Question

If $A = \begin{bmatrix}0&-1\\\ 1&0\end{bmatrix}$ then $A^{16}$ is equal to :

• 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. $\begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix}$

Answer 1

Answer: (D)

$\begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix}$

Answer 2

We have $A = \begin{bmatrix}0&-1\\\ 1&0\end{bmatrix}$ Now, $A^{2} = A.A = \begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}$ $= \begin{pmatrix}-1&0\\\ 0&-1\end{pmatrix} = - I$ where $I = \begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$ is identity matrix $\left(A^{2}\right)^{8} = \left(-I\right)^{8} = I$ Hence, $A^{16} = I$