Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be a $3\times 3$ matrix with real entries such that $\det(A)=6$ and $tr(A)=0$. If $\det(A+I)=0$ ($I$ denotes $3\times 3$ identity matrix), then the eigenvalues of $A$ are:
(i) $-1,2,3$;
(ii) $-1,2,-3$;
(iii) $1,2,-3$;
(iv) $-1,-2,3$.

If a,b,c are 3 eigenvalues then a+b+c=0 and abc=6 because sum of eigen values is trace and product is the determinant value. Then how to apply $\det(A+I)$?

share|improve this question
    
Sos de la facultad de ingeniería? –  Cure May 3 at 1:26

3 Answers 3

up vote 3 down vote accepted

Eigen values of $A+I$ are obtained by adding $1$ to the eigenvalues of $A$. So $\det(A+I)=0$ gives a third condition on them (besides $\det A = 6, \ \mathrm{tr\,}A=0$) and that should enable you to find the answer.

share|improve this answer
    
If a,b,c are eigen values of A then a+1,b+1,c+1 are eigen values of A+I.Am i right? then we have a+b+c=0, abc=6 and (a+1)(b+1(c+1)=0.IS it possible to solve these equation to find a,b,and c. –  mercy May 3 at 3:07
    
Expand the product $(a+1)(b+1)(c+1)=0$ fully then the solution will emerge. –  P Vanchinathan May 3 at 5:51

Hint:

Check the following: for a $\;3\times 3\;$ matrix $\;A\;$, and putting $\;\Delta:=\det A\;,\;\;\mathcal T:=tr. A\;$ , we have that its characteristic polynomial is

$$x^3-\mathcal T x^2+\left(\mathcal T^2-tr.\left(A^2\right)\right)x-\Delta$$

share|improve this answer
1  
@ DonAnatonio: Is the coefficient of $x^2$ really the determinant? –  P Vanchinathan May 3 at 2:44
    
Good catch, @PVanchinathan: it is minus the trace. Thanks. –  DonAntonio May 3 at 10:32

In the above question option (iv) is correct. Sum of eigenvalues is 0 product of eigenvalues is 6 (ie) a+b+c =0 and abc=6 and also (a+1)(b+1)(c+1)=0 since det(A+I)=0 Solving we get the roots.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.