, 3. F+3G, 4. 2B-5C

Answers:

1 3 2

2 3 0

0 1 5

B

4 10 5 0 2 30

3A-2C 3 2 3 2 1 3 4 15

6 9 4 7 10 13

1 2 2 5 5 173 3

8 7 1 6 11 25F G 2B-5C is not possible

Group work: now try

Consider 2

3,

35

21,

6

2

s

tC

r

pB

y

xA Find 2A + 2C, 2A+B

Example 6. Given the matrix

1

2 2 2 2

3 3 3 3

a b c

A b a c

c a b

. Find the determinant value of 4A .

Answer is zero.

Section: Matrix Multiplication

The multiplication of two matrices A and B is a unique matrix AB which is defined for A

of order nm , while B is of order pn . Note that the order of AB is pm .

Example 1. Given125

73A ,

43

25B and

105

65C . Find the following

results:

a) BC b) AB c) CABC

Example 2. Find x

4 5 4

2 1 5 0 1 1

4 35 3 0

x

x O

4 6 5 35 4 5 1

0

x

x x x O

0151235564

)54(335564

2 xxxx

Oxxxx

Now solve for x (Answer should be 4, -5/4)

16

Example 3. Multiply the matrices

)1(3

3

2)1(62)1(2

)1(26)1(22

0

3

22

62

1

1

a

a

aaaa

aaaa

aa

aa

Example 4. Let 25

34,

235

241,

36

12CBA . Determine the

following or explain why the answer does not exist.

a) Product AB. Answer: 6

6

339

57AB

b) Sum 3A + 2C. Answer: 123

71623 CA

c) Product BC. Answer: Does not exist

d) Sum A + 3B. Answer: Does not exist

Matrix Multiplication:

Matrix Multiplication is possible if number of column of first matrix is equal to number

of row of 2nd

matrix.

Example :

4 10 1 3 2 5 0

2 3 2 3 0 1 3

6 9 0 1 5 4 7

A B C

Find the following:

AB Not possible 3 2 3 3Because Ais and Bis

1(4) 3( 2) 2(6) 1(10) 3( 3) 2(9) 10 19

2(4) 3( 2) 0(6) 2(10) 3( 3) 0(9) 2 11

0(4) 1( 2) 5(6) 0(10) 1( 3) 5(9) 28 42

BA

Now try for BC and CB .

17

Properties of addition and scalar multiplication:

( ) ( )

0

( ) 0

( )

( )

( ) ( )

( )

( )

n n n

A B C A B C

A B B A

A A

A A

A A A

A B A B

AB C A BC

A B C AB BC

A B C AC BC

AI I A A where I is nby nidentitity matrix

Exercise 5. From the given matrices find ( )I A D , where I is the identity matrix.

0.2 0.5 200,

0.3 0.4 400A D

Example 6. Let 1 1

0 1A . Calculate

2 3 4, ,A A A , then look at the pattern and find nA ,

for n, a positive integer. Answer: 1

0 1

nn

A

Section 7.5 Rules for Matrix Multiplication (More examples Page # 282)

Example 1. Suppose P, and Q are n n square matrices such that 2PQ Q P . Prove that

2 6 2( )PQ Q P .

Example 2. For the following matrices verify that AB BA .

3 1 0 3,

6 2 2 0A B

Example 3. 1 4

0 1A , show that

2det(3 ) 3 det( )A A , find also 1det((3 ) )A

Example 4. Prove that if a b

Ac d

, then 2 ( ) ( )A a d A ad bc I , where I is a 2 by

2 identity matrix.

18

Section: The Inverse of a Matrix

For a given non-singular matrix A, the inverse matrix 1B A exists such that

AB BA I , where I is an identity matrix of order same as A or B.

A matrix A is non-singular iff det( ) | | 0A A

To find inverse of a nonsingular matrix using calculator:

Step 1. Input the matrix say A

Step 2. Call matrix A and hit 1x in your calculator then hit MATH and select 1 : Frac

to get the matrix along with determinant value.

Example 1. Find the inverse of a two by two matrix by hand:

a bA

c d then 1 1

d b

d b ad bc ad bcA

c a c aad bc

ad bc ad bc

Now verify that 1 1AA A A I

Example 2. Show that 1 1 1( )AB B A

Solution: We can consider that 1

1 1 1 1 1

1 1 1 1 1 1

1 1 1 1

1 1 1

( )

( ) , multiplying by

( ) , using and multiplying by

( )

( )

AB AB I

A AB AB A I A A

B IB AB B A A A I B

B B AB B A

AB B A

Example 3. Show that 1 1 1 1( )ABC C B A using 1 1 1( )AB B A

Example 4. Solve the following system of equations by matrix inverse:

12

2 2 3

2 6

x y z

x y z

x y z

Solution: We have the following matrix system

1

1 1 1 12 1 1 1

2 1 2 3 , check for det 2 1 2 4

1 2 1 6 1 2 1

27

1 1 1 12 3 1 1 12 41

2 1 2 3 4 0 4 3 64

1 2 1 6 5 1 3 6 45

4

x

y

z

x

y

z

19

Example 5. Show that the following system has no solution using inverse matrix method.

12

2 2 3

2 6

x y z

x y z

x y z

Section: Determinant of order 2 by 2 and Cramer’s Rule

We have discussed about it in the previous lecture. Check your note and also you may go

over your test book. Some more exercise problems are given below:

Exercise 1. Let ,a b c d

A Bb a d c

show that | | | |AB BA

Exercise 2. Let 1/ 2 1/ 2 1 1

,1 0 1 1

A B , calculate 2 2 2 2| | | | , | | | |A A B B

Exercise 3. Let

.2 .6 .2

0 .2 .4

.2 .2 0

A find 1 1| 2 | | ( 2 ) |A I A I

Exercise 4. Show that

5 2 1

2 1 0 (1 )( 6)

1 0 1

a

a a a a

a

Exercise 5. If

5 2 1

2 1 0 0

1 0 1

a

a

a

find all values of a. Answer: 0, 1, 6a

Exercise 6. If 22 3 1

4 1

x xA

x x determine det

dA

dx Answer: 4 12x

More examples:

Determinant:

2 by 2: 2 3

18 15 35 9

3 by 3: 2

2 13 2 3 2

2 3 2 1 ( 6) 2(2 3) 1(4 )2 1 1 2

1 2

aa a

a a a a a aa a

a

Another way of expansion (making two zeros by operation)

20

1 1 3

2 2 3

2

22

2 2 2

2 1 0 2 2 12 12 2 1

2 3 0 4 3 2 1 (1 )4 3 24 3 2

1 2 1 2

(1 )(6 4 ( 3 4)) (1 )(10 ) ( 1)( 10)

R R aRR R R

a a aaa a

a a a aa aa a

a a

a a a a a a a a a a

Now verify the following:

1.

a)

7 17 29

11 19 31 36

13 23 37

b)

265 240 219

240 225 198 0

219 198 181

c)

3 4 0

2 1 3 28

1 4 4

2. Prove the following:

a) 2 2 2( 2 )( )

0

a b ax by

b c bx cy ax bxy cy b ac

ax by bx cy

b)

1

1 0

1

x y z

y z x

z x y

c)

1 2 3

( 3 )

0

a a b a b a

a b

d) 2 2 2

3 3 3

1 1 1

( )( )( )( )a b c a b b c c a ab bc ca

a b c

e) 2 2 2 ( )( )( )( )

a b c

a b c a b b c c a a b c

b c c a a b

f) 2 2 2 2 2 2

3 3 3

1 1 1

( )( )( )( )a bc b ca c ab a b b c c a a b c

a b c

g) 4

b c a a

b c a b abc

c c a b

h)

2

2 3

2

( )

( ) 2 ( )

( )

a b ca bc

ca b c ab abc a b c

bc ab c a

21

i) 32 2( )

2

a b c a b

c b c a b a b c

c a c a b

3. Solve for all x, a, b, or c:

a) 2 2 2

1 1 1

0x a b

x a b

b) 2 2 2 0

a b c

a b c

b c c a a b

c)

1 2 3

0

0

a a b

a b

Answer: a) x a b b) 0a b c c) 0a b

Matrices:

4. Given 25 3

, find2 4

A A

5. Given

1

2, 3 2 4 5 find

3

4

A B BA

6. For a given matrix 5 3

2 4A , the transpose is

5 2

3 4A , first row becomes

first column and second row becomes second column. A matrix A is called

orthogonal (perpendicular) if AA A A I . Show that the matrix

1 2 21

2 2 23

2 2 1

A is orthogonal.

7. Given 9 1 1 5

,4 1 7 15

A B . If 3 5 2A B X O , where O is the zero

matrix, find the matrix X.

8. If 4 5 7 3x y , find x and y

9. Suppose that 2

2 0 1

2 1 3 , and ( ) 5 6

1 1 0

A f x x x find ( )f A

10. Given

4 4 8 4

1 1 2 1

3 3 6 3

x y z , find x, y and z

11. Solve the following system using both Cramer’s rule and matrix inverse:

22

a) 2 3 3

4 11

x y

x y b)

5 6 4 15

7 4 3 19

2 6 46

a b c

a b c

a b c

c)

2 3 110

3 5 2 190

3 120

x y z

x y z

z y x

d)

2 3 12 0

3 4 11 46

5 4 5

x y z

x y z

y z

e) 2 5

3 2 3

x y

x y

12. Given an input matrix

240 750

1200 1500

720 450

1200 1500

A and the demand matrix 210

330D .

Find the output matrix X, such that ( )I A X D

13. Find k so the system has no solution: 2 3 3

11

x y

kx y

14. How do you determine whether the system 2 3 3

1.5 11

x y

x y has infinitely many

solutions or no solution at all?

15. How do you determine whether the system 2 3 3

1.5 1.5

x y

x y has infinitely many

solutions or no solution at all?

16. Find an equilibrium vector for the transition matrix given below:

0.4 0.4 0.2

0.3 0.3 0.4

0.4 0.4 0.2

P

Solution: The equilibrium vector is a row matrix [ ], 1v x y z x y z such

that vP v . Use matrix multiplication and find v.

0.4 0.4 0.2

[ ] 0.3 0.3 0.4 [ ]

0.4 0.4 0.2

0.4 0.3 0.4

0.4 0.3 0.4Solve this system

0.2 0.4 0.2

1

4 4 30.36364 , 0.36364 , 0.27272

11 11 11

x y z x y z

x y z x

x y z y

x y z z

x y z

x y z

The required equilibrium vector is 4 4 3

11 11 11v which is a 1 by 3 matrix.