Question

[tex] If A= \left[\begin{array}{ccc}cos\frac{2π}{3}&-sin\frac{2π}{3}\\sin\frac{2π}{3}&cos\frac{2π}{3}\end{array}\right], then A³=.........

(a) \left[\begin{array}{ccc}0&1\\1&0\end{array}\right]

(b) \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

(c) \left[\begin{array}{ccc}1&1\\0&0\end{array}\right]

(d) \left[\begin{array}{ccc}0&0\\1&1\end{array}\right] [/tex]

Answer 1

Answer:

The correct answer is (b)

Step-by-step explanation:

The general matrix

$\left[\begin{array}{ccc}cos \theta &-sin \theta\\sin \theta&cos \theta\end{array}\right]$

is a rotation matrix. It represents a rotation of $\theta$ degrees counter-clockwise.

In this case, we have $\theta = \frac{2 \pi}{3}$, so when you do $A^3$ you are rotating three times by $\frac{2 \pi}{3}$, which gives a rotation of $2\pi$, that is, a full rotation that returns to the inicial point. So $A^3$ does nothing, that is, $A^3 = Id$, the identity matrix.