Question

If $A=\left(\begin{array}{cc}i& -i\\ -i& i\end{array}\right)$ and $B=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$, then $A^8$ equals

• A. $128B$
• B. $23B$
• C. $16B$
• D. $64B$

Answer 1

Answer: (A)

$128B$

$A=\left(\begin{array}{cc}i& -i\\ -i& i\end{array}\right)$ and $B=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$ ...Given

Now, $A=i\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$

$\therefore A=iB$So, $A^2=(iB{)}^2$

$\Rightarrow A^2=i^2{B}^2$

$\Rightarrow (A^2{)}^4=(i^2{B}^2{)}^4$

$\Rightarrow A^8=(i^2{)}^4.B^8$

$\Rightarrow A^8=(-1{)}^4.B^8$ ...$[i^2=-1]$

$\Rightarrow A^8=B^8$

Now, $B^2=B.B$

So, $B^2=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$$\Rightarrow B^2=\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$

$\Rightarrow B^2=2\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$

$\Rightarrow B^2=2B$

$\Rightarrow (B^2{)}^4=(2B{)}^4$

$\Rightarrow B^8=16B^4$

$\Rightarrow B^8=16(B^2{)}^2$

$\Rightarrow B^8=16(2B{)}^2$

$\Rightarrow B^8=64B^2$

$\Rightarrow B^8=128B$

So, $A^8=128B$