Question

If ω is a complex cube root of unity and if 1, ω and ω² are the cube roots of unity and $\left[\begin{array}{cc}\omega & 0\\ 0& \omega \end{array}\right]$, then $A^{50}=$

• A. ${\omega }^{2}A$
• B. $\omega A$
• C. $A$
• D. $0$

Answer 1

The correct option is B

$\omega A$

Explanation for the correct option

Step 1: Information required for the solution

We need to find $A^n$, where $A$ is the given matrix and $n$ is the number represents how many times the same matrix has been multiplied by itself.

Also, $\omega$ is a complex cube root of unity and $1,\omega ,\text{and}{\omega }^2$ are the cube roots of unity implies that ${\omega }^3=1$.

Step 2: Calculation for the correct option

Here, we need to multiply the matrix again and again to determine $A^n$.

Since, $A=\left[\begin{array}{cc}\omega & 0\\ 0& \omega \end{array}\right]$ then

$\begin{array}{rcl}\therefore A^2& =& A\times A\\ & =& \left[\begin{array}{cc}\omega & 0\\ 0& \omega \end{array}\right]\times \left[\begin{array}{cc}\omega & 0\\ 0& \omega \end{array}\right]\\ & =& \left[\begin{array}{cc}{\omega }^2& 0\\ 0& {\omega }^2\end{array}\right]\end{array}$

Now, we will find $A^3$,

$\begin{array}{rcl}\therefore A^3& =& A^2\times A\\ & =& \left[\begin{array}{cc}{\omega }^2& 0\\ 0& {\omega }^2\end{array}\right]\times \left[\begin{array}{cc}\omega & 0\\ 0& \omega \end{array}\right]\\ & =& \left[\begin{array}{cc}{\omega }^3& 0\\ 0& {\omega }^3\end{array}\right]\end{array}$

Similarly, $A^4$ will be $\left[\begin{array}{cc}{\omega }^4& 0\\ 0& {\omega }^4\end{array}\right]$ and $A^n$ will be $\left[\begin{array}{cc}{\omega }^n& 0\\ 0& {\omega }^n\end{array}\right]$.

Step 3: Determination of the ${\mathit{A}}^{\mathbf{50}}$

Now, $A^{50}$ is equal to $\left[\begin{array}{cc}{\omega }^{50}& 0\\ 0& {\omega }^{50}\end{array}\right]$.

Also, we can write ${\omega }^{50}$ as,

$\begin{array}{rcl}\therefore {\omega }^{50}& =& {\omega }^{48}\times {\omega }^2\\ & =& {\left({\omega }^3\right)}^{16}\times {\omega }^2\end{array}$

Since, ${\omega }^3=1$ then

${\omega }^{50}={\omega }^2$

From this, we have

$\begin{array}{rcl}\therefore A^{50}& =& \left[\begin{array}{cc}{\omega }^2& 0\\ 0& {\omega }^2\end{array}\right]\\ & =& \omega \left[\begin{array}{cc}\omega & 0\\ 0& \omega \end{array}\right]\\ & =& \omega A\end{array}$

Hence, the correct option is (B)