Question

Given $A$ and $B$ are two non-singular matrix such that $B\neq I,A^{5}=I$ and $AB^{2}=BA$ , 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 $AB^{2}=BA$
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
$A^{5}B^{2}A^{4}=A^{4}BA^{5}$
$\Rightarrow IB^{2}A^{4}=A^{4}BI$ $\left[\because A^{5} = I\right]$
$\Rightarrow B^{2}A^{4}=A^{4}B$ $\left[\because I A = A I = A\right]$
Pre-multiplying by $A$ on both the sides, we get
$AB^{2}A^{4}=A^{5}B$
$\Rightarrow AB^{2}A^{4}=IB$ $\left[\because A^{5} = I\right]$
$\Rightarrow AB^{2}A^{4}=B..........\left(i\right)$ $\left[\because I A = A I = A\right]$
Now, substituting the value of $B$ from equation $\left(i\right)$ in the LHS of equation $\left(i\right)$ itself, we get
$A\left(A B^{2} A^{4}\right)^{2}A^{4}=B$
$\Rightarrow A\left(A B^{2} A^{4}\right)\left(A B^{2} A^{4}\right)A^{4}=B$
$\Rightarrow A^{2}B^{2}A^{5}B^{2}A^{8}=B$
$\Rightarrow A^{2}B^{4}A^{3}=B.............\left(i i\right)$ $\left[\because A^{5} = I \& I A = A I = A\right]$
Now, substituting the value of $B$ from equation $\left(i\right)$ in the LHS of equation $\left(i i\right)$ itself, we get
$A^{2}\left(A B^{2} A^{4}\right)^{4}A^{3}=B$
$\Rightarrow A^{2}\left(A B^{2} A^{4}\right)\left(A B^{2} A^{4}\right)\left(A B^{2} A^{4}\right)\left(A B^{2} A^{4}\right)A^{3}=B$
$\Rightarrow A^{3}B^{2}A^{5}B^{2}A^{5}B^{2}A^{5}B^{2}A^{7}=B$
$\Rightarrow A^{3}B^{8}A^{2}=B.............\left(i i i\right)$ $\left[\because A^{5} = I \& I A = A I = A\right]$
From all the equations $\left(i\right),\left(i i\right)\&\left(i i i\right)$ , 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
$B^{31}\left(B B^{- 1}\right)=BB^{- 1}$
$\Rightarrow B^{31}=I$ $\left[\because A A^{- 1} = I \& A I = I A = A\right]$
So, the least value of $n$ for which $B^{n}=I$ is $31$ .