A and B are two square matrix such that A^2B = BA and If (AB)^{10} = A^K B^{10} then k is

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3 and 4 .Determinants and Matrices

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$A$ and $B$ are two square matrix such that $A^2B = BA$ and If $(AB)^{10} = A^K B^{10}$ then $k$ is

A

$10$

B

$1001$

C

$1023$

D

$1042$

Solution

$(\mathrm{AB})^{2}=(\mathrm{AB})(\mathrm{AB})=\mathrm{A}(\mathrm{BA}) \mathrm{B}=\mathrm{A}^{3} \mathrm{B}^{2}$

Now $(A B)^{3}=A B \,A \,B A\, B=A^{7} B^{3}$

so ${({\rm{AB}})^{\rm{n}}} = {({\rm{A}})^{{2^n} – 1}}{{\rm{B}}^{\rm{n}}}$

so $\mathrm{k}=2^{10}-1=1023$

Standard 12

Mathematics

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