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GATE-2012 ECE Q47 (math)

Posted By Krishna Sankar On November 13, 2012 @ 5:48 pm In GATE | No Comments

Question 47 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q47. Given that

and , the value of is

(A) 

(B) 

(C) 

(D) 

Solution

To answer this question, we need to refer to Cayley Hamilton Theorem. This is discussed briefly in Pages 310-311 of Introduction to Linear Algebra, Glibert Strang (buy from Amazon.com [1]buy from Flipkart.com [2])

From the wiki entry on Cayley Hamilton theorem [3],

If is a given  matrix, and  is the identity matrix, the characteristic polynomial of is defined as,

.

The Cayley Hamilton theorem states that substituting matrix for in this polynomial results in a zero matrix, i.e.

This theorem allows for  to be expressed as linear combination of the lower matrix powers of .

For a general 2×2 matrix the theorem is relatively easy to prove.

Let 

The characteristic polynomial is

Substituting by matrix  in the polynomial,

.

 

Now, applying Cayley Hamilton theorem to the problem at hand,

.

The characteristic polynomial is,

.

Substituting by matrix  in the polynomial,

.

Alternatively, .

Finding  in terms of by substituting for ,

Matlab example

>> A = [-5 -3 ; 2 0];
>> A^3
ans =

  -65  -57
   38   30

>> 19*A + 30*eye(2)
ans =

  -65  -57
   38   30

 

Based on the above, the right choice is (B) 

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf [4]

[2] Introduction to Linear Algebra, Glibert Strang (buy from Amazon.com [1]buy from Flipkart.com [2])

[3] wiki entry on Cayley Hamilton theorem [3]

 

 


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