GATE Overflow Test Series | Linear Algebra | Test 1 | Question: 15

Question

If $\omega$ is the one of the imaginary cube roots of unity, the value of the determinant $\begin{vmatrix}
1 & \omega & \omega^{2}\\
\omega & \omega^{2} & 1\\
\omega^{2} & 1 & \omega
\end{vmatrix} =$ _____
(Mark all the appropriate choices)

• A. $1$
• B. $\omega^{3}$
• C. $0$
• D. $1 + \omega + \omega^{2}$

Answer 1

Let $\Delta = \begin{vmatrix}
1 & \omega & \omega^{2}  \\
  \omega & \omega^{2} & 1  \\
 \omega^{2} & 1 & \omega
\end{vmatrix} $

Perform the operation $R_{1} \rightarrow R_{1} + R_{2} + R_{3},$ we get

$\Delta = \begin{vmatrix}
1 + \omega + \omega^{2} & \omega + \omega^{2} + 1 & \omega^{2} + 1  + \omega \\
  \omega & \omega^{2} & 1  \\
 \omega^{2} & 1 & \omega
\end{vmatrix} $

$ \implies \Delta = \begin{vmatrix}
0 & 0 & 0 \\
  \omega & \omega^{2} & 1  \\
 \omega^{2} & 1 & \omega
\end{vmatrix} \quad \left(\because 1 + \omega + \omega^{2} = 0\right)$

$\implies \Delta = 0.$

So, the correct answer is $C;D.$

References:

• https://math.vanderbilt.edu/sapirmv/msapir/proofdet2.html
• https://web.archive.org/web/20200615051706/https://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/least-squares-determinants-and-eigenvalues/properties-of-determinants/MIT18_06SCF11_Ses2.5sum.pdf