Question

If \(A=\begin{pmatrix}0&0&2\\1&0&0\\0&-1&0\end{pmatrix}\), then :

• A. \(A^{6}=2A^{3}\)
• B. \(2A^{6}=A^{3}\)
• C. \(A^{12}=4A^{6}\)
• D. \(A^{12}=2A^{6}\)

Hint:

Compute \(A^2\) and \(A^3\); the cube turns out to be a scalar matrix.

Answer 1

The Correct Option is C

Multiplying, \(A^2=\begin{pmatrix}0&-2&0\\0&0&2\\-1&0&0\end{pmatrix}\).
Then \(A^3=A^2A=\begin{pmatrix}-2&0&0\\0&-2&0\\0&0&-2\end{pmatrix}=-2I\).
Hence \(A^6=(A^3)^2=4I\) and \(A^{12}=(A^6)^2=16I\).
(1) \(2A^3=-4I\ne 4I=A^6\). (2) \(2A^6=8I\ne -2I=A^3\). (3) \(4A^6=16I=A^{12}\), true. (4) \(2A^6=8I\ne 16I=A^{12}\).
Answer: (3) \(A^{12}=4A^{6}\)