The actual question is to find (AB)−1. (AB)−1 is just A−1B−1 and we already know matrices A−1 and B−1 so taking the product should give us the matrix

2020年9月16日 · For me the fact that B−1A−1 B − 1 A − 1 "works" as a two-sided inverse to AB A B is both intuitive, and somewhat rigorous. Interestingly, as far as I know, the same identity, which holds in group theory more generally, is called the "socks and shoes identity".

The following question is problem Pinter's Abstract Algebra. And to put things in context: G is a group and a, b are elements of G. I want to show (ab)−1 = b−1a−1. I originally thought of proving the fact in the following manner:

2019年2月14日 · Is $y=ab $ you know $y^ {-1} $ exists, but you don't have *any* idea that if $y $ "breaks apart" to $a$ and $b $ you have no idea how $y^ {-1} $ breaks apart if it does at all.$\endgroup$

2016年4月25日 · Given $ (I + B (I - AB)^ {-1}A)$ to be inverse of $ (I + BA)$, how could we derive that the following alternative form holds $ (I + AB)^ {-1}A = A (I + BA)^ {-1}$.

I think you are missing the point of the proof. The middle line 1 ab = 1 a 1 b 1 a b = 1 a 1 b is precisely the thing you need to prove.

Give a counterexample to show that (AB)−1 (A B) − 1 doesn't equal A−1B−1 A − 1 B − 1 I'm not sure how to approach this, so I just used the idea that the matrix multiplication is not commutative. so it goes: AB doesn't equal BA now I just take the inverse of both sides if they are invertible (lets say they are) so I get B−1A−1 B − 1 A − 1 doesn't equal A−1B−1 A − 1 B ...

2010年9月2日 · 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, …

2024年9月18日 · In general for matrices A A and B B, it is not true that AB = BA A B = B A. Hence, if you have an equation X = Y X = Y, you can only conclude AX = AY A X = A Y or XA = YA X A = Y A. In the proof when you have

Hi I have a question about the following algebra rule A + AB = A My textbook explains this as follows A + AB = A This rule can be proved as such: Step 1: Dustributive Law: A + AB = A*1 = A(1+B) Huh...