Question

A and B are two non singular matrices so that $A^6=IandA{B}^2=BA(B\ne I).AvalueofKsothat{B}^K=Iis$

• A. 31
• B. 32
• C. 63
• D. 64

Answer 1

The correct option is C 63

$A^5\left(A{B}^2\right)=A^{5}BA.$
$\Rightarrow B^2=A^{5}BA$
$\Rightarrow B^4=\left(A^{5}BA\right)\left(A^{5}BA\right)=A^5{B}^{2}A=A^5\left(A^{5}BA\right)A$
$\Rightarrow B^4=A^{4}B{A}^2$
$\Rightarrow B^8=\left(A^{4}B{A}^2\right)$$\left(A^{4}B{A}^2\right)=A^4{B}^2{A}^2=A^4\left(A^{5}BA\right)A^2$
$\Rightarrow B^8=A^{3}B{A}^3$
$\Rightarrow B^{16}=\left(A^{3}B{A}^3\right)\left(A^{3}B{A}^3\right)=A^3{B}^2{A}^3=A^3\left(A^{5}BA\right)A^3=A^{2}B{A}^4$
$A^{32}=\left(A^{2}B{A}^4\right)\left(A^{2}B{A}^4\right)=A^2{B}^2{A}^4=A^2\left(A^{5}BA\right)A^4=AB{A}^5$
$A^{64}=\left(AB{A}^5\right)\left(AB{A}^5\right)=A{B}^2{A}^5=A\left(A^{5}BA\right)A^5=B$ $\Rightarrow A^{63}=I$