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Linear Equations and Matrices Part 2-Basic Mathematics-Assignment Solution, Exercises of Mathematics

This is solution to assignment of Basic Mathematics course. This was submitted to Karunashankar Sidhu at Institute of Mathematical Sciences. It includes: Images, Transformations, Maps, Matrices, Linear, Equations, Solution, Elementary, Row, Operation

Typology: Exercises

2011/2012

Uploaded on 08/03/2012

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Solution of Assignment # 2 (Lecture# 9 - 15) Of MTH501 (Spring
2012)
Maximum Marks: 30
Due Date: April 25, 2012
Question: 1 Marks: 10
Let 12 1 2
103 2
,, , .
01 5 7
e e y and y
 

 
  Let T: R2R2 be a linear transformation that maps e1
into y1 and maps e2 into y2. Find the images of 1
2
7.
6
x
and
x



Solution:
12 1 2
12
12
12
12 11 22
770
606
10
76
01
10
76 01
776
6
7(7 6 )
6
7() 6( )
76 () ()
32
76
57
21
e e e and e
ee
So
TTee
Te Te
y
y T e y and T e y
  

  
  
 

 
 
 

 
 





 






 

 
 
12
35 42
9
7
79
67
Therfore theimageof is







 
 
 
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Solution of Assignment # 2 (Lecture# 9 - 15) Of MTH501 (Spring

Maximum Marks: 30 Due Date: April 25, 2012

Question: 1 Marks: 10

Let (^1 2 1 )

e e y and y

Let T: R^2  R^2 be a linear transformation that maps e (^1)

into y 1 and maps e 2 into y 2. Find the images of 1 2

x and x

Solution:

1 2 1 2

1 2

1 2

1 2 1 2 1 1 2 2

e e e and e

e e

So

T T e e

T e T e y y T e y and T e y

  ^   

    ^  

Therfore the image of is

1 1 1 2 2 2 1 1 1 2 2 2 1 1 2 2

1 2

1 2 1 2 1 1 2 2 1 2

Similarly x x e x e x x T T x e x e x x T e x T e

x x

x x x x x x x Therfore the image of is x x x

 ^ 

 ^ ^  ^ 

 ^  

Question: 2 Marks: 10

Using elementary row operations, find the inverse of the following matrix, if it exists. 3 4 1 1 0 3 2 5 4

Solution:

Let

A

 ^ 

det 1 0 3 3( 15) 1( 11) 2(12) 45 11 24 10 2 5 4

A

As the given matrix is non-singular, therefore, inverse of the matrix is possible. We reduce it to reduced echelon form.

1 3 2 3

R R R R

  ^ 

  ^  

  ^ 

  ^ 

Hence the inverse of the original matrix is 1

A 

Question: 3 Marks: 5

Find an LU factorization of the matrix

Solution: We will reduce A to a row echelon form U and at each step we will fill in an entry of L in accordance with the four-step procedure. 3 6 3 6 7 2 1 7 0

A

 ^  

  • denotes an unknown entry of L. 1 2 1 6 7 2 1 1 7 0 3

multiplier

multiplier multiplier

multiplier

multiplier

U^ multiplier

 ^  

L

 ^ 

So

3 0 0 1 2 1 6 5 0 0 1 4 5 (^1 5 5 0 0 )

A LU

  ^  ^  