Invertibility of 3x3 matrix.

there is a 3 X 3 matrix ,
A = [ a b c]
[ b c a ]
[ c a b]
where a , b c are all distinct... can you prove that a + b + c is not 0 ?
the actual question is to prove whether it is invertible. If it is invertible, then its determiniant should not be 0 right? so I got |A| as -(a^3 + b^3 + c^3 ) + 3abc .. which is equal to -(a+b+c)(a^2 + b^2 + c^2 - ab - bc -ca ) which can be factored to
-1/sqrt(2) * (a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)

the second part cant be zero as a, b, c are distinct.... so I have to prove that a + b + c is also not 0..

Quote:

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Originally Posted by ice_syncer

there is a 3 X 3 matrix ,
A = [ a b c]
[ b c a ]
[ c a b]
where a , b c are all distinct... can you prove that a + b + c is not 0 ?
the actual question is to prove whether it is invertible. If it is invertible, then its determiniant should not be 0 right? so I got |A| as -(a^3 + b^3 + c^3 ) + 3abc .. which is equal to -(a+b+c)(a^2 + b^2 + c^2 - ab - bc -ca ) which can be factored to
-1/sqrt(2) * (a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)

the second part cant be zero as a, b, c are distinct.... so I have to prove that a + b + c is also not 0..

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No, what you say is that the matrix is invertible provided $\displaystyle a + b + c \neq 0$. Because, consider the example a = -3, b = 1 and c = 2 ....

question is if to prove if it is invertible... proving a + b + c is not 0 is equivalent..

Quote:

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Originally Posted by ice_syncer

question is if to prove if it is invertible... proving a + b + c is not 0 is equivalent..

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I have told you exactly what to give as the answer. It should be obvious that a + b + c is NOT zero in general and therefore what you originally posted (prove that a + b + c = 0) cannot possibly be proved.

can you prove the main quesiton that is to prove that the matrix given is invertible?

Quote:

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Originally Posted by ice_syncer

can you prove the main quesiton that is to prove that the matrix given is invertible?

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I think what Mr. Fantastic is trying to say is that (as you have stated) if a + b + c = 0 then the matrix is singular. If a + b + c is not zero, then it is not singular. There is nothing more to show. You are already done.

-Dan

hmm ok

You need to be more specific with questions, especially ones that involve undetermined quantities.

Does the question say something like:

"Assume $\displaystyle a,b,c$ are distinct real numbers. Determine whether or not the matrix $\displaystyle A=\left(\begin{array}{ccc}{a&b&c\\b&c&a\\c&a&b}\en d{array}\right)$ is invertible, or give necessary and sufficient conditions for the matrix to be invertible."

If this is what the question is asking, and if the determinant is indeed

$\displaystyle \mathrm{det}A=-\frac{1}{\sqrt{2}}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$,

then we know that

$\displaystyle A$ is invertible$\displaystyle \Leftrightarrow \mathrm{det}A\neq 0\Leftrightarrow -\frac{1}{\sqrt{2}}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)\neq 0$.

But $\displaystyle a,b,c$ distinct $\displaystyle \Rightarrow (a-b)^2,(b-c)^2,(c-a)^2>0\Rightarrow(a-b)^2+(b-c)^2+(c-a)^2>0$
$\displaystyle \Rightarrow \mathrm{det}A=0\Leftrightarrow a+b+c=0\Rightarrow A^{-1}$ exists$\displaystyle \Leftrightarrow a+b+c\neq 0$.