Let ω=-1/2+{i} {{3}}{2}, {i}={-1}, then the value of ≤ft|{array}{ccc}1 & 1 & 1 \\ 1 & -1-ω² & ω² \\ 1 & ω² & ω^{4}{array}| is

Correct Answer is : 3ω(ω−1)3 ω(ω-1)3ω(ω−1)

3 ω(1-ω) §SOLUTION：§ {aligned} ω² &= ≤ft(-1/2)²+≤ft({i} {{3}}{2})²-2 {i}≤ft({{3}}{4})=-1/2-{{3}}{2} {i} \\ ω³ &= 1/4+3/4=1 \\ ∴ 1+ω+ω² &= 0 \\ {vmatrix} 1 & 1 & 1 \\ 1 & ω & ω² \\ 1 & ω² & ω {vmatrix} &= 1≤ft(ω²-ω^{4})-1≤ft(ω-ω²)+1≤ft(ω²-ω) \\ &= ω²-ω^{4}-ω+ω²+ω²-ω \\ &= -ω^{4}+3ω²-2ω \\ &= -ω+3ω²-2ω \\ &= 3ω²-3ω \\ &= 3ω(ω-1). {aligned} §Correct Answer：B§

Let ω=-1/2+{i} {{3}}{2}, {i}={-1}, then the value of ≤ft|{array}{ccc}1 & 1 & 1 \\ 1 & -1-ω² & ω² \\ 1 & ω² & ω^{4}{array}| is

Correct Answer is : 3ω(ω−1)3 ω(ω-1)3ω(ω−1)

Test Index

MHTCET 11 May 2024 Shift 2 Paper

Question : 143 of 150

Marks: +1, -0

Let

\omega=-\frac{1}{2}+\mathrm{i} \frac{\sqrt{3}}{2}, \mathrm{i}=\sqrt{-1}

, then the value of

\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1-\omega^{2} & \omega^{2} \\ 1 & \omega^{2} & \omega^{4}\end{array}\right|

$3 \omega$
$3 \omega(\omega-1)$
$3 \omega^{2}$
$3 \omega(1-\omega)$
§SOLUTION：§
$\begin{aligned} \omega^{2} &= \left(-\frac{1}{2}\right)^{2}+\left(\mathrm{i} \frac{\sqrt{3}}{2}\right)^{2}-2 \mathrm{i}\left(\frac{\sqrt{3}}{4}\right)=-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i} \\ \omega^{3} &= \frac{1}{4}+\frac{3}{4}=1 \\ \therefore 1+\omega+\omega^{2} &= 0 \\ \begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^{2} \\ 1 & \omega^{2} & \omega \end{vmatrix} &= 1\left(\omega^{2}-\omega^{4}\right)-1\left(\omega-\omega^{2}\right)+1\left(\omega^{2}-\omega\right) \\ &= \omega^{2}-\omega^{4}-\omega+\omega^{2}+\omega^{2}-\omega \\ &= -\omega^{4}+3\omega^{2}-2\omega \\ &= -\omega+3\omega^{2}-2\omega \\ &= 3\omega^{2}-3\omega \\ &= 3\omega(\omega-1). \end{aligned}$
§Correct Answer：B§

Validate

Solution:

Concept:

Use the cube root of unity properties:

\omega^3=1

1+\omega+\omega^2=0

, and hence

\omega^4=\omega

Explanation:

Since

1+\omega+\omega^2=0

, we get

-1-\omega^2=\omega

. Also,

\omega^4=\omega

So the determinant becomes:

\begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix}

Expanding:

\begin{aligned} \Delta &= 1(\omega^2-\omega^4)-1(\omega-\omega^2)+1(\omega^2-\omega) \\ &= \omega^2-\omega^4-\omega+\omega^2+\omega^2-\omega \\ &= 3\omega^2-\omega^4-2\omega. \end{aligned}

Using

\omega^4=\omega

\Delta=3\omega^2-3\omega=3\omega(\omega-1).

Answer:

3\omega(\omega-1)

, i.e. option B.

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