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Linear Equations and Matrices Part 4-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: Eigenspace, Matrix, Linear, Equation, Augmented, System, Equivalent, Echelon

Typology: Exercises

2011/2012

Uploaded on 08/03/2012

roocky
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Solution of Assignment # 4 (Lecture# 31 - 35) Of MTH501 (Spring
2012)
Maximum Marks: 20
Due Date: June 27, 2012
Question: 1 Marks: 10
If
423
11 3
249
A






, then find a basis for the eigenspace corresponding to 3
.
Solution:
We form
4 2 3 100
31133010
249 001
123
123
246
AI












Now we make an augmented matrix
3|0AI
And convert into echelon form, we have
31 21
1230
1230
2460
1230
0000 2 3,
000 0
R
RRRR












The above system is equivalent to
123
230xxx
Let
2
3
x
s
x
t
Then
123
23 23
x
xx st 
So
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Download Linear Equations and Matrices Part 4-Basic Mathematics-Assignment Solution and more Exercises Mathematics in PDF only on Docsity!

Solution of Assignment # 4 (Lecture# 31 - 35) Of MTH501 (Spring

Maximum Marks: 20 Due Date: June 27, 2012

Question: 1 Marks: 10

If

A

 ^   

, then find a basis for the eigenspace corresponding to   3.

Solution:

We form 4 2 3 1 0 0 3 1 1 3 3 0 1 0 2 4 9 0 0 1 1 2 3 1 2 3 2 4 6

A I

  ^   ^  ^ 

 ^    

Now we make an augmented matrix A  3 I | 0 And convert into echelon form, we have

3 1 2 1

R R R R R

The above system is equivalent to x 1 (^)  2 x 2 (^)  3 x 3  0

Let 2 3

x s x t

Then x 1 (^)   2 x 2 (^)  3 x 3   2 s  3 t

So

1 2 3

x s t s t x s s s t x t t

   ^ ^   ^   ^     

Therefore the basis for the eigenspace is 2 1 0

^  

  and

^  

Question: 2 Marks: 10

If 1 5 2 3

A  ^ 

, then find the eigenvalues and a basis for each eigenspace in 2.

Solution:

The characteristic equation of A is

det( A   I )  0

2 2

Solving we have

  2  3 i

For

  2  3 i

The system is 1 3 5 0 2 1 3 0

i i

^   

2 1

i (^) i R R i i

 ^  

So, the given system is equivalent to ( 1  3 ) i x 1 (^)  5 x 2  0 2 1

let x t Then x t i t i