# Math Help - help needed!!! proof of invertible matrices

1. ## help needed!!! proof of invertible matrices

The question has been attached as an image.

any help or light on this question would be hugely appreciated!!

thank you

2. ## Re: help needed!!! proof of invertible matrices

Originally Posted by kaya2345
The question has been attached as an image.
any help or light on this question would be hugely appreciated!!
thank you

If $\delta _{j,k} = \left\{ {\begin{array}{rl} {1,} & {j = k} \\ {0,} & {j \ne k} \\ \end{array} } \right.$ and $I_n=\left[\delta _{j,k}\right]$.

Is $I_n$ invertible?

Can we take away any of those ones?

3. ## Re: help needed!!! proof of invertible matrices

Originally Posted by Plato
If $\delta _{j,k} = \left\{ {\begin{array}{rl} {1,} & {j = k} \\ {0,} & {j \ne k} \\ \end{array} } \right.$ and $I_n=\left[\delta _{j,k}\right]$.

Is $I_n$ invertible?

Can we take away any of those ones?
thank you. and yes In is invertible

4. ## Re: help needed!!! proof of invertible matrices

That shows, I think, that the minimum number of 1s in such an invertible n by n matrix is n. But the question was about the maximum number. We do know that if a matrix has two columns the same, it is not invertible. I think we can avoid that by having a 0, at a different row, in each column. That implies that the maximum number of 1s in such an invertible n by n matrix is $n^2- n$.

5. ## Re: help needed!!! proof of invertible matrices

Originally Posted by HallsofIvy
That shows, I think, that the minimum number of 1s in such an invertible n by n matrix is n. But the question was about the maximum number. We do know that if a matrix has two columns the same, it is not invertible. I think we can avoid that by having a 0, at a different row, in each column. That implies that the maximum number of 1s in such an invertible n by n matrix is $n^2- n$.
Thank you, for that. Could I prove this using proof by induction? or contradiction?

6. ## Re: help needed!!! proof of invertible matrices

Originally Posted by HallsofIvy
That shows, I think, that the minimum number of 1s in such an invertible n by n matrix is n. But the question was about the maximum number.
That mistake was once caused by my laptop not doing well with those attachments.
I wish we could outlaw the posting of questions in images.