Question

Let A & B are two non-singular square matrices such that $B\ne I$ and $A{B}^2=BA$. If $A^5=B^{-1}{A}^5{B}^K$ then value of K is

• A. 15
• B. 16
• C. 31
• D. 32

Answer 1

The correct option is D 32

$A^5=B^{-1}{A}^5{B}^K$
$B{A}^5=A^5{B}^K$
Solve this by making pattern for $B{A}^2,B{A}^3$,...so on
$B{A}^2=BA.A=A{B}^{2}A=A.B.BA$
$=ABA{B}^2$
$=AA{B}^2.B^2$
$B{A}^2=A^2{B}^4-\left(1\right)\Rightarrow B{A}^2=A^2{B}^{2^2}$
$B{A}^3=B{A}^2.A=A^2{B}^4.A$ (using (1))
$=A^2.B^3.A{B}^2$
$=A^2{B}^2.A{B}^2.B^2$
$=A^{2}B.A{B}^2.B^4$
$=A^2.A{B}^2{B}^6$
$B{A}^3=A^3{B}^8\Rightarrow A^3.B^{2^3}$
by obsobing pattern
$B{A}^5=A^5{B}^{2^5}$
So $A^5=B^{-1}{A}^5{B}^{2^5}$
So K =32