Question

If $a_r = cos \frac{2\pi}{9} + isin \frac{2\pi}{9}; r=1,2,3....i = \sqrt{-1}$ then the determinant $\begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{vmatrix}$ is equal to:

• A. $a_2a_6 - a_4a_8$
• B. $a_9$
• C. $a_1a_9 - a_3a_7$
• D. $a_5$

Answer 1

Answer: (C)

Option C: $a_1a_9 - a_3a_7$

Answer 2

We are given

$$a_r=\cos\frac{2\pi r}{9}+ i\sin\frac{2\pi r}{9} = \omega^r,\quad\text{with }\omega=\exp\left(\frac{2\pi i}{9}\right)$$

so that

$$a_1=\omega,\quad a_2=\omega^2,\quad a_3=\omega^3,\quad a_4=\omega^4,\quad \ldots,\quad a_9=\omega^9=1.$$

The $3\times3$ matrix is

$$M=\begin{pmatrix} \omega & \omega^2 & \omega^3 \\ \omega^4 & \omega^5 & \omega^6 \\ \omega^7 & \omega^8 & \omega^9 \end{pmatrix}.$$

Observation by factoring rows:

• Factor $\omega^0=1$ from the 1st row
• Factor $\omega^3$ from the 2nd row, since
  $\omega^4=\omega^3\cdot\omega,\; \omega^5=\omega^3\cdot\omega^2,\; \omega^6=\omega^3\cdot\omega^3$.
• Factor $\omega^6$ from the 3rd row, since
  $\omega^7=\omega^6\cdot\omega,\; \omega^8=\omega^6\cdot\omega^2,\; \omega^9=\omega^6\cdot\omega^3$.

After factoring, the determinant becomes

$$\det M = \omega^{0+3+6}\begin{vmatrix} \omega & \omega^2 & \omega^3 \\ \omega & \omega^2 & \omega^3 \\ \omega & \omega^2 & \omega^3 \end{vmatrix} = \omega^9 \cdot 0 = 0.$$

(The three rows become identical, hence the determinant is 0.)

Relating to options:
The given options include one choice which also evaluates to 0. Check option C:

$$a_1a_9 - a_3a_7=\omega\cdot\omega^9-\omega^3\cdot\omega^7 = \omega\cdot1-\omega^{10}.$$

Since $\omega^9=1$ we have $\omega^{10}=\omega^1=\omega$. Thus,

$$\omega-\omega = 0.$$

Thus, the determinant equals option C.