Determinants Matrices Question 90
Question: If $ A= \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix} $ , then $ A^{16} $ is equal to:
Options:
A) $ \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix} $
B) $ \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} $
C) $ \begin{bmatrix} -1 & 0 \\ 0 & 1 \\ \end{bmatrix} $
D) $ \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $
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Answer:
Correct Answer: D
Solution:
- [d] We have $ A= \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix} $ Now, $ A^{2}=A.A=( \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} )( \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} )=( \begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} )=-I $ where $ I=( \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} ) $ is identity matrix $ {{(A^{2})}^{8}}={{(-I)}^{8}}=I $ . Hence, $ A^{16}=I $
BETA
