Solution of Assignment # 3 of MTH501
(Fall2004)
Please read the following instructions before attempting the solution of this
assignment.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
In order to attempt this assignment you should have full command on
Lecture # 11 to 15 of MTH501.
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1
 6 1
Q1.
Given that the matrix A is symmetric if At   1 14
4  and

 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
3x  y  z 
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
3 x  y  z  2
1
So augmented matrix for the system is  2

3
1
matrix is 0

0
2
3
3
2
1 1
6
14  and echelon form of the this
 2 
3
1 2  x  1, y  2, z  3 , thus we get,
0 1
1
1
1
1  4  9
  6 1
 1


266
4
4  . Hence x  1, y  2, z  3 are

   1 14
 1
4
3  2  3  1
4  2
the required values.
2
4 1
9 0 1 
2



Q2.
If A  3 4 4 B  0  1 6  then find




8 0 
 2 8 0 
1
(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
1
1 1
that ( AB)  B A .
2 1
4 1  then show
8 0
Solution:
First of all we will find out the product AB then we will find out the
 4 14 3 
inverse of AB. So AB   4
32 0  and

6
2 
1
 0.0833  0.8000 
 0.5333

0.0417
0.1000  --------------- (i)
 AB   0.0667
 0.0833
0.6000 
 0.0667
Now we have
 0.3333  0.0833 0 
 0.8000  0.8000 0.2000
1
1


 A  0.3333 0.0833 1  and  B   0.1000 0.1000 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,
1
 sin   1  cos  
 sin    0   sin  
  cos 
0  cos 
T 

and
T


1  sin 
cos   0 sin  
cos   1   cos  
0 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 

1  cos(  0) 
0 
2 and this is also a
angle  as T    
similarly we have    


0 sin(  0 
1  sin  

2 
 

cos(  ) 
0 
2 . Since from the result of
rotation of an angle  as T    

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
 sin  
cos 
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 
Now as we now that
cos(   )  cos  cos   sin  sin 
cos(   )
 sin(   ) 
cos(   ) 
 cos  sin   sin  cos  
 sin  sin   cos  cos  
sin(   )  cos  sin   sin  cos  .
 sin(   ) 
 A  .
cos(   ) 
Hence we have, A A  
sin(   )
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
Where
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
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
1

2
1


 z1 
z 
2
Let Ux = z where z    and by solving Lz = b we get,
 z3 
 
 z4 


Thus we have U  



6
2
4
0
0
0
 z1  2
 1
 1
  z1   2  
1
  


1
0
0   z2   4   z2  4  z1  5


2
 2





z
8
3
 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
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