We know that $AA^{-1} = I$ and $A^{-1}A = I$, but is there a proof for the commutative property here? Or is this just the definition of invertibility?
If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.    I do not understand anything more than the following.   Elementary row operations. Linear dependence. Row reduced forms and their relations with the original matrix.   If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?  P.S.: Please avoid using the transpose and/or inverse of a matrix.
Are collective nouns always plural, or are certain ones singular? I'd say Microsoft have a way of bending the rules and I know that McLaren have won the championship. While this sounds strange, I believe it is correct English (sorry, I'm not native).   But when it's a small company, would you still use it this way? Is a company always plural, or are small companies singular? I.e., would you say Bakery Johnson makes fine bread or Bakery Johnson make fine bread? Is it My book seller, Woody's, have moved or is it has moved? Opt in "mentor" privilege at 7.5k I am suggesting a Mentor privilege at 7.5k. It would be on a per-user per-site opt in basis.  At some reputation level (perhaps 7.5k) there could be an opt in "mentor" privilege.  If someone opts in then when a new user signs up, they become their Mentor. The new user is the Mentee.  They will get notified of any posts or suggested edits they make, and can help and support them.  While this would by no means help everyone (I'm sure there are more new people signing up than there are 7.5k people willing to help them), it could help mitigate the "new user's bad posts" and such like.  By no means would being someone's mentor mean you are the only person who can edit and help them, and you aren't obliged to do so - but if you have opted in then you should help.  How would this work?   Any Mentor would have no more than 5 active Mentees at once. Once the Mentee gets to 125 rep, the downvote privilege (or any other level deemed sensible) they are no longer a Mentee. If they aren't active for a month, they are no longer a Mentee. If the Mentor has 5 Mentees (i.e. cannot mentor anyone else) then any Mentee not active for 2 weeks is no longer a Mentee. The Mentee can opt out at any time, the Mentor can opt out, but if they opt out a lot (say 5 people in 2 months) they are blocked from the privilege for 2 weeks. The Mentor gets no extra rep. They can still get rep from answering their Mentee's questions and similar. Once the new user gets 20 rep on the site, a chat room is created for them to talk. No user with the association bonus will become a Mentee. This would (maybe) be opt in for sites. For example, this might not be at all useful for this site, or for , but on  or  I imagine this could be quite good.    appears to be asking the same thing, based on the title, but that is asking about people who know each other offline mentoring each other.    Every single one of these suggestions (especially the numbers) is open to discussion. Maybe 125 is to low to stop being mentored, or maybe it's too high. What do you think? One thing that made me curious while watching a football game on TV was the numbers that appear on the ground. For instance, there's a "1st &amp; 10" beside one team, but there was a "3rd &amp; 6" a few minutes prior (and I think it was beside the other team). What do these numbers mean? Help proving the primitive roots of unity are dense in the unit circle. I'm having difficulty understanding how to prove that the primitive roots of unity are in fact dense on the unit circle. I have the following so far:  The unit circle can be written $D=\{x\in\mathbb{C}:|x|=1\}$. The set of primitive $m$-th roots of unity is $A_m=\{\zeta_k:\zeta_k^m=1,\zeta_k\text{ is primitive}\}$.  Hence, the set of all primitive roots $A$ is given by the union of $A_m$ over $m=1,2,3,\ldots$. But I can't seem to get started on how to prove that $A$ is dense in $D$. Number of total possibilities for an equation I need to find the number of possibilities for which the following equation exists:  $$x_1 + x_2 + x_3 + \cdots + x_{10} \leq 70$$  Each variable is a non-negative integer.  I tried simplifying the question to the point of finding the number of possible solutions to each equation seperatly:  $$x_1 + x_2 + x_3 + \cdots + x_{10} = 70,$$ which is ${79}\choose{10}$.  $$x_1 + x_2 + x_3 \cdots + x_{10} = 69,$$ which is ${78}\choose{10}$.  $$x_1 + x_2 + x_3 + \cdots + x_{10} = 68,$$ which is ${77}\choose{10}$, and so forth.  Then I thought about adding all of these together. This seems a bit too much work considering that the textbook solution is: ${80}\choose{10}$, and I can't seem to figure out the idiomatic way of approaching these kind of questions.   Any help is greatly appreciated. Full URL not working with php include Do Israeli security agents interview all those flying to Israel or is it restricted to El Al flights? The one thing I hate about flying to Israel are the security checks at the origin airports. I don't mind X-rays and deep luggage searches, but there's also an annoying interview you have to "pass". The security folks can't really turn you down (especially since I'm a citizen), but they always make it a pain in the butt, because for some reason I fit their profile of a suspicious person.  Can I avoid the personal interviews if I book on a non-Israeli airline? Or perhaps a low-cost airline? Humans don't digest fibre, and nor do any animals, but some creatures have micro-organisms (bacteria, I'm guessing) in their gut which digest fibre for them.  My question is, could humans take some sort of pill and establish our own fibre-digesting colonies in our intestines?  Has this been tried?  Supposing it could be done, would you suppose it would have any negative effects on our health?  (i.e.. would those whose gut digested fibre for them be stuck on a low-fibre diet?).    Purpose of question: just wondering.  It would be cool to be able to eat bark. SEO repeated keywords in domain and url Is there software on Macintosh for adding GPS data to RAW files? How do I resolve unmet dependencies after adding a PPA? Occasionally, when I'm installing stuff, I get an error like the following:  Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies:  package1 : Depends: package2 (&gt;= 1.8) but 1.7.5-1ubuntu1 is to be installed E: Unable to correct problems, you have held broken packages.   How can I resolve this? Choosing the best model from among different "best" models How do you choose a model from among different models chosen by different methods (e.g. backwards or forwards selection)?  Also what is a parsimonious model? Assign multiple keyboard shortcuts for one action I can run faster than _____. (1) him (2) he? LIKE uses index, CHARINDEX does not? Compare two objects in Java with possible null values Why is homology invariant under deformation retraction/homotopy equivalence? How can I add reflection to a C++ application? I'd like to be able to introspect a C++ class for its name, contents (i.e. members and their types) etc. I'm talking native C++ here, not managed C++, which has reflection. I realise C++ supplies some limited information using RTTI. Which additional libraries (or other techniques) could supply this information? Using Newton's method calculate $ \frac{1}{\sqrt{a}} $ without division Suggest algorithm for the numerical calculation $ \frac{1}{\sqrt{a}} \ a &gt; 0 $ without division, use Newton's method.  My idea is:  $$ \frac{1}{\sqrt{a}} = (\sqrt{a})^{-1} = a^{-\frac{1}{2}}$$ $$ x = a^{-\frac{1}{2}} \Rightarrow f(x)=x^2 - a^{-1}$$ And next step is calulate $ f'(x)=2x $, but after substitution I got: $$ x_{n+1} = \frac{x_n^2+a^{-1}}{2x_n}$$ And know I still have divion, I guess. Maybe you've better idea on this task? The arithmetic - geometric mean inequality states that $$\frac{x_1+ \ldots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}$$ I'm looking for some original proofs of this inequality. I can find the usual proofs on the internet but I was wondering if someone knew a proof that is unexpected in some way. e.g. can you link the theorem to some famous theorem, can you find a non-trivial geometric proof (I can find some of those), proofs that use theory that doesn't link to this inequality at first sight (e.g. differential equations …)?  Induction, backward induction, use of Jensen inequality, swapping terms, use of Lagrange multiplier, proof using thermodynamics (yeah, I know, it's rather some physical argument that this theorem might be true, not really a proof), convexity, … are some of the proofs I know.