Linear Algebra - Solved Assignments - Fall 2004 Semester

LINEAR ALGEBRA SOLVED ASSIGNMENTS SEMESTER FALL-2004

Solution of Assignment # 1 of MTH501 (Fall2004) Q1.

Consider the system of equations

x  y  2z  a x

z b

2 x  y  3z  c Show that the system to be consistent , the constant a, b and c must satisfy c = a + b. Ans wer: The augmented matrix correspond to the above system is 2 a 1 1 1 0 1 b  . We will find out the echelon form of this matrix which can be   2 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. 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 there a value of r so that the x = r, y = 2, z =1 is a solution to the following linear system? If there is a value find it. 3x  2z  4

x

 4 y  z  5

2 x  3 y  2 z  9 Ans wer:

We will try to find out the solution of the above system, so the augmented 0 2 4  3  1  5 and the echelon form of the matrix for the above system is  1  4  2 3 2 9  augmented matrix can be obtained by the following steps. 0 2 4  1  4 1  5  3  1 4   1  5   3 0 2 4  by Re placing R1 by R2   2 3 2 9   2 3 2 9  4 1  5  1   0 12 5 19  by R2  3R1 , R3  2 R1 5 4  1   0 1  0  0   1  0   0  1  0  0 

4

5

1 5 12 4

4

1

1

1 0 4 1 0

5 12 23 12 1 5 12 23

5  19  1 R2 12  12  1  5  19  5 R2  R3 12  83   12  5  19  12 R3 12  83

23 z = 83 z = 83/23 This doesn’t match with the value of z given in the question. So we can’t find the value of r from the given system of linear equations. Note: There are more then one method to solve that question as you can also put the given values x=r y=2 and z=1 in the system of linear equation and then you can conclude the same. But concept used in this question is that “If system of linear equations is consistent then it will have unique solution or infinite many solutions”. This is not possible that a system of linear equations have 2 or 3 or 10 solutions.

Q3.

An oil refinery produces low-sulfur and high-sulfur fuel. Each ton of low-sulfur requires 5 minutes in the blending plant and 4 minutes in the refining plant; each ton of high sulfur fuel requires 4 minutes in the blending plant and 2 minutes in the refining plant. If the blending plant is available for 3 hours and the refining plant is available for 2 hours, how many tons of each type of fuel should be manufactured so that the plants are fully utilized?

Ans wer: The data given in the question can be formed into system of linear equations as Low-sulfur High-sulfur Blending plant 5 4 Refining Plant 4 2 As we are given in the question that we have blending plant and refining plant available for 3 and 2 hours respectively. Let x tons of low-sulfur and y tons of high sulfur be the amount should be manufactured so that plants are fully utilized. Then from the above data we must have the system, 5x + 4y = 180 4x + 2y = 120 Augmented matrix for the above system 5 4 180  1 2 60   4 2 120    4 2 120 by R1  R2     is 2 60  1   0  6 120  So we have x + 2y = 60 and -6y = - 120, thus we have y= 20 tons and x = 20tons are the required manufactured tons of each low-sulfur and high-sulfur so that we can utilize both plants for the given time. Q4. (a) (b)

Find a linear equation in the variables x and y that has the general solution x = 5 + 2t, y= t. 1 5 Show that x=t, y  t  is also the general solution of the equation in 2 2 part (a).

Ans wer: (a) We have to find out linear equation for which the given parametric equations x = 5 + 2t, y= t define the coordinates of any point on that line. And in order to get the required linear equation we will simply find an equation independent of the parameter “t” Suppose x = 5 + 2t ---------- (1)

y = t--------------- (2) Put t=y in eq. (1) we get x= 5 + 2y This is the required linear equation. (b) We have a linear equation

x= 5 + 2y ----------- ( A)

Put x = t in eq.(A) we get t =5 + 2y --------------- (B) From (B) y = ( t – 5)/2 Y = t/2 – 5/2 1 5 y t 2 2 1 5 This shows that x=t, y  t  is also the general solution of the equation in part (a). 2 2

Q5.

5  1   k        Consider x   2  , y   2  and w   2  find the value(s) of k such that the 7  3   k  3 vector w is in the span of x and y.

Ans wer: We have to find the value(s) of k (if possible) under the condition that w is in the span of x and y W is in the span of x and y so we have to check whether x 1 x + x2 y = w have solution? Here x 1 and x 2 are constants. To answer this, row reduce the augmented matrix [ x y w]:

5 2  7 2 5  7 1 5  7 1 0  0 1  0  0 

k  2  k  3

1 2 3

2  k  R12 k  3

2 1 3 1 1 3

1  1 k  R1 2 k  3

 4  5  k   5 R1  R2 ,  7 R1  R3 4  10  k  1 1  5  k  1 1 R2 4  4 4  10  k  1 1 1  5  k  0 1 4 R2  R3 4   0 0  5   The third equation is 0x2 = -5 this is not possible which shows that the system has no solution. The vector equation x 1 x + x 2 y = w has no solution and so w is not in the span of x and y. Thus we can not able to find the value(s) of k. 1

1

Solution of Assignment # 2 of MTH501 (Fall2004) Q1. 11  9 5  12  7  9 4 8 7  3  If the columns of the matrix  span R4 then write  6  11  7 3 9    6 10 5 12   4 1  4 the vector   as linear combination of the columns of the above matrix. 5    7  Solution: We have to find out the values of the unknowns such that we can write the given 12   7  11   9   5  1  1   9   4  8   7  3  4  4           c c c  c      which is vector as c1  6  2  11 3  7  4  3  5  9  5  5                4   6   10   5  12  7  7  correspond to the system of linear equations whose augmented matrix is, 11  9 5 1  12  7  9 4 8 7  3 4    6 11  7 3  9 5    6 10 5 12 7   4

The Augmented Matrix is: 12 7 11 9 5  9 4 8 7 3   6 11 7 3 9   4 6 10 5 12

1 4  5  7  1 7 /12 11/12 9 /12 5 /12 1/12   9 4 8 7 3 4  1  R1 ~  6 11 7 3 9 5  12   6 10 5 12 7  4

1 7 /12 11/12 9 /12 5 /12 0 5 / 4 1/ 4 1/ 4 3/ 4 ~ 0 15 / 2 3 / 2 3 / 2 13 / 2  2 31/ 3 0 11/ 3 19 / 3 1 0 ~ 0  0 1 7 /12 0 1 ~ 0 0  0 0

7 /12 11/12 9 /12

1/12  R2  9 R1 19 / 4  R3  6 R1 11/ 2   R4  4 R1 20 / 3

1/12  1 1/ 5 1/ 5 3 / 5 19 / 5  4  R2 15 / 2 3 / 2 3 / 2 13 / 2 11/ 2  5  11/ 3 19 / 3 2 31/ 3 20 / 3  11/12 9 /12 5 /12 1/12  1/ 5 1/ 5 3 / 5 19 / 5  R3  (15 / 2) R2 0 0 2 34  R4  (11/ 3) R2  28 / 5 41/15 122 /15 109 /15 5 /12

12c1  7c2  11c3  9c4  5c5  1  c3  c4  3c5  19  2c5  34 84c3  41c4  122c5  109 There are four equations and five unknowns so one variable is taken to be arbitrary Let c2 = s On solving these four equations we get c5  17

c4  783 / 25 c3  967 / 25 c1 

1 10637 7047 [1  7 s    85] 12 25 25

As we are able to find the values of the unknowns so we can write the given 12   7  11   9   5  1  1   9   4  8   7  3  4  4           c c c c      vector as c1  6  2  11 3  7  4  3  5  9  5  5                4   6   10   5  12  7  7  Hence the given vector is the linear combination of columns of the given matrix.

2 x1  2 x2  x3

Q2.

Determine whether the Homogeneous system

 x5  0

 x1  x2  2 x3  3 x4  x5  0 x1  x2  2 x3

 x5  0

has

x3  x4  x5  0 non-trivial solution or not. If it has non-trivial solutions then write the solution in parametric form.

Solution: 0 1  2 2 1  1  1 2  3 1  Matrix of coefficients for the above system is  then  1 1  2 0  1   1 1 1  0 0  1 1  2 0 1   0 0 0  3 0  reduced echelon form of the matrix is  thus we have,  0 0 1 0 1   0 1 0  0 0 x1  x2  2 x3  x5  0  x1   x2  2 x3  x5 3 x4  0  x4  0 x3  x5  0  x3   x5 x4  0 Also by putting the value of x3   x5 in first equation we get, x1   x2  2 x5  x5  x1   x2  x5 which shows that x2 and x5 are free variables. So we can take x2  t and x5  s where s and t are arbitrary real numbers, thus the parametric form of the solution is given by the following equation.  x1   t  s   1  1 x     1   0  2  t       x3     s   t  0  s  1 .          x4  0   0  0  x   s   0  1   5

Q3. (a) (b)

Determine whether the given vectors (1, 2, 3, 5), (2, 7, 6, 0), (1, 1, 1, 1) and (3, 5, 7, 9) are linearly independent or not. Justify your answer. For what values of c the vectors (-1, 0, -1), (2, 1, 2) and (1, 1, c) are linearly dependent in R3 .

Solution: Consider the linear combination of the given vectors as

1  2  1 3   0  2 7 1 5   0        c1 c c  c      . Now we will solve that homogeneous 3  2  6  3 1 4  7   0           5  0 1 9   0  system and if we gat the non trivial solution of that system then the given four vectors will be linear dependent otherwise the four vectors will be linear independent. Matrix of coefficients for the above homogeneous system is, 3  1 2 1   1 2 1 3 0 1  1  1   2 7 1 5 3 3     and the echelon form of that matrix is  1 . 3 6 1 7  0 0 1   2     5 0 1 9   50  0 0 0   3  Since the matrix has pivot position in each row hence the homogeneous system has only trivial solution. Thus we can conclude that the given vectors are linearly independent. Also note that the homogenous system in Q2 has non trivial solution because second row of the matrix of coefficients in echelon form has no pivot position.

(b) In order to get the required value or values of c we will solve the homogeneous system of linear equations correspond to the linear combination  1  2 1   0     x1  0   x2 1   x3 1    0 and we will find out the value of c such that this  1   2   c   0 homogeneous system will have non-trivial solution. So the matrix of coefficients  1 2 1   1  2 1   0 1 1  1  . is   and the echelon form of that matrix is  0 1  1 2 c   0 0 c  1 Now if you choose c such that c -1 = 0 then the above system will have a free variable namely x3 and thus has non trivial solution, which implies that for this value of c we have three vectors linearly dependent in R3 thus the required value of c is 1. Q4. (a)

Show that the mapping T : R 2  R5 defined by T(x, y) = (x, x-y, 2x-4y, y, x + y) is a linear transformation.

Solution (a): We will show that the given transformation satisfies the conditions.

(i) (ii)

T (u + v) = T(u) + T(v) Where u = (a, b) and v = (c, d) are the arbitrary two elements of R2 . T(cu) = c T(u)

(i)

T(u + v) =T(a + c, b + d) since we have (u + v) =(a + c, b + d) now by definition of transformation we can write, T(a + c, b + d) = (a + c, a + c – b – d, 2a + 2c -4b – 4d, b + d, a + c + b + d) = (a, a – b, 2a -4b, b, a + b) + (c, c – d, 2c – 4d, d, c + d) = T(a, b) + T(c, d) Because by definition of transformation we have, T(a, b) = (a, a – b, 2a -4b, b, a + b) T(c, d) = (c, c – d, 2c – 4d, d, c + d) Thus we have T(a + c, b + d) = T(a, b) +T(b, d) or T (u + v) = T(u) + T(v) T(cu) = T(ca, cb) = (ca, ca – cb, 2ca -4cb, cb, ca + cb) = c (a, a – b, 2a -4b, b, a + b) =c T(a, b) T(cu) = c T(u) Thus both conditions are satisfied by the given transformation hence the transformation is linear. (b) If we have a linear transformation T:R3  R3 such that 1  1  0   1  0  1          T 0    2  , T 1    0  and T 0   0  then find the matrix of 0  0  0   1  1   0 (ii)

 x1    transformation T and find the image of  x2  .  x3  Solution (b): Since we know that matrix of linear transformation T:Rm  Rn is T (e1 ) T (e2 ) T (e3 ) . . . T (em )

Where e1  (1, 0, 0,..., 0), e2  (0,1, 0,..., 0), e3  (0, 0,1,..., 0)...em  (0, 0, 0,...,1) Thus matrix of the linear transformation is  x1   1  0  0   1 1 1              A  T  0  T 1  T  0     2 0 0  and hence the image of  x2  is given   0   x3   0  1    0  1 0   x1  1 1 1   x1   x1  x2  x3         2 x1 by T  x2    2 0 0   x2    .  x3  0  1 0   x3    x2 

Find a linear transformation which projects every element of R3 into xz 10  plane and then find the image of  15  . 100  (c) Three friends play a game in which there are always two winners and one loser. They have the understanding that the loser gives each winner an amount equal to what the winner already has. After three games, each just lost once and each has 24$. With how much money did each begin? Solution (a): Since we have to find out a linear transformation which projects every element of R3 into xz-plane that is the linear transformation when acts on an element of R3 made the y component of that vector 0. Thus we can say that  x1   x1      formula for that linear transformation is T  x2    0  and thus the image of  x3   x3  Q5.

(a)

 10   10   15  under such a linear transformation is  0  .     100  100  Solution (b): Let x, y and z be the starting money of the three friends A, B and C respectively. Now we will consider the three games separately. 1st game: Suppose that in the first game A having x money is loser then by the given conditions in the question after 1st game A, B and C will have x - y - z, 2y and 2z dollars respectively. 2nd game: Suppose that in the second game B having 2y money is loser then by the given conditions in the question after 2nd game A, B and C will have 2x -2 y -2z, 2y - 2z – x + y + z =3y – z – x and 4z dollars respectively. 3rd game: Suppose that in the third game C having 4z money is loser then by the given conditions in the question after 3rd game A, B and C will have 4x -4 y -4z, 6y – 2z - 2x and 4z -2x + 2y +2z -3y + z+ x=7z – x- y dollars respectively. Now as we are given in the question that after three games each A, B and C has 24$. So we have, 4x -4 y -4z = 24 6y – 2z - 2x= 24 7z – x - y = 24

24   4 4 4  2 24  and echelon form of that Augmented matrix for that system is  2 6  1  1 7 24  1 1  7  24   3 and this gives us z = 12$, y = 21$ and x = 39$. matrix is 0 0  2 0 0 1 12  You can also check that solution satisfies all the given conditions in the question.

Solution of Assignment # 3 of MTH501 (Fall2004) 1  6 1  4  and Q1. Given that the matrix A is symmetric if A   1 14  1 4  2  1 1   x  2 y  3z  A   1 2x  3y  2z 4  then find the values of x, y and z (if  1 4 3 x  y  z  t

possible). Solution: Since the given matrix is symmetric so we have A=At so we can write 1 1   6 1 1  x  2 y  3z  1   2x  3y  2z 4    1 14 4  As we know that two   1 4 3x  y  z   1 4  2  matrices will be equal if the order and corresponding entries of the matrices are equal. So by using this fact we can write, x  2 y  3z  6

2 x  3 y  2 z  14 3x  y  z  2 2 3 6 1 2  3 2 14  and echelon form of the this So augmented matrix for the system is  3 1  1  2  1 2 3    matrix is 0 1 2   x  1, y  2, z  3 , thus we get, 0 0 1

1 1  4  9  1 266   1 4 the required values.

  6    1 4   3  2  3  1 1

1 14 4

1 4  . Hence x  1, y  2, z  3 are  2 

4 1 9 0 1  2     Q2. If A  3 4 4  B  0  1 6  then find  2 8 0  1 8 0  (i) The element a32 of the matrix AB without calculating the matrix AB.

(ii)

The element a11 of the matrix BA without calculating the matrix BA

Solution (i): Since the a32 element of the product AB will be the sum of the corresponding elements of the third row of A and second column of B. Thus we          4                      have a32            1                         2 8 0    8    2  4  8  ( 1)  0  8   0   Solution (ii): Since the a11 element of the product BA will be the sum of the corresponding elements of the first row of B with the first column of A. Thus we have 4 1  9    2 a11          3     18  12  2   22          2    Q3.

 3 0 1 1   Consider the Matrices A  0 0 4  B  0 1 1 0  1

2 1 4 1  then show 8 0 

that ( AB)1  B 1 A1 . Solution: First of all we will find out the product AB then we will find out the 14 3  4  32 0  and inverse of AB. So AB   4 1 6 2 

 0.0833  0.8000   0.5333 0.0417 0.1000  --------------- (i)  AB    0.0667  0.0667  0.0833 0.6000  Now we have  0.3333  0.0833 0   0.8000  0.8000 0.2000  1 1   0.1000 0.1000   A   0.3333 0.0833 1  and  B   0.1000   0.4000 0 0.2500 0  0.6000  0.4000   0.0833  0.8000  0.5333  1 1 B A   0.0667 0.0417 0.1000  ----------------------- (ii)  0.0667  0.0833 0.6000  From (i) and (ii) it is easy to see that ( AB)1  B 1 A1 . 1

Q4.

Consider the Transformation T: R2  R2 defined by T(x) = A x where

 sin   cos  the show that, A   cos   sin  (i) The mapping T is a rotation through the angle  . A A  A  (ii) (iii) Find the inverse of A . Solution (i): First of all you should note that the transformation defined by the formula a  T(x) = A x is a linear transformation. Let x    and we can b  a  1  0 a 1   0 write x     a    b    T    aT    bT   then by definition of b  0 1  b   0 1  transformation we have,  sin   1  cos    sin   0   sin   1  cos  0 cos  T   and T          cos   0 sin   cos   1   cos   0 sin  1  sin  Now consider the geometric presentation of these two images, we can take the 1  0 1  unit vectors   and   along the unit circle obviously unit vector   makes an 0 1  0 0 angle 0 with the x-axis and vector   is along the y-axis which makes an 1   angle with the x-axis as shown in the fig below. 2

1  cos 0 Now as we can easily see that image of      is a rotation through an 0 sin 0    cos   0 1 cos(   0)       2 and this is also a angle  as T     similarly we have       1  sin   0 sin(  0   2 

   cos(  )   0  2 rotation of an angle  as T      . Since from the result of 1    sin(  )   2 

  trigonometry we have cos(  )   sin  and sin(  )  cos  . Thus in fact 2 2 the given matrix of transformation only rotates that point through an angle of . a  1  0 Now as every element x     a    b   is a linear combination of unit b  0  1  vectors this rotates through an angle  . Hence the transformation is a rotation through an angle  . Solution (ii): Since we have cos   sin    cos  thus A   A    cos   sin  sin  and we have

cos  A A   sin 

 sin   cos  cos   sin 

 sin   cos(   ) , A     cos   sin(   )

 sin   cos  cos   sin  sin   cos   sin  cos   cos  sin 

 sin(   )  cos(   )   cos  sin   sin  cos    sin  sin   cos  cos  

Now as we now that cos(   )  cos  cos   sin  sin  sin(   )  cos  sin   sin  cos  . cos(   )  sin(   )  Hence we have, A A     A  . sin(    ) cos(    )   Solution (iii): We know that if we have a b  1  a  b A then A1   ad  bc  c d  c d  Thus by using the above formula we sin    cos  1 A 1  as cos 2   sin 2   1 so   2 2 cos   cos   sin    sin  have sin    cos  A 1   cos     sin 

Q5.

Consider the system of linear equations 6u - 2v - 4w + 4y = 2 3u - 3v - 6w + y = - 4 -12u + 8v + 21w - 8y = 8 -6u - 10w + 7y = - 43 Find the solution of the above system using LU decomposition.

Solution: First of all we will find out the LU factorization of the matrix of 2 4 4  6  3 3 6 1   coefficients of the above system which is A   12 8 21  8   0  10 7  6 Step1: Zero out below the first diagonal entry of A. We get

0 0 0  1 4  1    0 2 4 1  1 0 0 1 R2  R1 , R3  2 R1 , R4  R1  2  0 4 13 0 2  2 * 1 0    0  2  14 11 * * 1  1 in the second matrix entries below first diagonal entry: 1 are the negative of the multipliers of the row operations which we did in the first matrix.      

6

2

4

Step2:

     

6

2

0

2

0 0

0 0

Zero out the entries below the second entry of the diagonal. 0 0 0  1 4 4  1   4  1  1 0 0 R3  2 R2 , R4  R2  2  5  2  2  2 1 0     10 12  1 * 1   1

Step3:

     

6

2

0

2

0 0

0 0

Zero out the entries below the third entry of the diagonal. 0 0 0  1 4 4  1   4  1  1 0 0 R4  2 R3  2 . 5  2  2  2  1 0    0 8 1 2 1  1

Where

0 0 0  1 4  1   0 2 4  1  1 0 0 L 2 . 0 0 5  2  2  2 1 0    0 0 0 8 1 2 1  1 Now our system becomes LUx = b where, 0 0 0  1 2 4 4  6 u   2  1       4   0 2 4 1  v 1 0 0    L 2 ,x  and b    ,U        0 0 5  2 w 8  2  2 1 0         0 0 0 8 y 43    1 2 1  1  z1  z  2 Let Ux = z where z    and by solving Lz = b we get,  z3     z4  0 0 0  z1  2  1  1   z1   2   1      1 0 0   z2   4   z2  4  z1  5   2  2   z3   8    2  2  1 0      z3  8  2 z1  2 z2  2   z  43   z  43  z  z  2 z  32 1 2 1  4    1 1 2 3  4   Thus we have U     

6

2

4

Now we will solve the system Ux = z which will give the solution of the above system.      

6

2

4

0

2

4

0

0

5

0

0

0

4  u   2   y  4  1   v   5   w  1.2    2   w  2  v  6.9     8   y   32  u  4.5

Solution of Assignment # 4 of MTH501 (Fall2004)

Q1.

3 2 

0 0 3  2 0 2 4 0 0    0 0 0 1 2 and V  0 1 2   0 0 2  3   4 1 0 0 4 1  Determine whether U and V are block diagonal matrices and find UV using block matrix multiplication. Solution 0 0 3  2 1 2 0 0 0  2 4 0 0  3 4 0 0 0    and V = 0 0 1 2 U=  0 0 5 1 2   0 0 2  3    1 0 0 3 4 0 0 4 1 0 0 0  1 2  0 0  5 1 4  Here A22 =  O 23 =  O22 =  B23 =      C22 = 0 0 0  3 4  0 0  3 4 1 0 2  0 1  2 0 0  0   0  D32 =  2  3 O22 =  O 32 =    4 0 0   0  4 1  0  Then U and V becomes 1 2 0 3 4 0 Let U   0 0 5  0 0 3

0

O23   A22 C22 U  and V    B23  O32 O22  A22C22  O23O32 UV   O22C22  O32 B23 2  3 2  1 A22C22   4   2 4  3 6 7  10  17 5 B 23 D32   3

1 4

1 2  2 1   4

9   1   5 7 6 0 7 17 10 0 UV   0 0 1  0 7 0 is the required result

Q2.

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

2  3  1

0 0  9  5 

Check whether the given system is diagonally dominant? -x 1 + 4x2 – x3 = 3 4x1 – x 2 = 10 - x 2 + 4x 3 = 6 Then solve the system after making appropriate changes by Jacobi’s Method. Also solve the above system by Gauss Seidal method. (Only three iterations and show your calculations)

Solution: From the given system we see that, 1  4  1 is not staisfied .

1  4

is not staisfied .

4  1 staisfied . Thus the given system is not diagonally dominant. But by interchanging first two rows we get the system 4x1 – x 2 = 10 -x 1 + 4x2 – x3 = 3 - x 2 + 4x 3 = 6

4  1 staisfied . Now you can check that 4  1  1 staisfied . hence we made the system diagonally

4  1 staisfied . dominant. Jacobi’s Method: 4x1 – x 2 = 10 -x 1 + 4x2 – x3 = 3 - x 2 + 4x 3 = 6

10  x2k 4 3  x1k  x3k x2k 1  4 6  x2k k 1 x3  4 Take initial iteration ( x10 , x20 , x30 )  (0, 0, 0) x1k 1 

For first iteration put k  0 10 x11  4 3 x12  4 6 x31  4 10 3 6 ( x11 , x12 , x31 )  ( , , ) 4 4 4 For sec ond iteration put k  1 10  3 / 4 43 x12   4 16 3  10 / 4  6 / 4 28 x22   4 16 6  3 / 4 27 x32   4 16 43 28 27 ( x12 , x22 , x32 )  ( , , ) 16 16 16 For third iteration put k  2 10  28 /16 188  4 64 3  43 /16  27 /16 118 x23   4 64 x13 

6  27 /16 123  4 64 188 118 123 ( x13 , x23 , x33 )  ( , , ) 64 64 64 x33 

Seidal method 4x1 – x 2

= 10

-x 1 + 4x2 – x3 = 3 - x 2 + 4x 3 = 6 10  x2k x  4 3  x1k 1  x3k x2K 1  4 K 1 6  x2 x3k 1  4 In int ial iteration ( x10 , x20 , x30 )  (0, 0, 0) k 1 1

For first iteration put k  0 10  0 10 x11   4 4 3  10 / 4  0 22 x12   4 16 6  22 /16 118 x31   4 64 10 22 118 ( x11 , x12 , x31 )  ( , , ) 4 16 64 For sec ond iteration put k  1 10  22 /16 182 x12   4 64 3  182 / 64  118 / 64 492 x22   4 256 6  492 / 256 2028 x32   4 1024 182 492 2028 ( x12 , x22 , x32 )  ( , , ) 64 256 1024 For third iteration put k  2 10  492 / 256 3052 x13   4 1024 3  3052 /1024  2028 /1024 8152 x23   4 4096 6  8152 / 4096 32728 x33   4 16384 3052 8152 32728 ( x13 , x23 , x33 )  ( , , ) 1024 4096 16384

0 7 3 5 4 0 0 2 0 0   3 6 4  8  then find Q3. Consider the matrix 7   0 5 2 3 5 0 0 9 1 2  (i) (3, 1)th Cofactor of the above matrix. (ii) Minor correspond to the element a55 . (iii) Determinant of the matrix using such a row or column least amount of computations. Solution: 0 7 0 2 (i) (3, 1)th Cofactor of the above matrix is (1)31 0 5 0 9

(ii)

(iii)

Minor correspond to the element a55 is

4 0 7

3

0 0

2

0

7 3 6

4

which involves the

3

5

0

0

2

3

1

2

.

5 0 5 2 From the given matrix you can see easily that second row and second column have only one non-zero entries, so if we expand the determinant by taking second row or column then we will involve fewer calculations. So expanding the given matrix by second 4 0 7 3 5 4 0 3 5 0 0 2 0 0 7 3 4 8 3 6 4  8  (1) 23 row 7 Now again 5 0 2 3 5 0 5 2 3 0 0 1 2 0 0 9 1 2 expanding the determinant by second column will reduce the calculations. 4 0 7 3 5 4 0 3 5 0 0 2 0 0 7 3 4 8 7 3 6 4  8  (1) 23 2 5 0 2 3 5 0 5 2 3 0 0 1 2 Again So 0 0 9 1 2

 4 3 5    2 2   2  (1) 3 5 2  3   0 1 2   expanding by the third row we

get 4 0 7 5 0

0

7

3

5

0

2

0

0

3

6

4

 8  ( 1) 23 2

0

5

2

3

0

9

1

2

4

0

3

5

7

3

4

8

5

0

2

3

0

0

1

2

 4 3 5   4 5 4     2  (1) 2 2 3 5 2  3   6 (1)3 2 (1)  (1)33 5 3 5   0 1 2    (6)  12  15    8  15   (6){(3)  (7)}  24

3  2

Q4. (i)

Determine the volume of the parallelepiped which has the following vectors as its adjacent sides,    a  3i  2 j  k , b  i  10 j  5k and c  i  2 j  3k .

(ii)

Consider the linear transformation T: R3 R3 defined as  x1  3x1  2 x2  x1  T  x2   0 x1  4 x2  5 x3   x3  0 x1  2 x2  2 x3  If S be the volume of the parallelepiped in R3 formed by the vectors (1, 1, 1) ,(2, 3, 7) and (3, 5, 7) find the volume of T(S).

Solution: (i)

(ii)

Volume of the parallelepiped is equal to the determinant of the given vectors which 3 2 1 10  5 1 5 1 10 2 1 120 16 8 128 is 1 10  5  3 2 3 1 3 1 2 1 2 3 First of all we will write the matrix of transformation which 3 2 1   5  Also we know that is 0 4 0 2  2 

Volume of T(S) = (Determinant of matrix of transformation of T) * (Volume of S) 1 1 1 3 7 2 7 2 3 1 1  14  7  1  6 but volume will be Volume of S = 2 3 7  1 5 7 3 7 3 5 3 5 7 positive so we have Volume of S = 6.

3

2

0

4

1 5 3

4 2

5  3(18)  54 . Thus we have, 2

0 2 2 Volume of T(S) = (-54) * 6 = -324, but we will take the positive value so, Volume of T(S) = 324

Q5. (i)

Let n  1. Then Rn consists of (a) (b) (c)

n real numbers. n-tuples of real numbers n-tuples of vectors

Solution: (b) is the correct answer as every element of Rn has the form (x 1 , x2 ,…, x n ) where x 1 , x2 ,.., xn are real numbers, also (x 1 , x2 ,…, x n ) =(y1 , y2 ,…, yn ) if and only if x1 ,= y1 , x2 ,= y2 …, xn ,= yn . (ii)

In a vector space V over a field F scalar multiplication is given by a map (a) (b) (c)

V V  F F V  V FF  F

Solution: The correct answer is (b) as we have scalar multiplications from F V  V to vector space V. (iii)

Which of the following is statement is not an axiom for the real vector space? (a) For all x, y V we have x + y = y + x (b) For all x, y, z V we have (x + y) + z = x + (y + z) (c) For all x, y, z V we have (xy)z = x(yz)

Solution: As we know that a Vector space V along with a field is a group under addition and satisfy the scalar multiplication properties but a vector space is associate with respect to multiplication is not necessary. Thus option (c) is incorrect. (iv)

Which of the following is true? If V is a vector space over the field F (a) (b) (c)

Solution:

x  y / x V , y V   V x  y / x V , y V   V V v / v V ,   F  F V

The option (a) is true as we know that for a vector space V we have x, y V then x  y V which means that x  y / x V , y V   V . (v)

Check whether the given subsets form subspaces of Rn (a) (b) (c)

U  x  Rn /, x1  x2  ...  xn 

U  x  Rn /, x12  x2 2 

U  x  R n /, x1  1

Solution: In order to check whether the given subsets are subspaces or not we will check the two conditions. (i) (ii) (a)

If x, y  U  x  y  U . If c  F   and v U then cv U

U  x  Rn /, x1  x2  ...  xn  is a subspace of Rn as for,

x, y  U where x  ( x1 , x2 ,...xn ) and y  ( y1 , y2 ,... yn ) also x1  x2  ...  xn , y1  y2  ...  yn Then we have, x  y  ( x1 , x2 ,...xn )  ( y1 , y2 ,... yn )  ( x1  y1 ,.x2  y2 ,..xn  yn ) U

because x1  y1  x2  y2  ...  xn  yn as x1  x2  ...  xn , y1  y2  ...  yn So first condition of subspace is satisfied by U. c  F   and v  U where v  ( x1 , x2 ,.., xn ) suchthat x1  x2  ..  xn Now take cv  (cx1 , cx2 ,.., cxn )  U because x1  x2  ..  xn  cx1  cx2  ..  cxn Hence the set given in the part (a) is a subspace of the vector space Rn .

(b)

U  x  Rn /, x12  x2 2 

U is not a vector space by providing a counter example U contains all those vectors in which the square of first two elements must be equal Let s,w  U S= (1, -1, 2, …n) W = (2, 2, 3, …n) S+w = (3, 1, 5,….2n) But in s + w the square of first two elements are not equal so s+w does not belongs to U so addition is not defined in U. For this reason U is not a vector space by the definition of vector space.

(c)

U  x  R n /, x1  1

U contains all those vectors in which the first element is 1

Let s and w belong to U Then s = (1,2,3,…n) and w = (1,3,4…n) s + w = (2, 5, 7, ….2n) s + w does not belong to U because the first element is not 1 So addition is not defined in U So U is not a Vector space under the given condition.

Solution of Assignment # 5 of MTH501 (Fall2004) Question # 1 a.

For the bases   {(4,3, 7), (2, 1, 0), (1, 0,13)}of R3 find the coordinate of the element (-7, 5, 1) relative to the bases  .

b.

For the bases   {2 x 2  x,3 x 2  1, x 2 } of  2 find the coordinate of the 16 x 2  5 x  9 relative to  .

Solution: (i) In order to get the coordinate of the element (-7, 5, 1) relative to the given bases   {(4,3, 7), (2, 1, 0), (1, 0,13)}of R3 we will solve the following system.

 4   2  1   7      x  3   y  1   z  0    5  The values of x, y and z will be the required coordinates.  7   0   13  1   4 2 1  7   5  and by solving it we get the Augmented matrix of that system is  3  1 0  7 0 13 1  values of unknowns as x = 2, y = 1, z = -1. (ii) In order to get the coordinate of the element 16 x 2  5 x  9 relative to the given bases   {2 x 2  x,3 x 2  1, x 2 } of  2 we will try to find out the values of a, b and c such that

a  2 x 2  x   b  3x 2  1  c  x 2   16 x 2  5 x  9 x 0

 2a  3b  c  x 2   a  x  bx 0  16 x 2  5x  9 x 0

comparing the coefficients of

2a  3b  c  16 x , x and x we get the following system a 2

0

 5 by solving it we get the values

b 9 of unknowns as a = -5, b = 9, c = -1. Thus the coordinate of the given polynomial is (5,9, 1)

Question # 2 (i)

Find the change of coordinate matrix that change   coordinates to  coordinates where   {x 2  x  1, x  1, x 2  1},    {x 2  x  4, 4 x 2  3 x  2, 2 x 2  3} are bases for  2 .

(ii)

1  2 7 5 Let b1 =   , b2 =   ,c1 =   , c2 =   and consider the bases 3 4  9  7  2 for R given by B = {b1 , b2 } and C = {c1 , c2 } a. Find the change of coordinates matrix from C to B b. Find the change of coordinates matrix from B to C.

Solution: (i) In that question we will find out the coordinates of elements of   using the bases  . As we did in Q1, thus

x2  x  4  a  x2  x  1  b  x  1  c  x 2  1 And solving as we did in above question we

 2 get a  2, b  3 and c  1 thus coordinate of  x  x  4     3  . Similarly we  1 1 1   2 2 have coordinate of  4 x  3 x  2     2  , coordinate of  2 x  3   1 .  3  1 2

 4 x 2  3x  2   2 x 2  3  P      x 2  x  4      1 1  2 Now as we know that  P     3  2 1 is the required matrix.  1 3 1 a. b1

Notice that PB C is needed rather than PC  B , and compute 1 b2 | c1 c2     3

 2 | 7 4

|9

 5 1  7  0

0

|5

1

|6

3 4 

3 5 So PB C   4  6 b. By part (a ) and u sin g the property ( with B and C int erchanged ) PC  B  ( PB C ) 1 

1 4 2  6

 3 2  5  3

3/ 2  5 / 2 

Question # 3 For dimension of Column space; Dimension of Row space and Rank of the following 1 2  1 1  2  3 . matrix  1 0  2 4 8 5  Solution:

In order to get the dimension of column space, dimension of row space and rank of the given matrix first of all we will find out the echelon form of the given 1 2  1 1  3  1  . From the reduced echelon form it is quite matrix which is  0 1  0 0 0 3  clear that the columns 1, 2 and four have the pivot positions thus the corresponding columns in the given matrix are linearly independent and form a bases for the column space of the given matrix. Hence dimension of column space of the given matrix is 3. Also from the echelon form it is clear that the non-zero rows of the given matrix are three so forms a bases for the row space thus dimension of row space as well as rank of the matrix is 3. Defective matrix: Let A be an n  n matrix. If there is an eigenvalue  of A such that the geometric multiplicity of  is less than the algebraic multiplicity of  , then A is called a defective matrix. Question # 4 Find all the eigenvalues and eigenvectors of the matrix A: 1 1 1  4  16 3 4 4  A  7 2 2 1   3 4  11 1 Also, show that the matrix A is defective. Solution: First of all we will get the eigenvalues of the given matrix for which we will solve the equation determined by the determinant det( A  tI )  0 now we have, det( A  tI )  t 4  5t 3  9t 2  7t  2  (t  1)3 (t  2) Thus det( A  tI )  0 gives us t = 1, 1, 1, 2 as roots of the characteristic polynomials which are the eigenvalues of the given matrix. As the eigenvalue 1 repeats three times so it has Algebraic multiplicity of eigenvalue 1 is 3 and Algebraic multiplicity of eigenvalue 2 is 1.

In order to get the geometric multiplicity we will find out the dimension of the eigenspaces correspond to these eigenvalues, for this we will solve the homogeneous system.

 A  1I   0 and the augmented matrix is,  6  16   7   11

1 1 2 1

1 4 0 3

0   6 1 1 1   0   48 3 12 12  0   42 12 0 6   0   66 6 18 12 0   6 1 1 1 0   0  5 4 4  0  0 0 3 3   0  0 0 0 0

1 4 1 2

0 0  0  0 0 0  0  0

1 1 1  6  0 5 4 4   0 5  7 1  7 1  0 5  3x4  3 3 5 5  x1   5       x   8x  8 8 2  4   x   x4   Thus the vector   form bases for the eigenspaces correspond  5  x3  5 5       x  1  1   x4   4   x4  1  1  to the eigenvalue 1. Thus the geometric multiplicity of the eigenvalue 1 is 1 which is less then the algebraic multiplicity of 1. Thus the matrix is by definition defective.  5  16  A  2 I    7   11 1  5 1  0 6 4   0 3 2  4  0 6

1 2 2

1 4 1

1 4 1

1

3

3

0  0     2 0    4 0  1 4

0   5 0   80  0   35   0   55 6 1 1 0 3 2 0 0

0 0

0 0

1 10 10 5

1 20 5 15

1 0 20 0  5 0  15 0 

1 0  2 0  0 0  0 0

 x3  x4     x1  1  1  3   x       2  x3  x4   x3  2  x4  2  2  x    3 3   3  0   x3   3    x       x4   3 0  3  x   4  Shows that dimension of eigenspaces is 2 which is the geometric multiplicity of the eigenvalue 2 thus the algebraic multiplicity of eigenvalue 2 is less then Geometric multiplicity of eigenvalue 2.

But eigenvalue 1 has geometric multiplicity less than algebraic multiplicity thus the given matrix is defective. Question # 5 1  (i) Find all 2  2 matrices for which   is an eigenvector correspond to the 2 eigenvalue 5. Solution: a b  We have to find out the matrix   such c d 

 a b  1  1   a  2b  5  c d   2  5  2  c  2d   10          that a  2b  5  a  5  2b thus the required matrix c  2d  10  c  10  2d b 5  2b is  where b and d are any real numbers. d  10  2d (ii) True or false; A square matrix A is invertible if and only if 0 is not an eigenvalue of A. Solution: The above statement is true. (iii) If 5 is an eigenvalue of matrix A then find the eigenvalue of A5 without any calculations and justify your answer. Solution: As we are given 5 is the eigenvalue of A then the eigenvalue of A5 will be 55 . Justification: Suppose  is an eigenvalue of A then we will show that A2 will have 2 eigenvalue  . Let x be the eigenvector correspond to that eigenvalue  then by Ax   x A  Ax   A   x   A2 x    Ax  definition we have, but Ax   x so,

A2 x     x  A2 x   2 x   2 is an eigenvalue of A2 . By using principle of Mathematical induction we can show that  k will be the eigenvalue of Ak for k  2,3, 4... .

(iv)

1 2 0 9  Find eigenvalues of the matrix A  0 0  0 0 0 0 calculation also find the eigenvalues of the that A is invertible.

3 5 0 4 2  1  3 1 2  without any  0 4  3 0 0 1  matrix A-1 . If you are given

Solution: Since we know that a triangular matrix has its eigenvalues as diagonal elements, so the eigenvalues of the given matrix are 1, 9, 3, 4. Also we are given A is 1 1 1 invertible so the eigenvalues of A-1 will be 1, , , . Where we use the fact that if  is an 9 3 4 1 eigenvalue of an invertible matrix A then will be an eigenvalue of A-1 .  1 k  (v) Consider the matrix A    , where k is any arbitrary constant. For 1 1  what values of k the matrix A will have two distinct eigenvalues? When is there no real eigenvalue? Solution: In order to get the answer of the asked questions we will solve the 1  k 2 A  I  0   0  1     k 1 1  equation

   1  From this you can easily decide that for k = 0 you will get the repeated eigenvalues and all other values give distinct eigenvalues. Also if the value of k is less then zero then you will get the complex eigenvalues so there will be no real eigenvalue if k is less then 0.

Question # 6

30   25  8   Show that A is diagonalizable, where A   24  7 30  .  12 4  14  Solution: First of all we will find out the eigenvalues of the given matrix which are 1  2  1 and 3  2 . A will be diagonalizable if it will have three linearly independent eigenvectors correspond to these eigenvalues. So now we will calculate the eigenvectors correspond to these eigenvalues. Eigenvector correspond to eigenvalue 1 are 1   4   4     u1  3  and u2   3  and eigenvector correspond to the eigenvalue 2 is u3   4  . 0   4  2 

4 1  4 1 0 0    3 4  and D  0 1 0  our claim is that A = PDP-1 so Take the matrix P  3 0 0 0 4  2  2  we will not calculate the inverse of the matrix P but equivalently we will show that AP = PD.

Question # 7 Determine whether the signals 2k , 4k , ( 5) k are solution of the difference equation yk 3  yk  2  22 yk 1  40 yk  0 . Also determine whether these signals forma basis for the solution space of the same equation. Solution: Take yk  2k then yk 3  2k 3 and so on. the difference equation becomes, 2k 3  2k  2  222k 1  402 k  0 2k  23  22  22(2)  40   0 Hence 2 k is a solution of the given difference equation. 00 Take yk  4k then yk 3  4k 3 and so on. the difference equation becomes, 4k 3  4k  2  224k 1  404 k  0 4k  43  42  22(4)  40   0 Hence 4 k is a solution of the given difference equation. 00 k k 3 Take yk   5 then yk 3   5 and so on. the difference equation becomes,

 5   5  22  5  40  5 k  5  125  25  22(5)  40   0 k 3

k 2

k 1

k

0 Hence  5  is a solution of the given k

00 difference equation. Now take the Casorati matrix of the given signals, and if the signals are linearly independent then these signals will form the bases for the solution space.  2k (4) k ( 5) k   k+1 k+1  (4) k+1  5   Take k  0 2  k+ 2 k+ 2  (4) k+ 2  5    2  1 1 1  1 1 1  1 1 1   2 4 -5    0 2  7    0 2  7         4 8 25   0 4 21  0 0 35 Since the echelon form of the matrix has pivot element thus the signals are linearly independent and form bases for the solution space of given difference equation.

Solution of Assignment # 6 of MTH501 (Fall2004)

Q1.

 p(3)   p(1)   then, Define T : P3   4 by T ( p)    p(1)     p(3) 

c. d.

Show that T is a linear transformation. Find the matrix for T relative to the basis {1, t, t2, t3} for P3 and the standard basis for 4 .

Solution: First of all we will show that the given transformation is linear for this we will show that (i) T( p + q) = T(p) + T(q) (ii) T(cp) = cT(p) Where p and q are arbitrary elements of polynomial space P 3 and c is scalar. (i)

Let

p  a0  a1 x  a2 x 2  a3 x3 and q  b0  b1 x  b2 x 2  b3 x 3 p  q  a0  b0   a1  b1  x   a2  b2  x 2   a3  b3  x 3

then we have,

 a0  b0   a1  b1  3  9  a2  b2   27  a3  b3      a0  b0   a1  b1  3   a2  b2    a3  b3   T ( p  q)     a0  b0   a1  b1    a2  b2    a3  b3    a  b   a  b  3  9  a  b   27  a  b   1 1 2 2 3 3   0 0  a0  3a1  9a2  27a3  b0  3b1  9b2  27b3  a  a  a  a  b  b  b  b  0 1 2 3  0 1 2 3    a0  a1  a2  a3  b0  b1  b2  b3       a0  3a1  9a2  27a3  b0  3b1  9b2  27b3   T ( p)  T (q)

Similarly you can show that the second condition satisfied by T. (ii)

First of all we will find out the image of bases of P3 and then we will find out the coordinate of these images in terms of standard basis of R4.

1 1 1 1 T (1)     Coordinate of T (1) in R 4 with s tan dard basis    1 1   1 1  3  3  1  1 4   T (t )   Coordinate of T (1) in R with s tan dard basis     1  1      3  3 9  9  1  1  T (t )     Coordinate of T (1) in R 4 with s tan dard basis    1  1      9  9   27   27   1   1  4    T (t )   Coordinate of T (1) in R with s tan dard basis    1  1      27   27  1  3 9  27  1  1 1  1  . So the required matrix is  1 1 1 1    3 9 27  1

Q2.

Find the B-matrix of the 4  14  transformation x  Ax where A   33 9  11  4  1   1   1  b3} where b1   2  , b2   1  , b3   2  .  1  1  0 

 14   31 , when B= {b 1, b 2, 11 

Solution: Let the transformation be T defined 4  14  by x  Ax where A  33 9  11  4

be

 14  31 then the required B-matrix will 11 

T (b1 )

T (b3 ) .

T (b2 )

4  1   14    T (  2  )   33 9  1  11  4 4  1   14    Now T (  1  )   33 9  1  11  4 4  1   14    T (  2  )   33 9  0   11  4

 14  1  14  8  14   8   31  2   33  18  31   15 11   1  11  8  11 8   14  1  14  4  14   4   31  1   33  9  31    7  11   1  11  4  11  4   14  1  14  8  0  6   31  2   33  18  0   15  11   0  11  8  0  3

4  8  So the required matrix is  15  7 8 4

Q3.

6 15  .  3 

Let T be the transformation whose standard matrix is given 0 5  0 3 by   1 4   1  2

0 0 3 0

0 0 . Find a basis for 4 with the property that T B is  0   3

diagonal. Solution: First of all we will find out the Eigenvalues of the given matrix, also we know that eigenvalues of triangular matrix are the elements in its diagonal so eigenvalues of the given matrix are 5,-3 and 3. Now we will find out the basis for eigenspaces of these eigenvalues. For eigenvalue 5 we will solve the homogeneous A  5I  0 augmented matrix is

0 0  0  0 2 0 system   1 4 8  0  1  2

0   0    0      8 

0  0    1 4 8 0     0 2  8  8   8   0 and basis for eigenspace is b1     1     1 0 0 0 2

0 0

0 1 0 0

0 0  basis 1 0 0  0  8  8 0 0 4 8

 0  3  Basis of the eigenspaces for eigenvalue 3 is b2    , basis for -3,  2     1  0 0  0 0 c1    , c2     1 0      0 1  0  8  0 3 Now taking P    1  2  1  1

0 0 1 0

0    0 and D    0   1 

5

0

0

0

3

0

0

0

3

0

0

0

0 0  . 0   3

Now using the Diagonal Matrix Representation Theorem on page # 325 the   required B-matrix of the transformation is D     

5

0

0

0

3

0

0

0

3

0

0

0

0 0  . 0   3

Q4. 1 1

For the matrix 

 2 , find an invertible matrix P and a matrix C 3 

such that A = PCP -1. Solution: First of all we will find out the eigenvalues of the given 1 

2 3

1

 0   2  4  3  2  0

  4  5  0 2

matrix.

Here we have a  1, b  4 and c  5 then u sin g quadratic formula.



4  16  20  2i 2

Now we will find out the eigenvector correspond to the eigenvalue 2 - i, for  1  i 1 

 2   x1  0  1  i   x2  0

which we will solve  1  i  x1  2 x2  0......(1)

x1  1  i  x2  0...........(2)

In order to get the nontrivial solution we should know that both equation 1 and 2 must determine the same relationship between x1 and x2.And we can use either equation to get the relationship between these two variables. Take equation 2 we get x1   1  i  by taking x2 = 1 then the eigenvector will 1  i  . 1 

be v  

a  b  where a  2, b  1 .So we have a  b

Now taking P   Re v Im v and C    1 P  1

1 2 ,C    0 1

 1 And you can check now that A = PCP -1.  2

Q5. Solve the initial value problem 3 x   Ax, x(0)  x0 where A  12  4

1 0 2

 1  7  5  and x0   3  16   1

Solution: First of all we will find out the eigenvalues of the matrix A now the eigenvalues of the matrix A are -1, 1, 2 and corresponding eigenvectors  1  3 1      u1   2  , u2   1  , u3  1  are .Apply the initial  2  7   2 and thus general solution is x(t )  a1e t u1  a2et u2  a3e 2tu3

 7  1  3 1   3   a  2   a  1   a 1  condition we get the system   1   2   3   by solving that 16   2   7   2

system you will get the values of constants. Q6.

7   1   Consider the vectors w  6  , x   5  13  3 

e. f.

Find distance between w and x. Compute a vector in the direction of w having unit length.

g.

    Compute  w.x  x and  w.w  w .  x 

 x 

Let w  spanv1, v2 ,..., vn  . Show that if x is orthogonal to each vj, for 1  j  n then x is orthogonal to every vector in w. Solution: Since we know that distance between w and x is

w x 

 7 1   6 5   13  2

2

3



2

 64 1 100 

w  x  165

7  w 1    6 . We will find out the unit vector in the direction of w which is w 254   13  w.x   1  w.x    x 62   5 . So we have   x   x  35    x   3  w.x  7  30  39  62, x  1  25  9  35 3

 w.w   254  2 . 254 Similarly we have  254   w  35 35  x 

Since each element y in w will be of the form y  c1v1  c2v2  ...  cnvn where c1 , c2 ,..., cn are real n umbers. . Now take x. y  x.  c1v1  c2v2  ...  cn vn   c1 x.v1  c2 x.v2  ...  cn x.vn Since we have xi .v j  0 for i  j x. y  c1 (0)  c2 (0)  ...  cn (0)  0

Hence proved as we take arbitrary element of w.

Q7.

5  1 2  1 1  4    Determine whether the QR factorization of the matrix  1 4  3  is   7  1 4  1 2 1 

possible or not? Justify! If it is possible then find Q and R. Solution: First of all we will show that columns of the given matrix are linearly independent, which will show that the matrix can be factored as QR. 5  1 2 5  1 2  1 1  4   0 3 1     1 4  3    0 6 2 From this it is quite clear that every column     7   0 6 2  1 4  1 2 1   0 0  4 

has a pivot position hence columns of the given matrix are linearly independent. So QR factorization is possible. Now we will find out the orthogonal basis for the column space of the given 1  2  5  1 1  4        matrix note that the vectors x1   1 , x2   4  , x2   3  form basis        1  4   7  1   2  1 

for the column space. We will use Gram Schmidt process to find out the orthogonal basis for the column space.

1  1   x .v v1   1 v2  x2  2 1 v1 v1.v1    1  1   3  0   x2 .v1  5 and v1.v1  5  v2   3     3  3 x .v x .v v3  x3  3 1 v1  3 2 v2 , x3 .v1  20, x3 .v2  12, v2 .v2  36 v1.v1 v2 .v2  2  0   v3   2     2  2 

Normalize these vectors we get,

     1  1    2 1  3  2     1  0  0  0 1    1   1    1  1 , v2   3     , v3    v1  6   2  5   2  1  3  1   1  1   3       2  2  1  1  2    2 

Now form the matrix Q which has columns as normalize basis which we       calculate above. So we have Q          

1 5 1 5 1 5 1 5 1 5

1 2 0 1 2 1  2 1 2

1 2   0  1 2  1 2  1   2

Find the matrix R by using the equation Q t A = R. Q8. (i) (ii)

3

8 

Let y    and u    . Compute the distance from y to the line 1  6 through u and the origin. Find the closest point to y in the subspace W spanned by v 1 and 3   3  1 1   1  1      ,v    v2 where y  , v1  5   1 2  1        1   1  1 

Solution: (i) Since we know that the distance from a vector y to a line through the line from u and origin is y  projection of y on u also we know projection of y on u 

that

y.u u u.u

y.u  30, u.u  100  projection of y on u  3 3 8  5.4  y  projection of y on u          1  10 6   2.8 y  projection of y on u  29.16  7.84  37

3 8  10 6 

(ii)

Since we know the closest point to y in the subspace W spanned by v1 y  y.v1 v  y.v2 v 1 2 v1.v1 v2 .v2 y.v1  9  1  5  1  6, v1.v1  9  1  1  1  12  3   y.v1 1  1 v1  v1.v1 2  1    1 y.v2  3  1  5  1  6, v2 .v2  4

 1   y.v1 3  1 v1  v1.v1 2  1    1  3  1  6  3   1  1     y  1    3    1  2    1  2  1 2  1  2  2   1 and v2 is          1  1  2   1 