[Linear Algebra] Matrix "proof" type problem involving determinants ( I think)

[malakai456]

[Linear Algebra] Matrix "proof" type problem involving determinants ( I think)

Problem
Consider three square nxn matrices A,B,C all different from O (theta matrix) but such that
ABC = O
Show it's impossible (hint: Proof by contradiction)
1) A and B both invertible 2) A and C both invertible 3) B and C both invertible

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Is it me or is this problem just wrong, or at least badly written? Here's my take on this, using determinants.
If ABC = O,
then detAdetBdetC = Det(O) = 0
Well, already you have a problem, because A and B CAN both be invertible, and let me prove this
By definition, det A and det B will give a non zero number, let's call them respectively n and m
well nmdet(C) = 0 -> nm0 = 0
So therefore, A and B can both be invertible, as long as C is not invertible.
[h=2]Basically, any 2 out of 3 of these matrices CAN be invert as long as the third ISN'T[/h] So am I right? If wrong, could I be clarified on this?

Thanks guys and please ask if you don't understand my proof

 

[barrick]

Hi malakai456,

If, for example, \(\displaystyle B\) and \(\displaystyle C\) are invertible, then, by definition, they have inverses \(\displaystyle B^{-1}\) and \(\displaystyle C^{-1}\) (after all, that is what "invertible" means).

You can then multiply the equation \(\displaystyle ABC=\mathbf{0}\) on the right by \(\displaystyle C^{-1}B^{-1}\). What do you get ?

The argument is similar if any two of the matrices are invertible (Try it !)

Nowever, if only one matrix is invertible, the argument breaks down. For example, take:

\(\displaystyle \displaystyle
A = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}
\qquad B = \begin{pmatrix}1&0\\0&0\end{pmatrix}
\qquad C=\begin{pmatrix}0&0\\0& 1\end{pmatrix}
\)

In this case, you will have \(\displaystyle ABC=\mathbf{0}\), but none of the matrices is \(\displaystyle \mathbf{0}\).