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

Institute of Mathematical SciencesMathematics

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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Download Linear Equations and Matrices Part 2-Basic Mathematics-Assignment Solution and more Exercises Mathematics in PDF only on Docsity!

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

  ^  ^  