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Assignment 4 (Solution)
50 x1  x2
 149
x1  100 x2  2 x3  101
Q1. Solve the system
2 x2  50 x3  98
using Jacobi’s method, with x = 0 and three iterations. Repeat the iterations until two
successive approximations agree within a tolerance of .001 in each entry.
(0)
Solution
Consider
x1 = (149 + x2)/50
x2 = (101 + x1 + 2x3)/100
x3 = (-98 – 2 x2)/50
x
(0)
0
 2.98 
3.0002 
3.0000 






(1)
(2)
(3)
 0 , x  1.01  , x  1.0006  , x  1.0000 
0
 1.96
 2.0004
 2.0000
Q2. Show that the Gauss Seidel method will produce a sequence converging to the
solution of the following system, provided the equations are arranged properly:
 x1  4 x2  x3  3
4 x1  x2
 10
 x2  4 x3  6
Solution
Consider
4 x1  x2
 10
 x1  4 x2  x3  3
 x2  4 x3  6
It implies,
x1 = (10 + x2)/4
x2 = (3 + x1 + x3)/4
x3 = (6 + x2)/4
x (0)
0
 2.5 
 2.84375 
3.0000 
 0 , x (1)   1.375  , x (2)   1.921875  ,..., x (6)  2.0000 
0
1.84375
1.98046875
2.0000 
Q3. Compute the determinant
1 3 1 0
0
2
2
4 1 6
2 6 2
3 7 3
3
8
9
7
3
2
7
5
5
Solution
Step 1: Transform the Matrix in Row Echelon Form







1
0
-2
3
3
3
2
-6
7
5
-1
-4
2
-3
5
0
-1
3
8
2
-2
-6
9
-7
7














1
0
0
0
0
3
2
0
-2
-4
-1
-4
0
0
8
0
-1
3
8
2
-2
-6
5
-1
13







add 2 times the 1st row to the 3rd row
the determinant does not change
add -3 times the 1st row to the 4th row
the determinant does not change
add -3 times the 1st row to the 5th row
the determinant does not change







1
0
0
0
0
3
1
0
-2
-4
-1
-2
0
0
8
0
-1/2
3
8
2
-2
-3
5
-1
13







multiply the 2nd row by 1/2
the determinant is multiplied by ½







1
0
0
0
0
3
1
0
0
0
-1
-2
0
-4
0
0
-1/2
3
7
0
-2
-3
5
-7
1







add 2 times the 2nd row to the 4th
row the determinant does not change
add 4 times the 2nd row to the 5th
row the determinant does not change







1
0
0
0
0
3
1
0
0
0
-1
-2
-4
0
0
0
-1/2
7
3
0
-2
-3
-7
5
1







interchange the 3rd row and the 4th
row the determinant is multiplied by
–1







1
0
0
0
0
3
1
0
0
0
-1
-2
1
0
0
0
-1/2
-7/4
3
0
-2
-3
7/4
5
1







multiply the 3rd row by -1/4
the determinant is multiplied by -1/4







1
0
0
0
0
3
1
0
0
0
-1
-2
1
0
0
0
-1/2
-7/4
1
0
-2
-3
7/4
5/3
1







multiply the 4th row by 1/3
the determinant is multiplied by 1/3
Step 2: Calculate the determinant



Let us denote the original matrix: A = 



1
0
-2
3
3



and the row echelon form obtained above: B = 



-1
-4
2
-3
5
0
-1
3
8
2
-2
-6
9
-7
7
1
0
0
0
0
3
1
0
0
0
-1
-2
1
0
0
0
-1/2
-7/4
1
0
then have the following relationship:
det(A) (1/2) (-1) (-1/4) (1/3) = det(B).
Since det(B) = 1 , we have,
det(A) = 24.
Compute the determinants of the matrices in question 4.
1 2
2 5
 1 2 0 0

 2 5 0 0
 1 3

Q4. (a) 
(b) 
 5 12 3 3
0 0


0 0
11 8 2 1 

0 0
Solution







3
2
-6
7
5
0 8
0 4
2 6
0 3
0 2
0 3
6 9 
7 5 
9 2 

0 0
1 0

8 4 
-2
-3
7/4
5/3
1



 We



1 2 0
2 5 0
(a)
5 12 3
11 8 2
1
2
1
(b)
0
0
0
2
5
3
0
0
0
0
0 1 2 3 3

 (5  4)(3  6)  3
3 2 5 2 1
1
0 8
0 4
2 6
0 3
0 2
0 3
6 9
7 5
1 2 0 3 0
9 2
= 2 5 0 2 1
0 0
1 3 2 3 8
1 0
8 4
0
0  2(5  4).3(4)  24 -+4
Q5. Find the volume of the parallelepiped with one vertex at the origin and adjacent
vertices at (1, 4, 0), (-2, -5, 2), (-1, 2, -1).
Solution
Volume of the Parallelopiped =
 1 4 0 


Abs  2 5 2   Abs  2(2  4)  1( 5  8)   15  15
 1 2 1 


Q6. Let T: R3  R3 be the linear transformation determined by the matrix
 a 0 0
A   0 b 0 , where a, b, c are positive numbers. Let S be the unit ball, whose
 0 0 c 
bounding surface has the equation x12  x2 2  x32  1 .
a. Show that T (S) is bounded by the ellipsoid with the equation
x12 x2 2 x3 2


 1.
a 2 b2 c 2
b. Use the fact that the volume of the unit ball is 4π / 3 to determine the volume of the
region bounded by the ellipsoid in part (a).
x12 x2 2 x3 2
Solution (a) We have to prove that the ellipsoid 2  2  2  1 is the image of the
a
b
c
2
2
2
unit ball x1  x2  x3  1 under the linear transformation T: R3  R3 determined by
the matrix
a 0 0 
A  0 b 0 
0 0 c 
where a, b and c are positive numbers.
 u1 
 x1 


If u  u 2 , x   x 2  , and x  Au , then u1 = x1/a, u2 = x2/b, u3 = x3/c. It follows that u
 
 
 u 3 
 x 3 
is in the unit ball with u12  u 22  u32  1 , iff x is in the ellipsoid with
x12 x2 2 x3 2


1
a 2 b2 c 2
(b) Volume of the ellipsoid = Volume of T (S)
 det  A  .{Volume of Unit ball}  abc.(4 / 3)π
 (4 / 3)πabc
In questions 7 & 8 a set of objects is given together with operations of addition and scalar
multiplication. Determine which sets are vector spaces under the given operations. For
those that are not, list all axioms that fail to hold.
Q7. The set of all positive real numbers with operations x + y = xy and kx = xk
Solution
Let V the set of all positive real number,
(i) if x and y are objects in v, then by definition x + y is in V
therefore V is closed under ‘+’.
(ii) Suppose x, y  V
then x + y = xy
=yx
= y +x
therefore V is commutative under ‘+’.
(iii)suppose x, y, z  V
then x + (y + z) = x + (yz)
= x(yz)
=(xy)z
=( x + y)z
therefore is associative under ‘+’
(iv) there is an object 1 in V such that
x + 1 = x.1 = x
and 1+ x = 1.x = x
for all x in V
therefore ‘1’ is an identity element in V
(v) for each x in V, there is an object 1/x in V , such that
x + 1/x = x.1/x = 1
and 1/x + x = 1/x .x = 1
therefore inverse of each element of V exists in V.
(vi) if k is any scalar and x is any object in v, then by definition kx is in V.
therefore V is closed under scalar multiplication.
(vii) If k is any scalar and x,y is any object in V, then k(x+y)= k(xy)= xyk = xkyk
(kx .ky) = kx + ky
(viii) If k and l be any scalar and x is any object in V, then (k +l )x = xk + l = xk xl =
(kx.lx)
kx + lx
= (k + l ).x
(ix) k(lx) = k (xl) = (xl)k = xkl = (kl)(x)
(x) 1x = x1 = x therefore V is the vector space under given operation
Q8. The set of all pairs of real numbers of the form (1, x) with the operations
(1, y )  (1, y)  (1, y  y) and k (1, y )  (1, ky)
Solution
i). By definition, V is closed under ‘+’.
ii). Let u, v  V. Then u + v = (1, y)  (1, y)  (1, y  y)  (1, y  y)  (1, y )  (1, y)  v  u
iii). Let u, v and w  V. Then
(u + v) + w = (1, y  y)  (1, y)  (1, ( y  y)  y)  (1, y  ( y  y))  (1, y )  (1, y  y)
 u  (v  w)
iv). e = (1, 0)
u  e  (1, y)  (1,0)  (1, y)  u
Similarly,
e+u=u
v). u  (1, y)
u  (u)  (1, y)  (1,  y)  (1, y  y)  (1,0)  e
Similarly,
(u)  u  e
vi). By definition, scalar multiplication is closed in V.
vii).
k(u  v)  k(1, y  y)  (1,k(y  y))  (1,ky  ky)  (1,ky)  (1,ky)
viii).
 k(1, y)  k(1, y)  ku  kv
(k  l)u  (k  l)(1, y)  (1,(k  l)y)  (1,ky  ly)
 (1,ky)  (1,ly)  k(1, y)  l(1, y)  ku  lu
k(lu)  k(1,ly)  (1,k(ly))  (1,(kl)y)  (kl)(1, y)  (kl)u
1u  1(1, y)  1,1.y)  (1, y)  u
ix).
x).
Hence V is the vector space under the given operation.
Q9. Determine which of the following are subspaces of Mnn.
(a) all n x n matrices A such that tr (A) = 0
(b) all n x n matrices A such that AT = -A
(c) all n x n matrices A such that the linear system Ax = 0 has only the trivial solution
(d) all n x n matrices A such that AB = BA for a fixed n x n matrix B
Solution (a) Let V = all n x n matrices A such that tr (A) = 0
 a11
a
21
Suppose A  


a n1
a11  a 22  ...  a nn
a1n 
 b11 b12

b
a 2n 
b 22
and B   21




an2
a nn 
 b n1 b n 2
 0 and b11  b22  ...  b nn  0
a12
a 22
 a11  b11 a12  b12
a  b
a 22  b 22
21
21
Now A  B  


a n1  b n1 a n 2  b n 2
b1n 
b 2n 
 V such that


b nn 
a1n  b1n 
a 2n  b 2n 
V


a nn  b nn 
Since
 a11  b11    a 22  b22   ...   a nn  b nn 
  a11  a 22  ...  a nn    b11  b 22  ...  b nn 
00 0
ka1n 
 ka11 ka12
 ka
ka 22
ka 2n 
21

 V since
Also, kA 




ka nn 
 ka n1 ka n 2
ka11  ka 22  ...  ka nn  k  a11  a 22  ...  a nn   k.0  0
Hence V is the subspace of Mnn.
(b) Let V = all n x n matrices A such that AT = -A
 a11 a12
a
a 22
21
Suppose A  


a n1 a n 2
a1n 
 b11 b12

b
a 2n 
b 22
and B   21




a nn 
 b n1 b n 2
AT = -A and BT = -B
 a11  b11 a12  b12
a  b
a 22  b 22
21
21
Now A  B  


a n1  b n1 a n 2  b n 2
a1n  b1n 
a 2n  b 2n 
V


a nn  b nn 
Since (A + B)T = AT + BT = (-A) + (-B) = - (A + B)
b1n 
b 2n 
 V such that


b nn 
 ka11 ka12
 ka
ka 22
21
Also, kA  


 ka n1 ka n 2
ka1n 
ka 2n 
 V since


ka nn 
(k A)T = k AT = k ( - A ) = - ( k A )
Hence V is the subspace of Mnn.
(c) Let V = all n x n matrices A such that the linear system Ax = 0 has only the trivial
solution
 a11 a12
a
a 22
21
Suppose A  


a n1 a n 2
det (A)  0 and det (B) 
 a11  b11
a  b
21
21
Now A  B  


a n1  b n1
a1n 
 b11 b12

b
a 2n 
b 22
and B   21




a nn 
 b n1 b n 2
b1n 
b 2n 
 V such that


b nn 
0
a12  b12
a 22  b 22
a n 2  bn 2
a1n  b1n 
a 2n  b 2n 
 V for some A and B


a nn  b nn 
because det (A + B) = 0 for some A and B.
Hence V is not the subspace of Mnn.
(d) Let V = all n x n matrices A such that AB = BA for a fixed n x n matrix B
 a11 a12
a
a 22
21
Suppose A1  


a n1 a n 2
a1n 
 b11 b12

b
a 2n 
b 22
and A 2   21




a nn 
 b n1 b n 2
A1B = BA1 and A2B = BA2 for a fixed n x n matrix B
 a11  b11 a12  b12
a  b
a 22  b 22
21
21
Now A1  A 2  


a n1  b n1 a n 2  b n 2
a1n  b1n 
a 2n  b 2n 
V


a nn  b nn 
Since (A1 + A2) B = A1 B + A2 B = B A1 + B A2 = B (A1 + A2)
b1n 
b 2n 
 V such that


b nn 
 ka11 ka12
 ka
ka 22
21
Also, kA  


 ka n1 ka n 2
ka1n 
ka 2n 
 V since


ka nn 
(k A) B = k (A B) = k (B A) = B (kA)
Hence V is the subspace of Mnn.
 s  3t 
 s -t 
 . Show that W is a subspace of
Q10. Let W be the set of all vectors of the form 
 2s - t 


 4t 
R4 .
Solution
3 
 s  3t   1 
 
 s - t  1

    s   1 t
 1
 2s - t   2 
 

  
 4t  0 
4 
v1
v2
W = Span {v1, v2}. Hence by definition W is subspace of R4.
 a 

 

  b  a - 2b = 4c 
Q11. Either uses an appropriate theorem to show that the set 
:
 , W is a
  c  2a = c + 3d 
  d 

vector space, or find a specific example to the contrary.
Solution W is a subspace of R4, because W is the set of solutions of the system
a–2b–4c=0
2a – c – 3 d = 0