Question

If w¹ 1 is cube root of unity and x + y + z ¹ 0 then $\left| \begin{matrix} \frac{x}{1 + \omega} & \frac{y}{\omega + \omega^{2}} & \frac{z}{\omega^{2} + 1} \\ \frac{y}{\omega + \omega^{2}} & \frac{z}{\omega^{2} + 1} & \frac{x}{1 + \omega} \\ \frac{z}{\omega^{2} + 1} & \frac{x}{1 + \omega} & \frac{y}{\omega + \omega^{2}} \end{matrix} \right|$ = 0 if

• A. x² + y² + z² = 0
• B. x + yw + zw² = or x = y = z
• C. x ¹ y ¹ z ¹ 0
• D. x = 2y = 3z

Answer 1

Answer: (B)

x + yw + zw² = or x = y = z

Answer 2

As 1 + w + w² = 0

$\left| \begin{matrix}

• \frac{x}{\omega^{2}} & - y & - \frac{z}{\omega} \
• y & - \frac{z}{\omega} & - \frac{x}{\omega^{2}} \
• \frac{z}{\omega} & - \frac{x}{\omega^{2}} & - y \end{matrix} \right|$ = x³ + y³ + z³ – 3xyz

= $\frac{1}{2}$ (x + y + z) {(x – y)² + (y – z)² + (z – x)²} = x + yw + zw² (x² + y² w² + z²w – xyw – yz – zxw²) The determinant varnishes

if x = y = z or x + yw + zw² = 0