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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Matrices
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If $A$ and $B$ are square matrices of the same order and A is non-singular then for a positive integer n,$(A^{-1}BA)^n$ is equal to

$\begin{array}{1 1}(A)\;A^{-n}B^nA^n&(B)\;A^nB^nA^{-n}\\(C)\;A^{-1}B^nA&(D)\;n(A^{-1}BA)\end{array}$

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1 Answer

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$(A^{-1}BA)^2=(A^{-1}BA)(A^{-1}BA)$
$\Rightarrow A^{-1}B(AA^{-1})BA$
$\Rightarrow A^{-1}BIBA$
$(A^{-1}BA)^3=(A^{-1}B^2A)(A^{-1}BA)$
$\Rightarrow A^{-1}B^2(AA^{-1})BA$
$\Rightarrow A^{-1}B^3A$
$\therefore (A^{-1}BA)^n=A^{-1}B^nA$
Hence (C) is the correct answer.
answered Apr 23, 2014 by sreemathi.v
 

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