- 11 | abba, where a and b are the digits in a 4 digit number.
Truly lost here, I know abba could look anything like 1221 or even 9999 However how do I prove 11 divides all of the possiblities?
- How to calculate total combinations for AABB and ABBB sets?
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: First one: AABB, BBAA,
- How many $4$-digit palindromes are divisible by $3$?
Hint: in digits the number is $abba$ with $2 (a+b)$ divisible by $3$
- If $A$ is positive definite, $B$ is self adjoint, and $AB+BA$ is . . .
Suppose $A$ and $B$ are complex linear operators of some finite-dimensional vector space $X$ In my definition, positive definite operators are only for self-adjoint
- matrices - When will $AB=BA$? - Mathematics Stack Exchange
Given two square matrices $A,B$ with same dimension, what conditions will lead to this result? Or what result will this condition lead to? I thought this is a quite
- How many ways can we get 2 as and 2 bs from aabb?
Because abab is the same as aabb I was how to solve these problems with the blank slot method, i e _ _ _ _ If I do this manually, it's clear to me the answer is 6, aabb abab abba baba bbaa baab Which is the same as $$\binom {4} {2}$$ But I don't really understand why this is true? How is this supposed to be done without brute forcing the
- $A^2=AB+BA$. Prove that $\det (AB-BA)=0$ [duplicate]
I get the trick Use the fact that matrices "commute under determinants" +1
- The commutator of two matrices - Mathematics Stack Exchange
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
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