Matrix

Question

Comments

is it 30/5=6

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anirudh can you explain a lil,,i mean with 3 different eigen val,how you getting 1 alpha?

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With 2 of value I.e. 1,1 get alpha cancle.. By only 2 I get value of alpha.

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Is $\alpha = 6$?

Answer 1

Let us do it though it through Cayley Hamilton's Theorem which states :

Every eigen value of the matrix satisfies its own characteristic equation.

To elaborate , if we have f(M) = 0 , where f(M) is the characteristic equation of matrix then the same will also be for f(λ) i.e. f(λ) = 0 , where λ is a corresponding eigen value of matrix..

So given ,

       α M^-1  = M²  - α M + 11 I₃   where I₃ is a 3*3 identity matrix , so for eigen value  λ , we can write :

       α  λ^-1  =  λ²  - α λ + 11

⇒     α /  λ =   λ²  - α λ + 11  [ as  λ^-1  = 1 / λ ]    ..........(1)

Now let us substitute the eigen values given one by one  in the equation (1) :

a)  λ  =  1 : 

       So we have ,  

        α /  λ =   λ²  - α λ + 11  

  ⇒    α     =   1 - α + 11

  ⇒   2 α   =   12

  ⇒    α     =   6

a)  λ  =  2 :  

        α /  λ =   λ²  - α λ + 11  

 ⇒    α / 2  =   4 - 2 α + 11

 ⇒    5α / 2  =  15

 ⇒    5α      =  30

  ⇒    α      =   6

So in both cases we have the value of  α      =   6 . So 6 is the correct answer

Comments

One confusion...see here characteristic eq is

λ³-α λ²+11α-α=0....so α is in constant term..and we know constant term is determinant of the matrix..right?

Then why not α=-(1*1*2)=-2??

May be I am doing a very silly mistake..please correct me

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The mistake u r doing is in writing is characteristic equation itself.. The term is 11λ instead of 11α in the characteristic equation..

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yes..thats fine

λ³-α λ²+11λ-α=0..even then constant term desnot change.Still (-α) is the determinant of the matrix??