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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 $\det(A+I)$?

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  • $\begingroup$ Sos de la facultad de ingeniería? $\endgroup$ – Cure May 3 '14 at 1:26
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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.

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  • $\begingroup$ 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. $\endgroup$ – mercy May 3 '14 at 3:07
  • $\begingroup$ Expand the product $(a+1)(b+1)(c+1)=0$ fully then the solution will emerge. $\endgroup$ – P Vanchinathan May 3 '14 at 5:51
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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.

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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

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

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    $\begingroup$ @ DonAnatonio: Is the coefficient of $x^2$ really the determinant? $\endgroup$ – P Vanchinathan May 3 '14 at 2:44
  • $\begingroup$ Good catch, @PVanchinathan: it is minus the trace. Thanks. $\endgroup$ – DonAntonio May 3 '14 at 10:32
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It is much simpler than this; $\det(A+I) = 0$ means that one of the eigenvalues must equal $-1$, so without loss of generality we can say $a = -1$; the other two bits of information then become $bc = -6, b+c=1$. So: $a=-1, b=-2, c=3$

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