Question

$A$ and $B$ are two square matrices such that $A^{2}B=BA$ and if $(AB{)}^{10}=A^k.B^{10}$ then the value of $k-1020$ is

Answer 1

$(AB{)}^2=AB.AB$
$=A.A^2{B}^2=A^3{B}^2$

$(AB{)}^3=(AB{)}^{2}AB$
$=A^3{B}^{2}AB=A^{3}B.A^{2}B.B=A^3.A^{2}B.ABB$
$=A^5.A^{2}B.B.B$
$=A^7{B}^3$
$=A^{2^3-1}.B^3$
So, $(AB{)}^{10}=A^{2^{10}-1}.B^{10}$
$\Rightarrow k=2^{10}-1$