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GATE-2012 ECE Q47 (math)

by Krishna Sankar on November 13, 2012

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.combuy from Flipkart.com)

From the wiki entry on Cayley Hamilton theorem,

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

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

[3] wiki entry on Cayley Hamilton theorem

 

 

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