Thread: Prove this matrix is invertible

is a square matrix such that =0. Prove that for any real number the matrix is invertible

Attempt: if =0 then is a singular matrix. A singular matrix times is still a singular matrix. so the idea is to prove that is an invertible matrix when A is singular.

I'm not sure where to go from here but this is something interesting I found:

C = D = C and D are both singular matrices.

1) C + = = invertible matrix

however,

2) D + = = singluar matrix

So I believe the idea is to figure out why D is not part of the matrix we are looking for, even though it is singular, and it probably has to do with the fact that

Thanks!

Here's an idea:

Try computing

Originally Posted by HappyJoe

Here's an idea:

Try computing

Genius (:

Is this a common type of question. How'd you think to do that?

I'm not sure if it's common. I haven't seen it before, but it's a fun result.

As for my approach, I was first inspired by the condition to try and square to see what that'd get me. In the result of , the term of course vanishes, but the cross-term doesn't, which is pretty useless. Then I just remembered that to get the cross-term out of the way, you can change the sign on in one of the factors. That's it.

Thanks Joe! Stay happy (: