Q101 / 180 · KCET 2018 · Mathematics
If \( A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \), then \( A^{n}=2^{k} A \), where \( K= \)
- A\( 2^{n-1} \)
- Bn \( +1 \)
- Cn-1
- Correct answerD\( 2(n-1) \)
Solution
Given that \( A=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \)
and \( A^{n}=2^{k} A \)
So, \( A^{2}=\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \)
\( A^{3}=\left[\begin{array}{cc}8 & -8 \\ -8 & 8\end{array}\right]=2^{2} A \)
\( =\left[\begin{array}{cc}32 & -32 \\ -32 & 32\end{array}\right]=2^{4} A \)
\( A^{4}=\left[\begin{array}{cc}32 & -32 \\ -32 & 32\end{array}\right]\left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right] \)
\( =\left[\begin{array}{cc}128 & -128 \\ -128 & 128\end{array}\right]=2^{6} A \)
So, \( k=2(n-1) \)
Hence, the correct answer is option: 'D'.