Upload
mrityunjoy-dutta
View
218
Download
1
Embed Size (px)
Citation preview
7/31/2019 Applied Maths i u i Solution
1/42
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
7/31/2019 Applied Maths i u i Solution
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
7/31/2019 Applied Maths i u i Solution
3/42
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 3
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
7/31/2019 Applied Maths i u i Solution
4/42
7/31/2019 Applied Maths i u i Solution
5/42
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 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
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.
7/31/2019 Applied Maths i u i Solution
6/42
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 6
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
7/31/2019 Applied Maths i u i Solution
7/42
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 7
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
Its characteristics equation is given by 0A I
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
7/31/2019 Applied Maths i u i Solution
8/42
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 8
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).
SOLUTION (Nov-Dec-2006)
(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
Its characteristics equation is given by 0A I
2 1 1
1 2 1 0
1 1 2
7/31/2019 Applied Maths i u i Solution
9/42
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 9
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
7/31/2019 Applied Maths i u i Solution
10/42
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 10
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.
7/31/2019 Applied Maths i u i Solution
11/42
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 11
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).
7/31/2019 Applied Maths i u i Solution
12/42
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 12
(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
7/31/2019 Applied Maths i u i Solution
13/42
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 13
Its characteristics equation is given by 0A I
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
7/31/2019 Applied Maths i u i Solution
14/42
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 14
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 (Nov-Dec-2007)
(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
7/31/2019 Applied Maths i u i Solution
15/42
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 15
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
Its characteristics equation is given by 0A I
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
7/31/2019 Applied Maths i u i Solution
16/42
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 16
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
Its characteristics equation is given by 0A I
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)
7/31/2019 Applied Maths i u i Solution
17/42
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 17
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
Its Auguemented matrix is
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
7/31/2019 Applied Maths i u i Solution
18/42
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 18
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
Its characteristics equation is given by 0A I
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
7/31/2019 Applied Maths i u i Solution
19/42
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 19
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
Its characteristics equation is given by 0A I
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
7/31/2019 Applied Maths i u i Solution
20/42
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 20
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).
7/31/2019 Applied Maths i u i Solution
21/42
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 21
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
7/31/2019 Applied Maths i u i Solution
22/42
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 22
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
Ans: The Given equation can be written in the matrix form as AX B Where
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
7/31/2019 Applied Maths i u i Solution
23/42
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 23
Its characteristics equation is given by 0A I
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:
7/31/2019 Applied Maths i u i Solution
24/42
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 24
(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.
7/31/2019 Applied Maths i u i Solution
25/42
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 25
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
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
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
This system is consistent and has unique solution.
2 3, 1,2 4x y z y z
1, 1, 2x y z (Ans)
SOLUTION (Nov-Dec-2009)
(a) Define rank of the matrix.
7/31/2019 Applied Maths i u i Solution
26/42
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 26
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
.
7/31/2019 Applied Maths i u i Solution
27/42
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 27
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
Eigen vectors corresponding to 3 is
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
Eigen vectors corresponding to 6 is
1
2
1
7/31/2019 Applied Maths i u i Solution
28/42
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 28
(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
Its characteristics equation is
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 .
7/31/2019 Applied Maths i u i Solution
29/42
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 29
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
7/31/2019 Applied Maths i u i Solution
30/42
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 30
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.
7/31/2019 Applied Maths i u i Solution
31/42
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 31
For 3 , 0 XIA
0
0
0
321
642
321
z
y
x
032,0642,032 zyxzyxzyx
032,032,032 zyxzyxzyx
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.
7/31/2019 Applied Maths i u i Solution
32/42
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 32
From the Last matrix 11629,1123,1233 zzyzyx
2,1,4 xyz 4,1,2 zyx (Ans).
SOLUTION (Nov-Dec-2010)
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
Its characteristics equation is given by 0A I
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
7/31/2019 Applied Maths i u i Solution
33/42
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 33
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
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
Its characteristics equation is given by 0A I
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)
7/31/2019 Applied Maths i u i Solution
34/42
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 34
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
7/31/2019 Applied Maths i u i Solution
35/42
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 35
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
7/31/2019 Applied Maths i u i Solution
36/42
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 36
| 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
7/31/2019 Applied Maths i u i Solution
37/42
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 37
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
7/31/2019 Applied Maths i u i Solution
38/42
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 38
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 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
SOLUTION (Nov-Dec-2011)
(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
7/31/2019 Applied Maths i u i Solution
39/42
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 39
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
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 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
and hence, find the matrix represented by
5 4 3 24 2 5 4A A A A A I .
7/31/2019 Applied Maths i u i Solution
40/42
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 40
Ans:
2 1 1
0 1 0
1 1 2
A
Its characteristics equation is given by 0A I
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
7/31/2019 Applied Maths i u i Solution
41/42
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 41
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 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
Eigen vectors corresponding to 3 is
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
7/31/2019 Applied Maths i u i Solution
42/42
UNIT I (I Semester)
Eigen vectors corresponding to 6 is
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