What is the Invertibility of 2N-I Matrix When N^2 = N?

[Precursor]

Homework Statement

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

Homework Equations

properties of invertible matrix

The Attempt at a Solution

I really don't know where to start here. I know that $$(2N-I)(2N-I) = I$$, but where do I go on from there?

 

[VeeEight]

You are correct in calculating (2N-I)(2N-I) and using N²=N in your calculations. If (2N-I)(2N-I)=I, then the desired result is achieved, as this implies [(2N-I)]^-1=(2N-I).

 

[Precursor]

You are correct in calculating (2N-I)(2N-I) and using N²=N in your calculations. If (2N-I)(2N-I)=I, then the desired result is achieved, as this implies [(2N-I)]^-1=(2N-I).

So the answer is simply $$(2N-I)(2N-I) = I$$? But how did I use $$N^{2} = N$$?

 

[VeeEight]

I just assumed you did :)
The result (2N-I)(2N-I)=I is the desired result. So take (2N-I)(2N-I) and expand it. After you are done that, use N²=N.

 

[Precursor]

I just assumed you did :)
The result (2N-I)(2N-I)=I is the desired result. So take (2N-I)(2N-I) and expand it. After you are done that, use N²=N.

ok, so I get:

$$2N^{2} - 2NI - 2NI + I^{2} = I$$

$$2N - 2N - 2N + I = I$$

But this doesn't work out.

 

[VeeEight]

The coefficient on N² is incorrect. It is not a 2.

 

[Precursor]

The coefficient on N² is incorrect. It is not a 2.

Ok I got it. Thanks for the help.

 

[VeeEight]

Cheers.