Question

a and b are 2 square matrices such that A^2B=BA ad if AB)^10 =?

Answer 1

A^{1023}B^{10}

Answer 2

Given that

$A^2B = BA$,

we can show by induction that

$(AB)^n = A^{2^n-1}B^n$.

Steps:

1. For $n=1$,

$(AB)^1 = A^{2^1-1}B^1 = A^1B$.

2. For $n=2$, write

$(AB)^2 = ABAB = A(BA)B$.

Using $BA = A^2B$, we get

$(AB)^2 = AA^2BB = A^3B^2$,

and since $3 = 2^2-1$, we have

$(AB)^2 = A^{2^2-1}B^2$.

3. Assume the formula holds for $n$; then for $n+1$,

$(AB)^{n+1} = (AB)^n(AB) = A^{2^n-1}B^n \cdot AB$.

Using the commutation relation repeatedly, it can be shown that

$(AB)^{n+1} = A^{2^{n+1}-1}B^{n+1}$.

For $n=10$:

$(AB)^{10} = A^{2^{10}-1}B^{10} = A^{1024-1}B^{10} = A^{1023}B^{10}$.

The key is to prove by induction that $(AB)^n = A^{2^n-1}B^n$ using the relation $A^2B = BA$. Then, setting $n=10$ gives the answer.