applied maths i u i solution

7/31/2019 Applied Maths i u i Solution

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APPLIED MATHS I

Sol u t i on :

CSVTU Ex a m i n a t i onPapers

Depa r tm en t of Ma th em a ticsDIMAT

MATRICE


2/42

UNIT I (I Semester)

De p a r t m e n t o f M a t h e m a t ics, D IM AT Page 2

APPLIED MATHS I

Time Allowed : Three hours

Maximum Marks : 80Minimum Pass Marks : 28

Note : Solve any two parts from each question. All questions carry equal marks.

UNIT I

MATRICES

SOLUTION (Nov-Dec-2005)

(a) Find the rank of matrix A by reducing to normal form:8 1 3 6

0 3 2 2

8 1 3 4

A

.

Ans:

8 1 3 6

0 3 2 2

8 1 3 4

A

1 2c c

1 8 3 6

3 0 2 2

1 8 3 4

A

2 2 1

3 3 1

3R R R

R R R

1 8 3 6

0 24 7 160 0 0 10

A

2 2 1

3 3 1

4 4 3

8

32

C C C

C C CC C C

1 0 0 0

0 24 7 2

0 0 0 10

A

73 3 224

14 4 212

C C C

C C C


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UNIT I (I Semester)


1 0 0 0

0 24 0 0

0 0 0 10

A

12 224

14 410

C C

C C

1 0 0 0

0 1 0 0

0 0 0 1

A

3 4c c

3

1 0 0 00

0 1 0 00 0

0 0 1 0

IA

Which is the normal form.

So, Rank of the matrix is 3.

(b) Test for consistency and solve:2 3

2 3 2 5

3 5 5 2

3 9 4

x y z

x y z

x y z

x y z

Ans: The Given equation can be written in the matrix form as AX B Where

1 2 1 3

2 3 2 5, ,

3 5 5 2

3 9 1 4

x

A B X y

z

The Auguemented matrix [ | ]C A B

1 2 1 3

2 3 2 5

3 5 5 2

3 9 1 4

2 2 1

3 3 1

4 4 1

23

3

R R RR R R

R R R

1 2 1 3

0 1 0 1

0 11 2 7

0 3 4 5

3 3 2

4 4 2

11

3

R R R

R R R

1 2 1 3

0 1 0 1

0 0 2 4

0 0 4 8

4 4 32R R R

1 2 1 3

0 1 0 1

0 0 2 4

0 0 0 0

This system is consistent and has unique solution.

2 3, 1,2 4x y z y z 1, 1, 2x y z (Ans)

(c) Find the characteristics roots and characteristic vectors of the matrix:8 6 2

6 7 4

2 4 3


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UNIT I (I Semester)


8 5 6 2 0

6 7 5 4 0

2 4 3 5 0

x

y

z

3 6 2 0

6 2 4 0

2 4 2 0

x

y

z

3 6 2 0, 6 2 4 0,2 4 2 0x y z x y z x y z

So, Eigen vector is

2

2

1

SOLUTION (Apr-May-2006)

(a)Test for consistency and solve:2 3 14

3 2 11

2 3 11

x y z

x y z

x y z

Ans:

2 3 14

3 2 11

2 3 11

x y z

x y z

x y z

1 2 3 14

3 1 2 11

2 3 1 11

x

y

z

Its Auguemented matix is

2 2 1

3 3 1

1 2 3 143

3 1 2 112

2 3 1 11

R R R

R R R

13 3 25

1 2 3 14

0 5 7 31

0 1 5 17

R R R

18 545 5

1 2 3 14

0 5 7 31

0 0

Here Rank of coefficient matrix and rank of auguemented matrix is3.So, system of equations is consistent and has unique solution.


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UNIT I (I Semester)


18 545 5

2 3 14, 5 7 31,x y z y z z

3, 2, 1z y x (Ans)

(b)Find the eigen values and eigen vectors of the matrix3 1 4

0 2 6

0 0 5

A

.

Ans: Given that

3 1 4

0 2 6

0 0 5

A

Its characteristics equation is given by 0A I

3 1 4

0 2 6 0

0 0 5

(3 )(2 )(5 ) 0 2,3,5 For 2 , 0A I X

1 1 4

0 0 6 0

0 0 3

x

y

z

4 0,6 0,3 0x y z z z 0,z x y

So, eigen vector is

1

1

0

For 5 ,

0A I X

2 1 4

0 3 6 0

0 0 0

x

y

z

2 4 0, 3 6 0x y z x z 2 , 0x z y

So, eigen vector is

2

0

1

For 3 , 0A I X

0 1 4

0 1 6 0

0 0 2

x

y

z

4 0, 6 0, 2 0y z y z z 0, 0z y


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UNIT I (I Semester)


So, eigen vector is

1

0

0

(c) If matrix1 0 2

0 2 1

2 0 3

A

, then verify Cayley Hamilton theorem. Hence find

1A .

Ans: Given that

1 0 2

0 2 1

2 0 3

A


1 0 2

0 2 1 02 0 3

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

2 3 22 8 2 4 0

3 26 7 2 0

3 26 7 2 0 -----------(1)

Now,

1 0 2

0 2 1

2 0 3

A

2

1 0 2 1 0 2 5 0 8

0 2 1 0 2 1 2 4 5

2 0 3 2 0 3 8 0 13

A

3

1 0 2 5 0 8 21 0 34

0 2 1 2 4 5 12 8 23

2 0 3 8 0 13 34 0 55

A

Now,3 2

6 7 2A A A I 21 0 34 5 0 8 1 0 2 1 0 0

12 8 23 6 2 4 5 7 0 2 1 2 0 1 0

34 0 55 8 0 13 2 0 3 0 0 1


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UNIT I (I Semester)


21 30 7 2 0 0 0 0 34 48 14 0

12 12 0 0 8 24 14 2 23 30 7 0

34 48 14 0 0 0 0 0 55 78 21 2

0 0 0

0 0 0 0

0 0 0

Which verifies Cayley Hamilton Theorem.

Now, 3 26 7 2 0A A A I

1 3 26 7 2 0A A A A I

2 16 7 2 0A A I A

1 21

6 72

A A A I

1

5 0 8 1 0 2 1 0 01

2 4 5 6 0 2 1 7 0 1 028 0 13 2 0 3 0 0 1

A

1

5 6 7 0 0 0 8 12 01

2 0 0 4 12 7 5 6 02

8 12 0 0 0 0 13 18 7

A

1 1 12 2

6 0 4 3 0 21

2 1 1 12

4 0 2 2 0 1

A

(Ans).


(a)Find the characteristics equation of the matrix2 1 1

1 2 1

1 2 2

A

and verify

that it is satisfied by A.

Ans:

2 1 1

1 2 11 1 2

A


2 1 1

1 2 1 0

1 1 2


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UNIT I (I Semester)


2(2 )((2 ) 1) 1( 2 1) 1(1 2 ) 0 2 3

8 12 6 2 1 1 0 3 2

6 9 4 0 3 26 9 4 0 ----------- (1)

Now,2

4 1 1 2 2 1 2 1 2 6 5 5

2 2 1 1 4 1 1 2 2 5 6 52 1 2 1 2 2 1 1 4 5 5 6

A

And 312 5 5 10 6 5 10 6 5 22 21 21

6 10 5 5 12 5 5 10 6 21 22 21

6 5 10 5 6 10 5 5 12 21 21 22

A

Now, 3 26 9 4A A A I

22 21 21 6 5 5 2 1 1 1 0 0

21 22 21 6 5 6 5 9 1 2 1 4 0 1 0

21 21 22 5 5 6 1 1 2 0 0 1

22 36 18 4 21 30 9 0 21 30 9 0

21 30 9 0 22 36 18 4 21 30 9 0

21 30 9 0 2130 9 0 22 36 18 4

0 0 0

0 0 0

0 0 0

Hence matrix A satisfies characteristics equation.

(b)Investigate the value of and so that the equations:2 3 5 9

7 3 2 8

2 3

x y z

x y z

x y z

have (i) no solution (ii) unique solution (iii) infinite solution.

Ans:

2 3 5 9

7 3 2 8

2 3

x y z

x y z

x y z

2 3 5 97 3 2 8

2 3

xy

z

Its Auguemented matrix is7

2 2 12

2 2 3

2 3 5 9

7 3 2 8

2 3

R R R

R R R


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UNIT I (I Semester)


15 39 472 2 2

2 3 5 9

0

0 0 5 9

From the Auguemented matrix:

(i) System has no solution if 5 0 9 0and 5 9and

(ii) System has unique solution if 5 0 and any value of 5 and any value of .

(iii) System has many solution if 5 0 9 0and 5 9and

(c)Find the rank of the following matrix:1 1 1

A b c c a a b

bc ca ab

.

Ans:2 2 1

3 3 1

1 1 1C C C

A b c c a a bC C C

bc ca ab

2 2 1

3 3 1

1 0 0( )

( )( ) ( )

R R b c Rb c a b a c

R R bc Rbc c a b b a c

2 2

3 3

1 0 0/ ( )

0/ ( )

0 ( ) ( )

C C a ba b a c

C C a cc a b b a c

3 3 2

1 0 0

0 1 1

0

R R cR

c b

3 3 2

1 0 0

0 1 1

0 0

C C C

b c

3 3

1 0 0

0 1 0 / ( )0 0

C C b c

b c

3

1 0 0

0 1 0 [ ]

0 0 1

I

, Which is the normal form

So, rank of the matrix = 3.


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UNIT I (I Semester)


SOLUTION (May-June-2007)

(a)Find the inverse of the following matrix:1 2 3

2 4 5

3 5 6

A

.

Ans:

1 2 3

2 4 5

3 5 6

A

2 2 1

3 3 1

1 2 3 1 0 0

22 4 5 0 1 03

3 5 6 0 0 1

R R RA IR R R

2 2 3

1 2 3 1 0 0

0 0 1 2 1 0

0 1 3 3 0 1

R R R

3 3 2

1 2 3 1 0 0

0 1 2 1 1 1

0 1 3 3 0 1

R R R

2 2 3

1 1 3

1 2 3 1 0 02

0 1 2 1 1 13

0 0 1 2 1 0

R R R

R R R

1 1 2

1 2 0 5 3 0

0 1 0 3 3 12

0 0 1 2 1 0R R R

1

1 0 0 1 3 2

0 1 0 3 3 1

0 0 1 2 1 0

I A

So,1

1 3 2

3 3 1

2 1 0

A

(Ans).


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UNIT I (I Semester)


(b)Prove that the following equations are consistent and solve:2 3

3 2 1

2 2 3 2

x y z

x y z

x y z

Ans:

2 3

3 2 1

2 2 3 2

x y z

x y z

x y z

1 2 1 3

3 1 2 12 2 3 2

x

yz

Its auguemented matrix is

2 2 1

3 3 1

1 2 1 33

3 1 2 12

2 2 3 2

R R R

R R R

63 3 27

1 2 1 3

0 7 5 8

0 6 5 4

R R R

5 207 7

1 2 1 3

0 7 5 8

0 0

From auguemented matrix, the rank of coefficient matrix is 3 and rank of auguemented

matrix is also 3 and order of matrix is 3.So, System is consistent and has unique solution.

5 207 7

2 3, 7 5 8,x y z y z z

4, 4, 1z y x (Ans)

(c)Find the characteristics roots and corresponding characteristics vectors ofthe matrix:

8 6 2

6 7 4

2 4 3

A

.

Ans: Given that

8 6 2

6 7 4

2 4 3

A


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UNIT I (I Semester)



8 6 2

6 7 4 0

2 4 3

(8 ){(7 )(3 ) 16} 6{( 6)(3 ) 8} 2{24 2(7 )} 0 3 218 45 0 ( 3)( 15) 0

0, 3, 15

Now for 0 , [ ] 0A I X

8 0 6 2 0

6 7 0 4 0

2 4 3 0 0

x

y

z

8 6 2 0

6 7 4 0

2 4 3 0

x

y

z

8 6 2 0, 6 7 4 0, 2 4 3 0x y z x y z x y z

So, Eigen vector is

1

2

2

Now for 3 , [ ] 0A I X

8 3 6 2 0

6 7 3 4 0

2 4 3 3 0

x

y

z

5 6 2 0

6 4 4 0

2 4 0 0

x

y

z

5 6 2 0, 6 4 4 0,2 4 0x y z x y z x y

So, Eigen vector is

2

1

2

Now for 5

,[ ] 0A I X

8 5 6 2 0

6 7 5 4 0

2 4 3 5 0

x

y

z


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UNIT I (I Semester)


3 6 2 0

6 2 4 0

2 4 2 0

x

y

z

3 6 2 0, 6 2 4 0,2 4 2 0x y z x y z x y z

So, Eigen vector is

2

2

1


(a)State and explain the application of Cayley Hamilton theorem.Ans: Cayley-Hamilton Theorem: - Every square matrix satisfies its own

characteristic equation.Let A is a square matrix of order n.

Its characteristic equation is:1 2

1 2( 1) ........ 0n n n n

nA I k k k

Then 1 21 2

( 1) ........ 0n n n nn

A k A k A k

(b)For what values of value of and do the system of equations:6

2 3 10

2

x y z

x y z

x y z

have (i) no solution (ii) unique solution (iii) infinite solution..

Ans:

6

2 3 10

2

x y z

x y z

x y z

1 1 1 6

1 2 3 10

1 2

x

y

z

Its Auguemented matrix is

2 2 1

3 3 1

1 1 1 6

1 2 3 10

1 2

R R R

R R R

3 3 2

1 1 1 6

0 1 2 4

0 1 1 6R R R


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UNIT I (I Semester)


1 1 1 6

0 1 2 4

0 0 3 10

(i) System has no solution if 3 0 10 0and 3 10and

(ii) System has unique solution if 3 0 and any value of 3 and any value of .

(iii) System has many solution if 3 0 10 0and 3 10and

(c)Find the eigen values and corresponding eigen vectors of the matrix:2 2 3

2 1 61 2 0

A

.

Ans:

2 2 3

2 1 6

1 2 0

A


2 2 3

2 1 6 0

1 2 0

2( 2 )( 12) 2( 2 6) 3( 4 1 ) 0 2 2 3

2 2 24 12 4 12 12 3 3 0 3 2

21 45 0 3 2

21 45 0 5, 3, 3

For 5 , [ ] 0A I X

7 2 3

2 4 6 0

1 2 5

x

y

z

7 2 3 0,2 4 6 0, 2 5 0x y z x y z x y z

, 2x z y z

So, eigen vector is

1

2

1


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UNIT I (I Semester)


For 3 , [ ] 0A I X

1 2 3

2 4 6 0

1 2 3

x

y

z

2 3 0, 2 4 6 0, 2 3 0x y z x y z x y z 2 3 0x y z

So, eigen vector is

1 3 2

1 , 0 , 1

1 1 0

.

(d)Find the characteristics equation of the matrix:2 1 1

0 1 0

1 1 2

A

and hence, find the matrix represented by

8 7 6 5 4 3 25 7 3 5 8 2A A A A A A A A I .

Ans:

2 1 1

0 1 0

1 1 2

A


2 1 1

0 1 0 0

1 1 2

2(2 )( 3 2 0) 1(0 0) 1(0 1 ) 0 2 3 2

2 6 4 3 2 0 1 0 3 25 7 3 0 3 2

5 7 3 0 ----------(1)By Cayley Hamilton theorem A satisfies (1).

So, 3 25 7 3 0A A A I

Now, 8 7 6 5 4 3 25 7 3 5 8 2A A A A A A A A I

3 2 5 25 7 3A A A I A A A A I 2

0 A A I 2A A I

2 1 1 2 1 1 2 1 1 1 0 00 1 0 0 1 0 0 1 0 0 1 0

1 1 2 1 1 2 1 1 2 0 0 1

5 4 4 2 1 1 1 0 0

0 1 0 0 1 0 0 1 0

4 4 5 1 1 2 0 0 1

8 5 5

0 3 0

5 5 8

(Ans)


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UNIT I (I Semester)


SOLUTION(May-June-2008)

(a)Define rank of a matrix.Ans: Rank of a matrix is the largest order of any non-vanishing minor of a matrix.

(b)For what value of parameter will the following equation fail to have uniquesolution:

3 1

2 2

2 1

x y z

x y z

x y z

with the equation have any solution for these values of .

Ans:

3 1

2 2

2 1

x y z

x y z

x y z

3 1 1

2 1 1 2

1 2 1

x

y

z


2 2 1

3 3 1

3 1 13 2

2 1 1 23

1 2 1

R R R

R R R

3 3 2

3 1 1

0 5 3 2 55 7

0 7 4 4R R R

3 1 1

0 5 3 2 5

0 0 6 21 55

-------------- (1)

From the auguemented matrix

The system is fail to have unique solution is 72

6 21 0 .

For 72

, auguemented matrix at (1) becomes


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UNIT I (I Semester)


72

3 1 1

0 5 10 5

0 0 0 55

which has no solution. (Ans)

(c)Find the characteristics roots and corresponding characteristics vectors ofthe matrix:

8 6 2

6 7 4

2 4 3

A

.

Ans: Given that

8 6 2

6 7 4

2 4 3

A


8 6 2

6 7 4 0

2 4 3

(8 ){(7 )(3 ) 16} 6{( 6)(3 ) 8} 2{24 2(7 )} 0

3 218 45 0

( 3)( 15) 0

0, 3, 15

Now for 0

8 0 6 2 0

6 7 0 4 0

2 4 3 0 0

x

y

z

8 6 2 0

6 7 4 0

2 4 3 0

x

y

z

8 6 2 0, 6 7 4 0, 2 4 3 0x y z x y z x y z

So, Eigen vector is

1

2

2

Now for 3


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UNIT I (I Semester)


8 3 6 2 0

6 7 3 4 0

2 4 3 3 0

x

y

z

5 6 2 0

6 4 4 0

2 4 0 0

x

y

z

5 6 2 0, 6 4 4 0,2 4 0x y z x y z x y

So, Eigen vector is

2

1

2

Now for 5

8 5 6 2 0

6 7 5 4 0

2 4 3 5 0

x

y

z

3 6 2 0

6 2 4 0

2 4 2 0

x

y

z

3 6 2 0, 6 2 4 0,2 4 2 0x y z x y z x y z

So, Eigen vector is

2

2

1

(d)Find the characteristic equation of the matrix:2 1 1

1 2 1

1 1 2

A

. And verify that it is satisfied by A. Hence find 1A .

Ans:

2 1 1

1 2 1

1 1 2

A


2 1 11 2 1 0

1 1 2

2(2 )((2 ) 1) 1( 2 1) 1(1 2 ) 0 2 3

8 12 6 2 1 1 0 3 2

6 9 4 0


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UNIT I (I Semester)


3 26 9 4 0 ----------- (1)

Now, 24 1 1 2 2 1 2 1 2 6 5 5

2 2 1 1 4 1 1 2 2 5 6 5

2 1 2 1 2 2 1 1 4 5 5 6

A

And 312 5 5 10 6 5 10 6 5 22 21 21

6 10 5 5 12 5 5 10 6 21 22 21

6 5 10 5 6 10 5 5 12 21 21 22

A

Now, 3 26 9 4A A A I

22 21 21 6 5 5 2 1 1 1 0 0

21 22 21 6 5 6 5 9 1 2 1 4 0 1 0

21 21 22 5 5 6 1 1 2 0 0 1

22 36 18 4 21 30 9 0 21 30 9 0

21 30 9 0 22 36 18 4 21 30 9 021 30 9 0 2130 9 0 22 36 18 4

0 0 0

0 0 0

0 0 0

Which verifies Cayley Hamilton Theorem.

So, 3 26 9 4 0A A A I

1 3 26 9 4 0A A A A I 2 1

6 9 4 0A A I A

1 21 6 94

A A A I

1

6 5 5 2 1 1 1 0 01

5 6 5 6 1 2 1 9 0 1 04

5 5 6 1 1 2 0 0 1

A

1

6 12 9 5 6 0 5 6 01

5 6 0 6 12 9 5 6 04

5 6 0 5 6 0 6 12 9

A

13 1 111 3 1

41 1 3

A

3 1 14 4 4

1 31 14 4 4

31 14 4 4

A

(Ans).


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UNIT I (I Semester)


SOLUTION(Dec-Jan-2008-2009)

(a)The sum of eigen values of the matrix1 1 3

1 5 1

3 1 1

is ____.

Ans: 7.

(b)Reduce to normal form and find the rank of the matrix:1 2 3 4

2 1 4 3

3 0 5 10

.

Ans:

1 2 3 4

2 1 4 3

3 0 5 10

2 2 1

3 3 1

1 2 3 42

0 3 2 53

0 6 4 22

R R R

R R R

2 2 1

3 3 1

4 4 1

21 0 0 030 3 2 5

0 6 4 22 4

C C CC C C

C C C

3 3 2

1 0 0 02

0 3 2 5

0 0 0 12

R R R

4 4 3 2

1 0 0 0

0 3 2 0

0 0 0 12

C C C C

2 2 3

1 0 0 02

0 1 2 0

0 0 0 12

C C C

3 3 2

1 0 0 02

0 1 0 0

0 0 0 12

C C C

4 41 0 0 0

/ 120 1 0 0

0 0 0 1

C C

3 4

1 0 0 0

0 1 0 0

0 0 1 0

C C


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UNIT I (I Semester)


3 0I which is the normal form.. So Rank of the givenmatrix is 3.

(c)Test the consistency and solve the following system of equations:2 3 4 11

5 7 15

3 11 13 25

x y z

x y z

x y z


2 3 4 11

1 5 7 , , 15

3 11 13 25

x

A X y B

z

Its Auguemented matix is C A B

2 3 4 111 5 7 15

3 11 13 25

1 2

1 5 7 15

2 3 4 11

3 11 13 25

R R

2 2 1

3 3 1

1 5 7 152

0 7 10 193

0 4 8 20

R R R

R R R

3 3 2

1 5 7 15

0 7 10 19 7 4

0 0 16 64

R R R

As here rank of coefficient matrix = 3, rank of auguemented matrix = 3 andorder of coefficient matrix is 3. So system is consistent and has unique solution.

16 64, 7 10 19, 5 7 15

2, 3, 4

z y z x y z

x y z

(d)Find the eigen values and corresponding eigen vectors of the matrix:1 6 40 4 2

0 6 3

A

.

Ans: Here

1 6 4

0 4 2

0 6 3

A


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UNIT I (I Semester)



1 6 4

0 4 2 0

0 6 3

(1 ){(4 )( 3 ) 12} 0 2(1 ){ 12 4 3 12} 0

2(1 ){ } 0 (1 )( 1) 0 0, 1, 1

Now for 0

1 0 6 4 0

0 4 0 2 0

0 6 3 0 0

x

y

z

1 6 4 0

0 4 2 0

0 6 3 0

x

y

z

6 4 0, 4 2 0, 6 3 0x y z y z y z 6 4 0, 2 0x y z y z

2 , 2z y x y

So for 0 eigen vector is

2

1

2

Now for 1

1 1 6 4 0

0 4 1 2 00 6 3 1 0

x

yz

0 6 4 0

0 3 2 0

0 6 4 0

x

y

z

6 4 0,3 2 0, 6 4 0y z y z y z

3 2y z

So, for 1 eigen vector are

1 1

2 , 2

3 3

Solution (Apr-May-2009)

(a)Select the correct answer.By Applying elementary transformation to matrix its rank:


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UNIT I (I Semester)


(i) Increases.(ii) Decreases.(iii) Does not change.(iv) Multiplied by a constant.

Ans: Does not change.

(b)Reduce to normal form and find the rank of the matrix:

2 1 3 6

3 3 1 2

1 1 1 2

A

.

Ans:

2 1 3 6

3 3 1 2

1 1 1 2

A

1 2c c

1 2 3 6

3 3 1 2

1 1 1 2

A

2 2 1

3 3 1

3R R R

R R R

1 2 3 6

0 9 8 16

0 1 4 8

A

2 2 1

3 3 1

4 4 3

2

3

6

C C C

C C C

C C C

1 0 0 0

0 9 8 16

0 1 4 8

A

2 2 3C C C

1 0 0 0

0 1 8 16

0 3 4 8

A

3 3 13R R R

1 0 0 0

0 1 8 16

0 0 28 56

A

3 3 2

4 4 2

8

16

C C C

C C C

1 0 0 0

0 1 0 0

0 0 28 56

A

3 3/ 28R R

1 0 0 0

0 1 0 0

0 0 1 2

A

4 4 32C C C

3

1 0 0 00

0 1 0 00 0

0 0 1 0

IA

Which is the normal form.


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UNIT I (I Semester)


So, Rank of the matrix is 3.

(c)Test the consistency and solve the following system of equations:2 3

2 3 2 5

3 5 5 2

3 9 4

x y z

x y z

x y z

x y z


1 2 1 3

2 3 2 5, ,

3 5 5 2

3 9 1 4

x

A B X y

z

The Auguemented matrix [ | ]C A B

1 2 1 3

2 3 2 5

3 5 5 2

3 9 1 4

2 2 1

3 3 1

4 4 1

2

3

3

R R R

R R R

R R R

1 2 1 3

0 1 0 1

0 11 2 7

0 3 4 5

3 3 2

4 4 2

11

3

R R R

R R R

1 2 1 3

0 1 0 1

0 0 2 4

0 0 4 8

4 4 32R R R

1 2 1 3

0 1 0 1

0 0 2 4

0 0 0 0


2 3, 1,2 4x y z y z

1, 1, 2x y z (Ans)


(a) Define rank of the matrix.


26/42

UNIT I (I Semester)


Ans: -Rank of a matrix is the largest order of any non-vanishing minor of a matrix.

(b) For what value of k the equations2

1

2 4

4 10

x y z

x y z k

x y z k

have solutions and solve

them completely in each case?

Ans: -

2

1

2 4

4 10

x y z

x y z k

x y z k

2

1 1 1 1

2 1 4

4 1 10

x

y k

z k

Its Auguemented matrix is2 2 1

3 3 12

1 1 1 12

2 1 44

4 1 10

R R Rk

R R Rk

3 3 22

1 1 1 1

0 1 2 2

30 3 6 4

k

R R Rk

2

1 1 1 1

0 1 2 2

0 0 0 3 2

k

k k

Here rank of coefficient matrix = 2.For consistency rank of auguemented matrix = 2.

For rank = 2 in auguemented matrix2 3 2 0 1,2k k k

For k = 1

Auguemented matrix is

1 1 1 1

0 1 2 1

0 0 0 0

It has infinite many solution.

1, 2 1x y z y z

So solution is 3 , 1 2 ,x k y k z k

For k = 2

Auguemented matrix is

1 1 1 1

0 1 2 0

0 0 0 0

It has infinite many solution.1, 2 0x y z y z

2 , 1 3y z x z

So solution is 1 3 , 2 ,x k y k z k

(c) Find the eigen values and eigen vectors of the matrix1 1 3

1 5 1

3 1 1

.


27/42

UNIT I (I Semester)


Ans: - Let

1 1 3

1 5 1

3 1 1

A

Its characteristics equation is

1 1 3

0 1 5 1 0

3 1 1

A I

(1 )((5 )(1 ) 1) 1(1 3) 3(1 15 3 ) 0 2(1 )(4 6 ) 1( 2) 3( 14 3 ) 0

2 2 34 6 4 6 2 42 9 0 3 27 36 0

3 27 36 0 is the characteristics equation.2,3,6

For 2

2 0A I X

3 1 3 3 3 0

1 7 1 0 7 0

3 1 3 3 3 0

x x y z

y x y z

z x y z

3 3 0, 7 0

, 0

x y z x y z

x z y

Eigen vectors corresponding to 2 is

1

0

1

For 3

3 0A I X

0 1 3 3 01 2 1 0 2 0

3 1 0 3 0

x y zy x y z

z x y

, 3x z y z


1

3

1

For 6

6 0A I X

5 1 3 5 3 0

1 1 1 0 0

3 1 5 3 5 0

x x y z

y x y z

z x y z

, 2x z y z


1

2

1


28/42

UNIT I (I Semester)


(d) Find the characteristics equation of the matrix1 2 2

1 1 1

1 3 1

A

. Hence find

1A .

Ans: -

1 2 21 1 1

1 3 1

A


1 2 2

0 1 1 1 0

1 3 1

A I

2(1 )((1 ) 3) 2(1 1) 2(3 1 ) 0 2(1 )( 2 2) 2( ) 2(2 ) 0

2 3 2

2 2 2 2 2 4 2 0

3 2

3 6 0 3 23 6 0 is the characteristics equation.

By Cayley Hamilton theorem3 2

3 6 0A A I 2 13 6 0A A A

1 21

36

A A A

2

1

1 2 2 1 0 01

1 1 1 3 0 1 06

1 3 1 0 0 1

A

11 2 2 3 6 613 6 0 3 3 3

65 8 2 3 9 3

A

1

1 3 2 6 2 61

3 3 6 3 0 36

5 3 8 9 2 3

A

1

2 8 41

0 3 36

2 1 1

A

82 46 6 6

1 3 36 6

2 1 16 6 6

0A

(Ans).

Solution (May-June-2010)

(a) If

23

41A , find the value of IAAAA 71574

234 .


29/42

UNIT I (I Semester)


Ans: - Characteristics equation is 0 IA

0103122323

4122

By Cayley Hamilton theorem A satisfies its own characteristics equation.

So, 01032

IAA Now, IAAAIAAIAAAA 75))(103(71574

22234

1715

2012

10

017

23

41575071574

234IAIAAAA

(Ans).

(b) By elementary row transformation, find the inverse of

014

112

425

A .

Ans: - 311100

010

001

014

112

425

RRRIA

133

122

4

2

100

010

101

014

112

431

RRR

RRRIA

322 25504

212

101

16130

950

431

RRRIA

233 13504

052

101

16130

1310

431

RRRIA

)185/(56530

052

101

18500

1310

431

33

RRIA

311322

4

13

37

1

37

13

37

6 052

101

1001310

431

RRR

RRR

IA


30/42

UNIT I (I Semester)


211 3

37

1

37

13

37

637

13

37

16

37

437

33

37

52

37

13

100

010

031

RRRIA

]|[

37

1

37

13

37

637

13

37

16

37

437

6

37

4

37

1

100

010

0011

AIIA

37

1

37

13

37

637

13

37

16

37

437

6

37

4

37

1

1

A (Ans)

(c) Find the eigen values and eigen vectors of the matrix

021

612

322

A .

Ans: - Characteristics equation is 0 IA

0

21

612

322

0)14(3)62(2)12)(2(2

039124122422322

0452123

3,3,5

For 5 , 0 XIA

0

0

0

521

642

327

z

y

x

052,0642,0327 zyxzyxzyx

052,032,0327 zyxzyxzyx

So, Eigen vectors corresponding to 5 are

1

2

1

,

1

2

1

2

k

k

k

etc.


31/42

UNIT I (I Semester)


For 3 , 0 XIA

0

0

0

321

642

321

z

y

x



So, Eigen vectors corresponding to 3 are

1

2

1

,

1

1

123

2

1

12

k

k

kk

etc.

(d) Show that the equation532;2310;42;1233 zyxzyyxzyx are consistent and

hence obtain the solutions for zyx ,,

Ans: - 532;2310;42;1233 zyxzyyxzyx

5

2

4

1

132

3100

021

233

z

y

x

Its Auguemented matrix is144

122

23

3

5

2

4

1

132

3100

021

233

RRR

RRR

244

233

153

103

13

2

11

1

7150

3100

230

233

RRR

RRR

344 5129

204

116

11

1

5100

2900

230

233

RRR

0

116

11

1

000

2900

230

233

---------------------- (1)

Here rank of matrix A and rank of Auguemented matrix are equal to 3 and thatrank is same as number of variables. So It has unique solution.


32/42

UNIT I (I Semester)


From the Last matrix 11629,1123,1233 zzyzyx

2,1,4 xyz 4,1,2 zyx (Ans).


1.(a) State Cayley Hamilton theorem.

Ans: Cayley-Hamilton Theorem: - Every square matrix satisfies its own

characteristic equation.Let A is a square matrix of order n.

Its characteristic equation is:1 2

1 2( 1) ........ 0n n n n

nA I k k k

Then1 2

1 2( 1) ........ 0n n n n

nA k A k A k

(b) Show that if 5 , the system of equations:356,232,343 zyxzyxzyx have unique solution. If

5 , show that the equations are consistent. Determine the solution in eachcase.

(c) Find the characteristics roots and characteristics vectors of the matrix:.

342

476

268

.

Ans: Given that

8 6 2

6 7 4

2 4 3

A


8 6 2

6 7 4 0

2 4 3

(8 ){(7 )(3 ) 16} 6{( 6)(3 ) 8} 2{24 2(7 )} 0

3 218 45 0 ( 3)( 15) 0 0, 3, 15

Now for 0

8 0 6 2 0

6 7 0 4 0

2 4 3 0 0

x

y

z

8 6 2 0

6 7 4 0

2 4 3 0

x

y

z

8 6 2 0, 6 7 4 0,2 4 3 0x y z x y z x y z


33/42

UNIT I (I Semester)


So, Eigen vector is

1

2

2

Now for 3

8 3 6 2 06 7 3 4 0

2 4 3 3 0

xy

z

5 6 2 06 4 4 0

2 4 0 0

xy

z

5 6 2 0, 6 4 4 0, 2 4 0x y z x y z x y So, Eigen vector

is

2

1

2

Now for, 15 [ ] 0A I X

0

0

0

1242

486

267

0

0

0

15342

41576

26158

z

y

x

z

y

x

01242,0486,0267 zyxzyxzyx

So, Eigen vector is

2

2

1

(d) Find the characteristics equation of the matrix:2 1 1

0 1 0

1 1 2

A


8 7 6 5 4 3 25 7 3 5 8 2A A A A A A A A I .

Ans:

2 1 1

0 1 0

1 1 2

A


2 1 1

0 1 0 0

1 1 2

2(2 )( 3 2 0) 1(0 0) 1(0 1 ) 0 2 3 2

2 6 4 3 2 0 1 0 3 25 7 3 0 3 2

5 7 3 0 ----------(1)


34/42

UNIT I (I Semester)


By Cayley Hamilton theorem A satisfies (1).

So, 3 25 7 3 0A A A I

Now, 8 7 6 5 4 3 25 7 3 5 8 2A A A A A A A A I

3 2 5 25 7 3A A A I A A A A I 2

0 A A I

2

A A I 2 1 1 2 1 1 2 1 1 1 0 0

0 1 0 0 1 0 0 1 0 0 1 0

1 1 2 1 1 2 1 1 2 0 0 1

5 4 4 2 1 1 1 0 0

0 1 0 0 1 0 0 1 0

4 4 5 1 1 2 0 0 1

8 5 5

0 3 0

5 5 8

(Ans)

Solution (April-May-2011)

1.a) State and explain the application of Cayley Hamilton theorem.

Ans: Cayley-Hamilton Theorem: - Every square matrix satisfies its own

characteristic equation.Let A is a square matrix of order n.Its characteristic equation is:

1 2

1 2( 1) ........ 0n n n n

nA I k k k

Then1 2

1 2( 1) ........ 0n n n n

nA k A k A k

b) Find the characteristics roots and characteristic vectors of the matrix:6 2 2

2 3 1

2 1 3

Let

6 2 2

2 3 1

2 1 3

A


35/42

UNIT I (I Semester)


The characteristic equation of the matrix A is

|

| |

|

[ ]

|

| =

6

2 2

2 3

1

2 1 3 =0

= 1 2 + 3 6 3 2 = 0 (-8) ( 2 ) = 0 =8 ,2 ,2

Therefore the eigen value of A are 2,2,3(i) Eigen vector corresponding to eigen value =8 is given by

| 8 | = 0 6 8 2 22 3 8 1

2 1 3 8 =

00

0

2

2 2

2 5 12 1 5 = 0

00

Operating , ,2 2 20 3 3

0 3 3 =

00

0

2

2 2

0 3 30 0 0 = 000

The coefficient matrix of these equation is of rank 22 2 2 = 0..(1)3 3 = 0.(2)From equation (2) ,we get = Suppose

= 1 , then

=

1

Then substituting the values in equation (1) ,we get\2 + 2 2 = 0..(1)=>2=0=0Hence =0, = 1 , = 1

(ii) Eigen vector corresponding to eigen value =2 is given by


36/42

UNIT I (I Semester)


| 2 | = 0 6 2 2 22 3 2 1

2

1 3

2

=

000

4 2 22 1 12 1 1

= 000

2 1 14

2 2

2 1 1

=

00

0

2 1 10 0 00 1 0

= 000

+ 2, + ince the rank of coefficient matrix is 1 ,so the following equation will havwe3-1=2 ,linearly independent solution .

2 + = 0 ..(1)From equation (1) ,we have If=1, = 0 ,then from equation (1)

= And if,=1, = 0 ,then from equation (1)

= 2Hence the eigen vector corresponding to eigen value =2 are 102 ,

12

0


37/42

UNIT I (I Semester)


c) Find the rank of the following matrix:1 1 1

A b c c a a b

bc ca ab

.

Ans:

2 2 1

3 3 1

1 1 1 C C CA b c c a a b

C C Cbc ca ab

2 2 1

3 3 1

1 0 0( )

( )( ) ( )

R R b c Rb c a b a c

R R bc Rbc c a b b a c

2 2

3 3

1 0 0/ ( )

0/ ( )

0 ( ) ( )

C C a ba b a c

C C a c

c a b b a c

3 3 2

1 0 0

0 1 1

0

R R cR

c b

3 3 2

1 0 0

0 1 1

0 0

C C C

b c

3 3

1 0 0

0 1 0 / ( )

0 0

C C b c

b c

3

1 0 0

0 1 0 [ ]

0 0 1

I

, Which is the normal form

So, rank of the matrix = 3.

d) For what values of value of and do the system of equations:6

2 3 102

x y z

x y zx y z

have (i) no solution (ii) unique solution (iii) infinite solution.

Ans:

6

2 3 10

2

x y z

x y z

x y z


38/42

UNIT I (I Semester)


1 1 1 6

1 2 3 10

1 2

x

y

z


2 2 1

3 3 1

1 1 1 61 2 3 10

1 2

R R R

R R R

3 3 2

1 1 1 6

0 1 2 4

0 1 1 6R R R

1 1 1 6

0 1 2 4

0 0 3 10

(iv) System has no solution if 3 0 10 0and 3 10and

(v) System has unique solution if 3 0 and any value of 3 and any value of .

(vi) System has many solution if 3 0 10 0and 3 10and


(a)Product of the Eigen Value of a matrix 1 1 3

1 5 1

3 1 1

...= - 36

(b)Test for consistency and solve:2 3

2 3 2 5

3 5 5 2

3 9 4

x y z

x y z

x y z

x y z

Ans.

The Given equation can be written in the matrix form as AX B where


39/42

UNIT I (I Semester)


1 2 1 3

2 3 2 5, ,

3 5 5 2

3 9 1 4

x

A B X y

z

The Auguemented matrix [ | ]C A B 1 2 1 3

2 3 2 5

3 5 5 2

3 9 1 4

2 2 1

3 3 1

4 4 1

2

3

3

R R R

R R R

R R R

1 2 1 3

0 1 0 1

0 11 2 7

0 3 4 5

3 3 2

4 4 2

11

3

R R R

R R R

1 2 1 3

0 1 0 1

0 0 2 4

0 0 4 8

4 4 32R R R

1 2 1 3

0 1 0 1

0 0 2 4

0 0 0 0


2 3, 1,2 4x y z y z

1, 1, 2x y z (Ans)

(c)Find the eigen values and eigen vectors of the matrix:.

3 6 6

0 2 0

3 12 6

.

(d)Find the characteristics equation of the matrix:2 1 1

0 1 0

1 1 2

A


5 4 3 24 2 5 4A A A A A I .


40/42

UNIT I (I Semester)


Ans:

2 1 1

0 1 0

1 1 2

A


2 1 1

0 1 0 0

1 1 2

2(2 )( 3 2 0) 1(0 0) 1(0 1 ) 0 2 3 2

2 6 4 3 2 0 1 0 3 25 7 3 0 3 2

5 7 3 0 ----------(1)By Cayley Hamilton theorem A satisfies (1).

So, 3 25 7 3 0A A A I

Now, 8 7 6 5 4 3 25 7 3 5 8 2A A A A A A A A I

3 2 5 25 7 3A A A I A A A A I 2

0 A A I 2A A I

2 1 1 2 1 1 2 1 1 1 0 0

0 1 0 0 1 0 0 1 0 0 1 0

1 1 2 1 1 2 1 1 2 0 0 1

5 4 4 2 1 1 1 0 0

0 1 0 0 1 0 0 1 0

4 4 5 1 1 2 0 0 1

8 5 5

0 3 0

5 5 8

(Ans)

Solution (April-May-2012)(a)Define the rank of Matrix.

Ans: Rank of a matrix is the largest order of any non-vanishing minor of a matrix

(b)Find the eigen values and eigen vectors of the matrix1 1 3

1 5 1

3 1 1

Ans: - Let


41/42

UNIT I (I Semester)


1 1 3

1 5 1

3 1 1

A


1 1 3

0 1 5 1 0

3 1 1

A I

(1 )((5 )(1 ) 1) 1(1 3) 3(1 15 3 ) 0 2(1 )(4 6 ) 1( 2) 3( 14 3 ) 0

2 2 34 6 4 6 2 42 9 0 3 27 36 0

3 27 36 0 is the characteristics equation.2,3,6

For 2

2 0A I X

3 1 3 3 3 0

1 7 1 0 7 0

3 1 3 3 3 0

x x y z

y x y z

z x y z

3 3 0, 7 0

, 0

x y z x y z

x z y


1

0

1

For 3

3 0A I X 0 1 3 3 0

1 2 1 0 2 0

3 1 0 3 0

x y z

y x y z

z x y

, 3x z y z


1

3

1

For 6

6 0A I X 5 1 3 5 3 0

1 1 1 0 0

3 1 5 3 5 0

x x y z

y x y z

z x y z

, 2x z y z


42/42

UNIT I (I Semester)


1

2

1

(c)Reduce following matrix to normal form and find its rank2 3 4 5

3 4 5 6

4 5 6 7

9 10 11 12

(d)Find the characteristics equation of the matrix2 1 1

1 2 1

1 1 2

A

Ans:- The characteristic equation of the matrix A is

| |=02 1 11 2 1

1 1 2 =0 2 [ 2 ] + 1 + 2 + 1 + 1 2 + 2 [ 4 4 + + 1 ] + [ 3 ] + [ 1 ] =0 2

[

4

+

+ 5 ] + 3

+

1=0

2 8 + 1 0 + 4 5 + 2 = 0 + 6 1 3 + 1 2 = 0 6 + 1 3 1 2 = 0 (-3)( 3 + 4 ) = 0 (-3)(+1)(-4)=0 =3,-1,4

applied maths i u i solution

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