Linear Algebra - Solved Assignments - Fall 2005 Semester

Solution of Assignment # 1 of MTH501 (Fall2005) Total Marks: 50

Q1.

Consider the system of equations

x  y  2z  a x

z b

2 x  y  3z  c Find the relations between a, b and c so that system is (i) Inconsistent (ii) Consistent.

Marks:10

Solution:

1 1   2

(i) (ii)

The augmented matrix correspond to the above system is 1 2 a 0 1 b  . We will find out the echelon form of this matrix which can be 1 3 c 

obtained by the following steps. 2 a 1 1 1 0 1 b    2 1 3 c  2 a  1 1   0  1  1 b  a  by R2  R1 , R3  2 R1 0  1  1 c  2a  2 a  1 1  0 1 1 a  b  by  R2 , R3  R2 0 0 0 c  a  b  Now as you know that if augmented matrix of a linear system in echelon form have any row of the form [0, 0, 0, …, b] where b is non zero then that system will be inconsistent. So the system of linear equation will be inconsistent if we have c-a-b  0 or c  a  b . Now in the above matrix we have third row as [0, 0, 0, c-a-b] and in order to system be consistent we must have c –a –b =0 which is equivalent to c = a + b.

Q2.

Is x = 0, y = 2, z =-2 is a solution to the following linear system? Justify! 3x  2z  4

x

 4 y  z  5

2 x  3 y  2 z  9 Can a system of linear equations have exactly two solutions? Give reasons for your answer. Marks: 5 + 5 Solution: If x = 0, y = 2, z =-2 is a solution of the given system of linear equations then these values will satisfy each equation of the system, and if we have one equation which didn’t satisfy by these values it will imply that these values are not solution of the given system. So put x = 0, y = 2, z =-2 in the first equation of the system we see that we will get 4 = 4, hence first equation satisfied. But consider the second equation and put the values we will get -10 = -5 which is not true hence the given values are not solution of the system. Any system of linear equation can never have exactly two solutions there are three possibilities for any system of linear equations which are (i) System has no solution. (ii) System has unique solution (iii) System has infinite many solutions.

Q3.

A plastic manufacturer makes two types of plastic: regular and special. Each ton of regular requires 2 hours in plant A and 5 hours in plant B; each ton of special plastic requires 2 hours in plant A and 3 hours in plant B. If plant A is available 8 hours per day and plant B is available 15 hours per day, how many tons of each type of plastic should be made daily so that the plants are fully utilized? Marks: 10

Solution: The data given in the question can be formed into system of linear equations as

Plant A Plant B

Regular-Plastic 2 5

Special-Plastic 2 3

As we have given in the question that Plant A and Plant B are available for 8 and 15 hours per day respectively. Let x1 tons of Regular-Plastic and x2 tons of Special-Plastic be the amount should be manufactured so that plants are fully utilized. Then from the above data we must have the system,

2 x1  2 x2  8 5 x1  3x2  15 2 2 8  Augmented matrix for the above system is   we will find out the 5 3 15 echelon form of that matrix and for getting the echelon form we will apply the row operation on the augmented matrix. 2 2 8  5 3 15   1  5

1 3

4  1 by R1 15 2

1 1 4    by R2  5 R1  0  2  5 5 x1  4   x1  x2  4  2   2 x2  5 5 x2  2

3 2    

Hence 3/2 tons of regular plastic and 5/2 of special plastic will fully utilize the capacity of the plants.

Q4.

(i)

Solution:

1  2 0 2  3  1 If A   1 3 2  0 1 1 row equivalent to A

2 5  find a matrix C in reduced echelon form that is 5  2

1  2 0 2  1  2 0 2   2  3  1 5  0 1  1 1   R2  2 R1   by  R3  R1 A 1 3 2 5  0 5 2 3  R  R1     0 2  0 3 0 0   4 1 1 1  2 0 2  0 1  1 1  R  5 R2  by  3  0 0 7  2   R4  3R2   0 0 3  3  1  2 0 2  0 1  1 1   by int erchanging 3 and 4 row.  0 0 3  3    0 0 7  2  1  2 0 2  0 1  1 1   by 1 R  3 0 0 1  1  3   0 0 7  2  1  2 0 2  0 1  1 1   by R4  7 R3  0 0 1  1    0 0 0  9  Which is the echelon form which is row equivalent to the given matrix A to get the reduced echelon form.

1 0  0  0 1 0  0  0

0 2

4 1  1 1  by R1  2 R2 0 1 1   0 0  9 0 0 2 1 0 0  by R1  2 R3 , R2  R3 0 1 1   0 0  9 1 0 0 2  0 1 0 0   by  1 R  4 0 0 1  1 9   0 0 0 1 1 0 0 0  0 1 0 0   by R3  R4 , R1  2 R4  0 0 1 0    0 0 0 1 Which is the required reduced echelon form of the matrix A.

(ii)

Write three vectors which are linear combination of the vectors 1   3 Marks: 7 + 3 x1    , x2    .  2  1

Solution: Any vector of the form y  c1 x1  c2 x2 where c1 , c2   is said to be a linear combination of the vectors x 1 and x 2 .since there are infinite real numbers so we have infinite many these combinations and we have to only write any three of them. Let x 1 =4 1   3  7  y1  4         for c1  4, c2  1  2   1  7  1   3  13 and x 2 = 1 then we have y2     4      for c1  1, c2  4 2  1  2  1   3  4  y3          for c1  1, c2  1  2   1  1 Q5. 6  2 0 10    5  , b  3  and let W be the set of all linear combinations of Let A   1 8  1  2 3  1 

the columns of A. Is b in W? Justify!

Solution:

 2  0  6     Let x1   1  , x2   8  x3   5  then b will be in W if we can find the  1   2   1  solution of the system of linear equations correspond to the equation c1 x1  c2 x2  c3 x3  b .  2  0  6  10      c1  1   c2  8   c3  5   3   1   2   1  3  2c1  0c2  6c3  10 c1  8c2  c3  3 c1  2c2  c3  3 0 6 10   2  Augmented matrix is  1 8 5 3   1  2 1 3  We will find out the echelon form of that matrix for which we will follow the following steps. 0 6 10   1 0 3 5  2 1  1 8   5 3    1 8 5 3 by R1  2  1  2 1 3   1  2 1 3

0 3 5  1   0 8 8 8  by R2  R1 , R3  R1  0  2  2  2  0 3 5  1 1 1   0 1 1 1  by R2 ,  R3 8 2  0 1 1 1  0 3 5  1   0 1 1 1  by R3  R2  0 0 0 0  Which shows that system of linear equations have infinite solutions and we can find infinite many combinations of the unknowns such that equation (i) satisfied consequently b is in W.

Solution of Assignment # 2 of MTH501 (Fall2005) Total Marks: 40 Q1. Find the general solution of system of linear equation

x1  x2  x3  9 x1  x2

2

in

Parametric Form? Marks: 10 Solution: System of linear equations is

x1  x2  x3  9 x1  x2

2

The augmented matrix is 1 1 9  1 1 1 0 2  Transform the augmented matrix in echelon form by applying row operations 1 1 9 1  R1  R2 0 0 1  7   The augmented matrix is now in triangular form. To interpret it correctly, go back to equation notation. x1 +x 2+x3 = 9 -x 3 = -7 x3 = 7 x1 + x 2 + 7 = 9 x1 + x 2 = 9 – 7 = 2 x1 + x 2 = 2 Now we have x1 + x 2 = 2 x3 = 7 Here x 2 is free variable. x1 = 2 – t x2 = t x3 = 7 Suppose we have two equation and three variables then we must take one variable arbitrary as we take x 2 = arbitrary = t Now the general solution of the system in parametric vector form is

 x1   2  t  x   x2   t   x3  7   2   t   0   t  7   0   2   1  0   t 1  7  0  Q2. (i)

 4  1 6     Show that the vectors v1   7  , v2   5 , v3  3  are linearly dependent  9   3  3  0 by showing that the system c1v1  c2v2  c3v3  O Where O  0  has non 0  trivial solution.

Solution: The augmented matrix correspond to the system of linear equations c1v1  c2v2  c3v3  O is,

1 6 4 -7 5 3    9 -3 3 Swap rows 1 and 3 that is interchange row 1 and 3. -3 3 9 -7 5 3    4 1 6  1 By R1 9 -1/3 1/3 1 -7 5 3    4 1 6  By R2  7 R1 , R3  4 R1

1 0  0

-1/3 8/3 7/3

3 By R2 8 -1/3 1 0 1  0 7/3 1 By R1  R2 3 0 1 0 1  0 0

1/3  16/3  14/3 

1/3  2  14/3 7 R3  R1 3 1 2  0 

Which shows that the there is a free variable and hence set of vectors are linear dependent. (ii) Without calculations show that the following set of vectors are linear dependent. Justify you answer. (i) 0,0,0 , 1,0,1 , 1,1,1 (ii) (iii)

5,7,11 , 1,0,0 , 0,1,1 , 3,4,10  2,8,0 , 3,12,0

Marks: 7 + 3

Solution:

(i) The set  0,0,0 , 1,0,1 , 1,1,1 contain the Null vector namely (0, 0, 0) and any set of vectors containing Null vector is always linearly dependent. (ii) The set 5,7,11 , 1,0,0 ,  0,1,1 , 3,4,10 is linear ly dependent because the vectors in that set are four and the each vector of that set has three entries. In short any set containing more than three vectors in R3 will be linearly dependent. (iii) The set  2,8,0 , 3,12,0 is linear dependent as these vectors are multiple of each other. Q3.

1  1  1  2 Let T : R 2  R 2 be Linear Transformation such that T      , T      1 2  1  3  using the properties of Linear Transformation find  1 T   ? (i)  5

 0 T  ?  4 (iii) Find the matrix of transformation T. (ii)

Marks: 3 + 3 +4 Solution:

 1 1  1 First of all note that we can write    2    3   now as we  5 1  1  are given that T is a linear transformation so we must have from the properties of linear transformation that T  c1v1  c2v2   c1T  v1   c2T  v2 Where c1, c2 are real numbers. (i)

 1 1  1  1   2 T    2T    3T    2    3    5 1  1  2  3  Thus we have  1 8 T    5   5 0  1  1 (ii) Once again note that we can write    2    2   Thus we  4 1  1 0  1  1  1  2 T    2T    2T    2    2    4 1  1  2  3 have 0  6  T    4  2 1 1 (iii) Note that the vectors are linearly independent e1    , e2    so 1  1  1 2 the matrix of transformation will be T (e1 ), T (e2 )   . 2 3 

Q4.

Use Theorem 11 on page # 88(Third Edition) of your book to Justify whether the  x1  1  2  1   x1     3   x2  is one to one or Linear Transformation defined by T  x2    2 4  x3  3 0  2   x3  not? Marks: 10

Solution: Theorem states that a linear transformation will be one to one if the homogeneous system T(x) = 0 has trivial solution only.

 x1  0  1  2  1   x1  0  T  x2   0    2 4 3   x2   0   x3  0  3 0  2   x3  0  1  2  1 0  1  2 Augmented matrix  2 4 3 0   0 8 3 0  2 0  0 6 0 1  2  1 1   0 1 5 / 8 0  by R2 8 0 6 1 0 

1 5 1

0 0  by R2  2 R1 , R3  3R1 0 

0 1  2  1   0 1 5 / 8 0  by R3  6 R2 0 0 7 / 2 0  Which shows that the homogeneous system T(x) = 0 has trivial solution thus the given transformation is one to one.

Solution of Assignment # 3 of MTH501 (Fall2005) Total Marks: 80 Q1. Determine whether the matrices are conformable for matrix multiplication AB 0 2 3  2 3   5  , B  1 4  Justify! and BA where A  1 4 0 1 0 1  3  Without obtaining the complete matrix AB find a32 element of AB. Marks: 10

Solution:

Since order of A is 3  3 and order of B is 3  2 also we know that AB is possible if number of columns of A must be equal to the number of rows of B. From the order of A and B it clear that number of columns of A = number of rows of B so matrix multiplication AB is possible. For the matrix multiplication BA number of columns of B must be equal to the number of rows of A but from the order of A and B it is clear that number of columns of B is 2 and number of rows of A are 3, hence BA is not possible. Now we have to find out the a32 element of AB without computing the complete matrix AB, for this note that a32 element will be obtained by multiplying the third row of A with                     the second column of B. So we have      .   a32    0  3  1 4  3  1   7  Q2. x  2 y  3z  9 Consider the following system of linear equations 2 x  y  z  8 find the inverse

3x z 3 of Coefficient matrix for that system and use that inverse to get the solution of linear system. Hint: Use the formula x  A1b to get the solution. Marks: 10

Solution:

1 Coefficient matrix  2 3

1 2 3  2 -1 1  3 0 -1

1 0 0

0 1 0

3 -1 1 0 -1 0  1 2 3 1 0   0 -5 -5 -2 1  0 -3 -7 -3 2

0 1 0

0 0  by R2  2 R1 , R3  3R1 1 

  1 2 3 1 0 0   2 -1 1  0 1 1 0  by  R2 , R3  3R2   5 5 2   -21 -3 0 0 -4 1 5 5   Similarly apply the row operations to make the identity matrix in place of A and when you get the identity matrix in place of A then the matrix which replaces identity matrix will be the inverse of A in the end you will get the result. 0 0.0500 0.1000 0.2500  1 0 0 1 0 0.2500 -0.5000 0.2500   0 0 1 0.1500 0.3000 -0.2500  0.0500 0.1000 0.2500    Also you check by multiplying A and B  0.2500 -0.5000 0.2500  you will get the 0.1500 0.3000 -0.2500  result as identity matrix I. Since we are given the hint that solution of the system is  x  0.0500 0.1000 0.2500  9   2  1 x  A b   y   0.2500 -0.5000 0.2500  8   -1  z  0.1500 0.3000 -0.2500  3   3  Q3.

Compute the determinant by using as less operations as possible 6 3 2 4 0

9 0 4 1 0 8 5 6 7 1 . 3 4 Solution:

0 2

0 3

0 0 2 0

Expanding the determinant by fourth row as we have only one non zero entry in this row. So we get

6

3

2

4 0

3 2 9 0 4 1 0 0 4 4 1 8 5 6 7 1   1 3 5 6 3 0 0 0 0 2 3 4 2 3 2 0  3 2 4 3 2 3 4    3  1 1 0  4 1  3 0  4  2 3 2  2 3  3 3  8  3  2  2  16    9

4

0

1

0

7 2

1 0

Expanding by 4th column

4 1 2

Marks: 10

Q4.

2 1  1  3 3 . Find LU factorization of the matrix A   2  3  10 2  Marks: 10

Solution:

 U   

First of all we will zero out the elements below the a11 so we have, 1 2 1 0 0  1   0  1 1  , multiplier  2 and 3 L   2 1 0  0  4 5  3 * 1 

Now zero out the element in U below a22 2 1  1  1   U   0  1 1 , multiplier  4 L   2  0  3 0 1

0 1 4

0 0  1 

So the required LU are, 1 2 1   1 0 0   U  0  1 1  , L   2 1 0  0 0 1   3 4 1  Q5.

3  2 0 2 4   0 0 0 and V  0 1 2  0  0 4 1 0 0 Find UV using block matrix multiplication. 1 2 0 3 4 0 Let U   0 0 5  0 0 3

0

0 0 1 2 4

0 0  2   3 1 

Marks: 10 Solution: 1 2 0 3 4 0 U=  0 0 5  0 0 3

1 Here A2x2 =  3 3 2 

0 1 4

2 0 0 O = 2x3 0 0 4  

0 0 O 3x2 =  0  0  0 Then U and V becomes O23   A22 C22 U  and V    B23  O32 O22

 2 0 O2x2 =   4 0

 A22C22  O23O32 UV   O22C22  O32 B23 2  3 1 A22C22   4   2 3 7  17 5 B 23 D32   3

3 2  0  0 0

2 4 0 0 0

0 0 0 0  1 2  2  3 4 1 0 0 0  5 1 4  O = B = 2x2 2x3 0 0  3 4 1 C2x2 = 0     0 2  1   0  D3x2 =  2  3 0   4 1  0 0  and V = 2  1

0

O22  D32  A22O22  O23 D32   O22O22  B23 D32 

2  4 

6 10  1 4

 1  7 6 7 17 10 UV   0 0  0 0

1 2  2 1   4

2  3  1

9   5 0 0 1 7

0 0  9  5 

is the required result

Q6. Find the solution of the following system of equations using Jacobi’s iterative 83x  11y  4 z  95 method for the first five iterations. 7 x  52 y  13z  104 Taking initial solution as

3x  8 y  29 z  71

(0, 0, 0)t . Marks: 10 Solution: 95 11 4  y z 83 83 83 104 7 13 y  x z 52 52 52 71 3 8 z  x y 29 29 29 x

 x (1)   1.1446    Since we are given initial solution as (0, 0, 0)t so we have  y (1)    2.0000   z (1)   2.4483     

x (2)  1.1446  0.1325 y (1)  0.0482 z (1)   Second approximation is y (2)  2.0  0.1346 x (1)  0.25 z (1)  (2) (1) (1)  z  2.4483  0.1035 x  0.2759 y   x (2)   0.9976   (2)     y    1.2339   z (2)   1.7424     

Variables y

Itera t io n number r

x

z

1 2

1.1446 0.9976

2.0000 1.2339

2.4483 1.7424

3 4

1.0651 1.0517

1.4301 1.3555

2.0046 1.9435

5

1.0587

1.3726

1.9655

Q7. Find the area of the image of a parallelogram whose vertices are given by (0, 0), (3, 0) (3, 4) (0, 4) in R2 under the transformation T: R2  R2 defined by  x   y T     . Marks: 10  y  2 x  Solution: Area of the parallelogram with the given vertices=

0

4

3

0



3

0

3

4



0

4

3

4

 12

 0 1 Matrix of transformation is A    A  2 0  2 Area of the image of that parallelogram=Area of the parallelogram with the given vertices x determinant of Matrix of transformation of T. Area of the image of that parallelogram=-24 but area is always positive so we have, Area of the image of that parallelogram==24 2 1   2 4    2 2  Q8. Check whether the vector  2  is in the Null space of the matrix  8  3   1  4 1  Marks: 10 Solution: 2  2 4 1    2 2  and b =  2  if Ab = 0 then the vector b will be in Let A =  8  1  4  3  1  Null space of the matrix A otherwise it will not. 2  1   2  8  6  0   2 4  2 2   2    8  4  6    2  which is not null vector so vector b is not Ab =  8  1  4 1  3  1  8  3   4  in the Null space of the matrix A.

Solution of Assignment # 4 of MTH501 (Fall2005) Total Marks: 40 Q1. Find bases for the Column space and Null space of the following matrix A= 1 265 0         . Also find the dimension for Column Space and       Null space. Marks: 10 Solution: To find the bases for Column space, rows reduce the given matrix 1 2650        to an echelon form. A=       

1 0  0  0 1 0  0  0

2 1 1

6 7 4

5 8 2

3

7

9

2 1

6 7

5 8

1 3

4 7

2 9

1 0  0  0

2 1 0 0

6 7  11  28

5 8  10  33

1 0  0  0

2 1

6 7

5 8

0 0

1  28

1 0  0  0

2

6

1 0 0

7 1 0

8 10 /11  83 /11

1 0  0  0

2 1 0

6 7 1

5 8 10 /11

0

0

1

1 0 A B 0  0

10 /11  33 5

0 3   R1  R2 ,  R1  R3 ,  R1  R4 1   1 0  3   R2 1   1 0   3   R2  R3 , 3R2  R 4 2   10  0   3  1 R3  2 /11 11  10  0   3  28 R3  R4  2 /11   54 /11 0   3  11 R4  2 /11 83   54 / 83 

2

6

5

1

7

8

0

1

10 /11

0

0

1

  3   2 /11   54 / 83  0

For column space, observe from B that the pivots are in columns 1, 2, 3, 4. Hence the columns 1, 2, 3, 4 of A (not B) forms a basis for Col A

 1  2 6  5             1 3 1 3  Basis for Col A =    ,   ,   ,     1  1  2  3    1 1   1  4  Dim Col A = 4

For Null space, we need the reduced echelon form. Further row operations on B yields 1 0 A BC  0  0

0

0

0

1

0

0

0

1

0

0

0

1

 44 / 83  55 / 83  34 / 83    54 / 83 

The equation Ax = 0 is equivalent to Cx = 0, that is

1 0  0  0

0 1 0 0

0 0 1 0

0 0 0 1

x1  44 / 83 x5  0 x2  55 / 83 x5  0 x3  34 / 83 x5  0 x4  54 / 83 x5  0 Take x5  t free var iable  x1   44 / 83 t   44 / 83  x       2  55 / 83 t  55 / 83   x3    34 / 83 t   t  34 / 83        x4  54 / 83 t  54 / 83   x  t  1   5

 x1   44 / 83   x2  55 / 83     x3   0 34 / 83     x  54 / 83   4  x   5

 44 / 83  55 / 83    Basis for Null A =  34 / 83   54 / 83  1  Dim Null A = 1 Q2. Find the vector x if Coordinate matrix of x is  x B

 1    2  with respect to the bases  5 

 1   0  1     B   0  ,  1  , 0   .  3   1  0         

Marks: 10 Solution:

1   0 1        Let b1  0  , b2   1  , b3  0  then by definition of coordinate matrix of 3   1  0  vector with we have x   1 b1  2b2  5b3 thus we have 1   0 1   1  0  5   4      x  1 0   2  1   5 0   0  2  0    2  3   1  0   3  2  0   5

Q3.

 3a  6b  c       6a  2b  2c   : a, b, c    Find the bases and dimension for the subspace H     9a  5b  3c     3a  b  c   Marks: 10 Solution:

 3  6  1   6  2   2  v1    , v2    , v3    then H = Span{v1,v2,v3}  9   5  3        3   1  1 6  1  3 6  1  3  6  2  2   0  14   R2  2 R1 0    by  R3  3R1  9 5 3   0 23 0  R  R1     1  0 7 0  4  3 1 6  1  1 R2  3  0   14 1 0   by  R3  23R2  0 0 0  R  7 R2   0  4  0 0   3  6  6  2    , v    are linearly independent and thus form This shows that the vectors v1   9  2  5       3   1 bases for the given subspace.  6  1   2   2    , v    also form basis for the set H, dimension of the space is 2. Similarly v2   5 3  3      1  1 Q4. Let D  d1 , d2 , d3 and B  b1 , b2 , b3 be bases for a vector space V, and suppose b1  2d1  d 2  d3 , b2  3d 2  d3 , b3  3d1  2d3 (a) Find the change of coordinate matrix from B to D. (b) Find  xD for x  3b1  2b2  b3 . Solution:  2 0  3 P   1 3 0   DB  1 1 2 

 x D

 2 0   1 3  1 1

 3  3  3  0   2    9  2   1  3  Marks: 5 + 5

Solution of Assignment # 5 of MTH501 (Fall2005) Total Marks: 40 Q1. Determine whether the signals (-2)k, (3)k , (7)k are linearly independent or not.

Solution:  1  2   4

1  1 1 1  1   3 7 0 5 9    0 9 49   0 5 45  0 The Casorati matrix is invertible for k = 0 so are linearly independent. 1

1 5

1  9  36 

0 whether the signals (-2)k, (3)k , (7)k

Q2.

2 1  0 Find Characteristic Polynomial for the matrix 1  4  4 Eigenvalues and eigenvectors of the same matrix. 2 1  1 1 0 1    4  4 5 1 

2

1

1



1

4

4

5

    1 (5   )  4   2  (5   )  4   1 4  4      1  5   2  4   2  (5   )  4   1 4  4 

Eigenvalues are 1, 2, 3.

2 1  1  A  1 0 1   4  4 5 For Eigen value t = 1 We have to solve the homogenous system (A – tI) x = 0 (A – I) x = 0

1  1  , and then find the 5

 1 2  0  1  4  4  0 1   4 The

 1  1 0 0    1   0 1 0   x  0 5 0 0 1   2 1  x1  0  1 1   x2   0  4 4   x3  0  augrmented matrix is

 0 2 1 1 1 1   4 4 4 2 x2  x3  0

0  0 0   1 0  1

2

1

1

1

1

1

0  0 0   1 0  0

2

1

1

1

0

0

x1  x2  x3  0 Take x3  u

free var iable

 x1   u / 2   1/ 2   x   u / 2   u 1/ 2   2      x3  u  1   1/ 2  1/ 2  be the eigen vector corresponding to eigen value t  1   1  For Eigen value t = 2 We have to solve the homogenous system (A – tI) x = 0 (A –2 I) x = 0

0 0  0 

 1 2  0  1  4  4 

 1  2 0 0   1    0 2 0   x  0 5  0 0 2    1 2 1  x1  0   1 2 1   x   0    2    4 4 3   x3  0  The augrmented matrix is 0   1  1 2 1  1 2 1 0   0   4 4 3 0   4  x1  2 x2  x3  0

2 0 4

1 0 3

0   1 0   0 0  0

2 0 4

4 x2  x3  0 Take x3  s  free var iable x1  2 x2  x3 x2  x3 / 4 x3  s  x1    s / 2   1/ 2   x    s / 4   s 1/ 4   2      x3   s  1   1/ 2  1/ 4  be the eigen vector corresponding to eigen value t  2   1  For Eigen value t = 3 We have to solve the homogenous system (A – tI) x = 0 (A – 3I) x = 0

1 0 1

0 0  0 

 1 2  0  1  4  4 

 1  3 0 0   1   0 3 0   x  0 5 0 0 3   2 2 1  x1  0   1 3 1   x   0    2    4 4 2   x3  0  The augrmented matrix is 3  2 2 1 0  1  1 3 1   0    2 2   4 4 2 0   4 4 3 1 0  1 1   0 4 1 0   0 0 4 1 0  0

1 1 2 3 4 0

0  1 0   0 0  0 1 1 0

3

1

4 8

1 2

0 0  0 

0 0  0 

Q3. (i)

Without any calculation find the Eigenvalues of the matrix 0 0 1 0 5 2 0 0   . Justify your answer. 3 0 3 0   2 1  0 1 Solution: 0 0 1 0 5 2 0 0   is a lower triangular matrix 3 0 3 0   2 1  0 1 So by using the theorem The Eigen values of a triangular matrix ( lower or upper ) are the entries on its main diagonal. 1 5 So  3  0

0

0

2

0

0

3

1

2

0 0  have Eigen values 1, 2, 3, -1. 0  1 

(ii)

7  4  3    Is  3  an eigenvector of  4  5  1  2 4 corresponding eigenvalue.

9 1  ? If yes then find the 4 

7  4  3    By definition x =  3  is an eigenvector of A =  4  5  1  2 4 is a scalar 7  3  First of all note that  4  5  2 4

9 1  . If Ax =  x where  4 

9   4  0 1   3   0  4   1 0 

 4 Thus  3  is the eigenvector of the matrix and the corresponding eigenvalue is 0.  1 Q4. 2 3 For the matrix A   find an invertible matrix P such that A=P-1 DP where D is  0 4  diagonal matrix. Solution: Eigenvalues are 2, 4 0 3  1  0 2   x2  0 Eigenvector is 0      Eigenvectors are 3  2 3  3    0 0   x1  2 x2 Eigenvector is  2    1 

 1 P  0 2 D 0 Take Now we 2 AP   0  1 PD    0

3 2   1 0  4  will check AP  PD  3  1 4   0 3 2 2   0 1 

3 2 2    0 1 

6 4 

0   2    4    