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AP EAPCET 23-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 3 of 160

Marks: +1, -0

If

\omega

is a root of the equation

x+\frac{1}{x}+1=0

\begin{vmatrix} 1 & 1+\omega & 1+\omega+\omega^{2} \\ 3 & 4+3\omega & 5+4\omega+3\omega^{2} \\ 6 & 9+6\omega & 11+9\omega+6\omega^{2} \end{vmatrix}=

$1$
$-1$
$0$
$1+\omega$

Solution:

\omega

is root of

x+\frac{1}{x}+1=0

, i.e.

x^{2}+x+1=0

This gives roots

\frac{-1 \pm \sqrt{1-4}}{2} = \frac{-1 + i \sqrt{3}}{2}

\therefore \omega = \frac{-1 + \sqrt{-3}}{2}

and

\omega^{2} = \frac{-1 - \sqrt{-3}}{2}

are roots of

x^{2}+x+1=0

Also,

\omega^{3} = \omega \cdot \omega^{2} = 1

and

1+\omega+\omega^{2}=0

Then, A =

\begin{vmatrix} 1 & 1+\omega & 1+\omega+\omega^{2} \\ 3 & 4+3\omega & 5+4\omega+3\omega^{2} \\ 6 & 9+6\omega & 11+9\omega+6\omega^{2} \end{vmatrix}

is written as

\left| A \right| = \begin{vmatrix} 1 & 1+\omega & 1+\omega+\omega^{2} \\ 3 & 4+3\omega & 3(1+\omega+\omega^{2})+2+\omega \\ 6 & 9+6\omega & 6(1+\omega+\omega^{2})+5+3\omega \end{vmatrix}

\left| A \right| = \begin{vmatrix} 1 & 1+\omega & 0 \\ 3 & 4+3\omega & 2+\omega \\ 6 & 9+6\omega & 5+3\omega \end{vmatrix}

Expand along row 1,

\left| A \right| = (4+3\omega)(5+3\omega)

- (2+\omega)(9+6\omega)

- (1+\omega) \{15+9\omega-12-6\omega\}

=20+27\omega+9\omega^{2}

-18-21\omega-6\omega^{2}

-3-6\omega-3\omega^{2}=-1

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