Invertibility of 3x3 matrix.

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February 9th 2011, 11:44 PM
ice_syncer

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..
February 10th 2011, 03:03 AM
mr fantastic

Quote:

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..

No, what you say is that the matrix is invertible provided $a + b + c \neq 0$ . Because, consider the example a = -3, b = 1 and c = 2 ....
February 10th 2011, 04:57 AM
ice_syncer

question is if to prove if it is invertible... proving a + b + c is not 0 is equivalent..
February 10th 2011, 11:50 AM
mr fantastic

Quote:

Originally Posted by ice_syncer

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

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.
February 11th 2011, 02:45 AM
ice_syncer

can you prove the main quesiton that is to prove that the matrix given is invertible?
February 11th 2011, 08:31 AM
topsquark

Quote:

Originally Posted by ice_syncer

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

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
February 11th 2011, 10:09 AM
ice_syncer

hmm ok
February 12th 2011, 04:15 PM
topspin1617

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

Does the question say something like:

"Assume $a,b,c$ are distinct real numbers. Determine whether or not the matrix $A=\left(\begin{array}{ccc}{a&b&c\\b&c&a\\c&a&b}\en d{array}\right)$ 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

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

then we know that

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

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