@VivaanDaga Why should it be undeleted? If a user has posted a question and someone has deleted it, it is that user's responsibility to improve the question by editing instead of asking a new one. Repeated practice of this can even give you a suspension
I encountered above doubt while solving following question.
If $A$ and $B$ are $n-$ square matrices such that $AB=I+BA$ , find number of such possible pairs?
I started with $$AB-BA=I$$
I decided to compare their elements
$$\displaystyle \sum_{r=1}^n(a_{ir}b_{rj}-b_{ir}a_{rj})=0 \text{ when i} \...
@LalitTolani That duplicate solves your question, and I agree with lulu there: you claim that $0=1$ but with no justification. Indeed, if you made the justification correctly, it will be the same as this answer.
Try clicking on the "show all 4242 posts" option if it is there? (It is there for me). Or let me roughly know which comment you are looking for and I can try to find it.
@Peter OP's background seems to be very basic : I'm concerned that any answer to this will end up being too broad. I posted a comment asking OP for their background, based on two comments given. If there's no reply to this in some time, I would (if on-site then!) close it for lacking context, at least the context I want.
If they've understood the comments, then I would probably ask them to self-answer. If their background is so weak that they failed to understand this, then any answer would have to cover all these materials from scratch (unless it was very link-heavy , which may actually not be a big problem) but it could end up being a very broad question.
In that case, I would close it for being too broad. For now, I hope I will get a response.
Maybe this could work. If not, then this one or one of the duplicates associated to it, which contain near-about the same proof either in the question or in the answer.
I encountered above doubt while solving following question.
If $A$ and $B$ are $n-$ square matrices such that $AB=I+BA$ , find number of such possible pairs?
I started with $$AB-BA=I$$
I decided to compare their elements
$$\displaystyle \sum_{r=1}^n(a_{ir}b_{rj}-b_{ir}a_{rj})=0 \text{ when i} \...
