[Linear Algebra] For which a is 0 an eigenvalue?

Ortix
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Homework Statement



I have to find for which "a" an eigenvalue for the following system is 0.

The system:

1 -1 1
-1 2 -2
0 a 1

Homework Equations


My characterstic equation:
(1-λ)(2-λ)(1-λ)+2a -(1-λ) -a = 0

The Attempt at a Solution



I then proceed:
(1-λ)(λ2-3λ-2+a) = 0

but then I'm kind of clueless.. Now what?
 
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You wrote the equation for the eigenvalues of the system. Now, if you want 0 to be an eigenvalue, then it better satisfy that equation.
 
Well I get a=2 but the answer is a=-1

Can't seem to find my error. I've tried it a bajillion times (3 times actually) :)
 
Ortix said:
(1-λ)(λ2-3λ-2+a) = 0
Where did this come from?
 
Can you check for the 4th time, what happens if you plug in λ=0 to (1-λ)(2-λ)(1-λ)+2a -(1-λ) -a = 0 ? :) You just calculated something wrong somewhere along the way. (I'm assuming the characteristic equation is correct)
 
You really don't need to find the entire eigenvalue equation to answer this. A matrix has 0 as an eigenvalue if and only if it is NOT invertible (since there must be a non-zero v such that Av= 0) and that is true if and only if its determinant is 0. Set the determinant, which depends on a, equal to 0 and solve for a.
 
HallsofIvy, you tha man! Solved it! :D
 
Ortix said:

Homework Statement



I have to find for which "a" an eigenvalue for the following system is 0.

The system:

1 -1 1
-1 2 -2
0 a 1

Homework Equations


My characterstic equation:
(1-λ)(2-λ)(1-λ)+2a -(1-λ) -a = 0


The Attempt at a Solution



I then proceed:
(1-λ)(λ2-3λ-2+a) = 0

but then I'm kind of clueless.. Now what?

The equation in (a) is not consistent with that in (b). You don't need (b); just plug λ=0 into (a).

RGV
 
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