We have the following group: aabb a a b b It is commutative, so abab is the same as aabb. I have figured out this is a combinatorics question. Because abab is the same as aabb. I was how to solve …

The Problem Let A A be the matrix (a c b d) (a b c d), where no one of a,b,c,d a, b, c, d is 0 0. Let B B be a 2×2 2 × 2 matrix such that AB+ BA= (0 0 0 0) A B + B ...

Apr 19, 2022 · Although both belong to a much broad combination of N=2 and n=4 (AAAA, ABBA, BBBB...), where order matters and repetition is allowed, both can be rearranged in different ways: …

Oct 4, 2016 · The algorithm is normally created by taking AB, then inverting each 2-state 'digit' and sticking it on the end (ABBA). You then take this entire sequence and repeat the process (ABBABAAB).

Let A, B A, B be two 3 × 3 3 × 3 matrices with complex entries, such that A2 = AB + BA A 2 = A B + B A. Prove that det(AB − BA) = 0 det (A B B A) = 0 Nice problem, and I want to find a solution. AB − BA = …

{AABB, ABAB, ABBA, BBAA, BAAB, BABA} {A A B B, A B A B, A B B A, B B A A, B A A B, B A B A} So, I need to know what this w(n) = w(ni,nj) w (n) = w (n i, n j) is. Some references on aproaches on how …

Jun 23, 2022 · Question: prove Γ(a)Γ(b) = Γ(a + b)B(a, b) Γ (a) Γ (b) = Γ (a + b) B (a, b) using polar transformation: my attempt: L.H.S =

The 4 4 -digit palindrome abba a b b a is divisible by 101 iff a = b a = b. The 5 5 -digit palindrome abcba a b c b a is divisible by 101 iff c = 2a c = 2 a. The 6 6 -digit palindrome abccba a b c c b a is divisible …

The commutator [X, Y] of two matrices is defined by the equation $$\begin {align} [X, Y] = XY − YX. \end {align}$$ Two anti-commuting matrices A and B satisfy $$\begin {align} A^2=I \qu...

For example a palindrome of length $4$ is always divisible by $11$ because palindromes of length $4$ are in the form of: $$\\overline{abba}$$ so it is equal to $$1001a+110b$$ and $1001$ and $110$ are