Question

Let $A$ and $B$ be two non-singular matrices such that $A^6=I$ and $A{B}^2=BA(B\ne I)$. If $B^m=I$ (where $I$ is the identity matrix), then the least value of $m$ is

Answer 1

Step- 1: Simplify the given conditions

Given, $A^6=I,A{B}^2=BA,B^m=I$

Consider $A,B$ to be $n\times n$ matrices which are non singular $\Rightarrow \left|A\right|\ne 0,\left|B\right|\ne 0$

Consider equation

$A{B}^2=BA$ $...\left(i\right)$

Multiply equation $\left(i\right)$ by $A^5$

$A{B}^2{A}^5=B{A}^6$ $...\left(ii\right)$

Now,

Pre multiply equation $\left(ii\right)$ by $A^5$

$A^6{B}^2{A}^5=A^{5}B{A}^6$ $...\left(iii\right)$

Now, as $A^6=I$, equation $\left(iii\right)$ can be reduced to

$B^2{A}^5=A^{5}B$ $...\left(iv\right)$

Again, Pre multiply equation $\left(iv\right)$ by $A$

$A{B}^2{A}^5=A^{6}B$

$\Rightarrow$ $B=A{B}^2{A}^5$ $...(\because A^6=I)$$..\left(v\right)$

Step- 2: Solve the RHS of equation $\left(v\right)$

$B=A{\left(A{B}^2{A}^5\right)}^2{A}^5$

$\Rightarrow$ $B=AA{B}^2{A}^{5}A{B}^2{A}^5{A}^5$

$\Rightarrow$ $B=A^2{B}^2{A}^6{B}^2{A}^{10}$

$\Rightarrow$ $B=A^2{B}^4{A}^6{A}^4$ $...(\because A^6=I)$

$\Rightarrow$ $B=A^2{B}^4{A}^4$ $...(\because A^6=I)$$...\left(vi\right)$

Substitute value of $B$ from equation $\left(v\right)$ in RHS of equation $\left(vi\right)$

$B=A^2{\left(A{B}^2{A}^5\right)}^4{A}^4$

$\Rightarrow$$B=A^{2}A{B}^2{A}^{5}A{B}^2{A}^{5}A{B}^2{A}^{5}A{B}^2{A}^5{A}^4$

$\Rightarrow$ $B=A^3{B}^2{A}^6{B}^2{A}^6{B}^2{A}^6{B}^2{A}^9$

$\Rightarrow$ $B=A^3{B}^8{A}^3$ $...(\because A^6=I)$$...\left(vii\right)$

Step- 3: Analyse the pattern of equations obtained

The three equations are $\left(v\right),\left(vi\right),\left(vii\right)$

We can observe that $B=A^l{B}^m{A}^n$ where,

$l=$increasing by $1$

$m=$ a GP of $2,4,8,..$

$n=$ decreasing by $1$

When $l=6,m=64,n=0$

$B=A^6{B}^{64}{A}^0$

$\Rightarrow$$B=B^{64}$ $...(\because A^6=I)$$...\left(viii\right)$

Multiply both sides of equation $\left(viii\right)$ by $B^{-1}$

$B{B}^{-1}=B^{64}{B}^{-1}$

$\Rightarrow$ $B^{63}=I$ $...(\because B{B}^{-1}=I)$

On comparing $B^m=B^{63}$ we get,

$m=63$

Hence, the least possible value of $m$ is $63$.