Show that the square matrix $\displaystyle 2N - I$ is its own inverse if $\displaystyle N^{2} = N$

I really don't know where to start here. All I know is that I have to show that $\displaystyle 2N - I = (2N - I)^{-1}$.

Any help is appreciated.

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- Feb 28th 2010, 12:59 PMtemaireInvertible matrices problem
Show that the square matrix $\displaystyle 2N - I$ is its own inverse if $\displaystyle N^{2} = N$

I really don't know where to start here. All I know is that I have to show that $\displaystyle 2N - I = (2N - I)^{-1}$.

Any help is appreciated. - Feb 28th 2010, 01:15 PMDefunkt
- Feb 28th 2010, 01:31 PMtemaire
- Feb 28th 2010, 02:29 PMDefunkt
- Feb 28th 2010, 02:32 PMtemaire
Ok I think I got it.

If I expand it, I will get the identity matrix.

Thanks for the help.