Question

Suppose $A$ and $B$ are two $3\times 3$ non-singular matrices such that $(AB{)}^k=A^k{B}^k$
$fork=2008,2009,2010$, then

• A. $A{B}^{-1}{A}^{-1}=B^{-1}$
• B. $A^{-1}{B}^{-2}A=B^2$
• C. $AB=BA$
• D. $A^{-2}{B}^2{A}^2=(A^{-1}BA{)}^2$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $A{B}^{-1}{A}^{-1}=B^{-1}$
C $AB=BA$
D $A^{-2}{B}^2{A}^2=(A^{-1}BA{)}^2$

${\left(AB\right)}^{2010}=A^{2010}.B^{2010}$

$A.B.{\left(AB\right)}^{2009}=A^{2010}.B^{2010}$

$A^{-1}.A.B.{\left(AB\right)}^{2009}=A^{-1}.A.A^{2009}.B^{2010}$

$B.{\left(AB\right)}^{2009}=A^{2009}.B^{2010}$

$B.A^{2009}.B^{2009}=A^{2009}.B^{2010}$

$B.A.A^{2008}.B^{2008}.B=A^{2009}.B^{2009}B$

$B.A.A^{2008}.B^{2008}.B.B^{-1}=A^{2009}.B^{2009}.B.B^{-1}$

$B.A.{\left(AB\right)}^{2008}={\left(AB\right)}^{2009}$

$=A.B.{\left(AB\right)}^{2008}$

$B.A.{\left(AB\right)}^{2008}{\left({\left(AB\right)}^{2008}\right)}^{-1}=A.B.{\left(AB\right)}^{2008}{\left({\left(AB\right)}^{2008}\right)}^{-1}$

$\Rightarrow B.A=A.B-\left(C\right)$ option

We know $A.B^{-1}=A.B^{-1}$

$A.B^{-1}.A^{-1}=A.B^{-1}.A^{-1}$ $(\because {\left(AB\right)}^{-1}=B^{-1}.A^{-1})$
$=A.{\left(AB\right)}^{-1}$
$=A.{\left(BA\right)}^{-1}$ $(\because BA=AB)$
$=A.A^{-1}.B^{-1}$
$=B^{-1}$
$\therefore A.B^{-1}.A^{-1}=B^{-1}-\left(A\right)$ option
$A^{-2}.B^2.A^2={(A.B.A^{-1})}^2$ $[\because {\left(ABC\right)}^n=C^n.B^n.A^n]$
$A.B=B.A$

$A^{-1}.A.B=A^{-1}.B.A$
$B=A^{-1}.B.A$
$B.A^{-1}=A^{-1}.B-\left(1\right)$
So as $AB=BA$

$A.B.A^{-1}=B.A.A^{-1}$
$=B$

$=B.A^{-1}.A$

$=A^{-1}B.A$ $[\because B{A}^{-1}=A^{-1}B]$ from $\left(1\right)$

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