Question

Given $A$ and $B$ are two non-singular matrix such that $B\ne I,A^5=I$ and $A{B}^2=BA$ , then the least value of $n$ for which $B^n=I$ is

• A. $63$
• B. $64$
• C. $31$
• D. $32$

Answer 1

Answer: (C)

Given $A{B}^2=BA$
Pre-multiplying by $A^4$ on both the sides, we get
$A^5{B}^2=A^{4}BA$
Post-multiplying by $A^4$ on both the sides, we get
$A^5{B}^2{A}^4=A^{4}B{A}^5$
$\Rightarrow I{B}^2{A}^4=A^{4}BI$ $\left[\because A^5=I\right]$
$\Rightarrow B^2{A}^4=A^{4}B$ $\left[\because IA=AI=A\right]$
Pre-multiplying by $A$ on both the sides, we get
$A{B}^2{A}^4=A^{5}B$
$\Rightarrow A{B}^2{A}^4=IB$ $\left[\because A^5=I\right]$
$\Rightarrow A{B}^2{A}^4=B..........\left(i\right)$ $\left[\because IA=AI=A\right]$
Now, substituting the value of $B$ from equation $\left(i\right)$ in the LHS of equation $\left(i\right)$ itself, we get
$A{\left(A{B}^2{A}^4\right)}^2{A}^4=B$
$\Rightarrow A\left(A{B}^2{A}^4\right)\left(A{B}^2{A}^4\right)A^4=B$
$\Rightarrow A^2{B}^2{A}^5{B}^2{A}^8=B$
$\Rightarrow A^2{B}^4{A}^3=B.............\left(ii\right)$ $\left[\because A^5=I\&IA=AI=A\right]$
Now, substituting the value of $B$ from equation $\left(i\right)$ in the LHS of equation $\left(ii\right)$ itself, we get
$A^2{\left(A{B}^2{A}^4\right)}^4{A}^3=B$
$\Rightarrow A^2\left(A{B}^2{A}^4\right)\left(A{B}^2{A}^4\right)\left(A{B}^2{A}^4\right)\left(A{B}^2{A}^4\right)A^3=B$
$\Rightarrow A^3{B}^2{A}^5{B}^2{A}^5{B}^2{A}^5{B}^2{A}^7=B$
$\Rightarrow A^3{B}^8{A}^2=B.............\left(iii\right)$ $\left[\because A^5=I\&IA=AI=A\right]$
From all the equations $\left(i\right),\left(ii\right)\&\left(iii\right)$ , we can say that for the form $A^p{B}^q{A}^r=B$ , $p$ is increasing by $1$ , $q$ is increasing in a GP as $2,4,8$ and so on and $r$ is decreasing by $1$ .
In the same manner, we find that when $p=5$ , we get $q=2^5=32$ and $r=0$ .
So, the expression becomes $A^5{B}^{32}{A}^0=B$
$\Rightarrow B^{32}=B$ $\left[\because A^5=I\right]$
Post-multiplying $B^{-1}$ on both the sides, we get
$B^{31}\left(B{B}^{-1}\right)=B{B}^{-1}$
$\Rightarrow B^{31}=I$ $\left[\because A{A}^{-1}=I\&AI=IA=A\right]$
So, the least value of $n$ for which $B^n=I$ is $31$ .