Q92 / 160 · TS EAMCET 2012 · Mathematics

$A=\left[\begin{array}{rr}i & -i \\ -i & i\end{array}\right], B=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right] \Rightarrow A^8$

  • A
    $4 B$
  • B
    $8 B$
  • C
    $64 B$
  • Correct answer
    D
    $128 B$

Solution

Given, $A=\left[\begin{array}{rr}i & -i \\ -i & i\end{array}\right]$ and $B=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right]$ $\begin{aligned} \therefore A^2 & =\left[\begin{array}{rr}i & -i \\ -i & i\end{array}\right]\left[\begin{array}{rr}i & -i \\ -i & i\end{array}\right] \\ & =\left[\begin{array}{rr}i^2+i^2 & -i^2-i^2 \\ -i^2-i^2 & i^2+i^2\end{array}\right] \\ & =\left[\begin{array}{rr}-1-1 & 1+1 \\ 1+1 & -1-1\end{array}\right]=\left[\begin{array}{rr}-2 & 2 \\ 2 & -2\end{array}\right]=-2 B \\ B^2 & =\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right] \\ & =\left[\begin{array}{rr}1+1 & -1-1 \\ -1-1 & 1+1\end{array}\right]=\left[\begin{array}{rr}2 & -2 \\ -2 & 2\end{array}\right] \\ & =2 B\end{aligned}$ $\begin{aligned} & \text { Now, } A^8=\left(A^2\right)^4=(-2 B)^4 \\ & =16 B^4=16\left(B^2\right)^2 16 \cdot(2 B)^2 \\ & =16 \times 4 \times B^2=16 \times 4 \times 2 B \quad \text { [from Eq. (i)] } \\ & =128 B\end{aligned}$

Hence, the correct answer is option: 'D'.

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