Let = - 1 2 + i 3 2 , then the value of the determinant | 1 1 1 1 - 1 - 2 2 1 2 4 | is [2002]

Question

Let $ω = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$ , then the value of the determinant $| \begin{matrix} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{matrix} |$ is [2002]

Smart Achievers · Accepted Answer

(2)

$Applying R_{1} \to R_{1} + R_{2} + R_{3}, we get$

$| \begin{matrix} 1 & 1 & 0 \\ 1 & - 1 - ω & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{matrix} |$ $=$ $| \begin{matrix} 3 & 0 & 0 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{matrix} |$

$= 3 [- ω - 1 - ω]$ $= 3 (ω^{2} - ω)$

Solution

Q.
Let $ω = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$ , then the value of the determinant $| \begin{matrix} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{matrix} |$ is [2002]

1 $3 ω$

2 $3 ω (ω - 1)$

3 $3 ω^{2}$

4 $3 ω (1 - ω)$

Ans.
(2)

$Applying R_{1} \to R_{1} + R_{2} + R_{3}, we get$

$| \begin{matrix} 1 & 1 & 0 \\ 1 & - 1 - ω & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{matrix} |$ $=$ $| \begin{matrix} 3 & 0 & 0 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{matrix} |$

$= 3 [- ω - 1 - ω]$ $= 3 (ω^{2} - ω)$

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Solution

Q. Let ω=-12+i32, then the value of the determinant |1111-1-ω2ω21ω2ω4| is [2002]

1 3ω

2 3ω(ω-1)

3 3ω2

4 3ω(1-ω)

Ans. (2) Applying R1→R1+R2+R3, we get |1101-1-ωω21ω2ω4|=|3001-1-ω2ω21ω2ω4| =3[-ω-1-ω]=3(ω2-ω)