Question

Let ${\Delta }^{=}$ $\left|\begin{array}{ccc}a& a+b& a+b+c\\ 3a& 4a+3b& 5a+4b+3c\\ 6a& 9a+6b& 11a+9b+6c\end{array}\right|$ where $a=1,b=\omega$ and $c={\omega }^2$ then ${\Delta }^{=}$

• A. $\omega$
• B. $-\omega 2$
• C. $1$
• D. $-1$

Answer 1

The correct option is D $-1$

Given $a=1b=\omega$ and $c={\omega }^2$
$\omega$ is to be root of whity, so,
${\omega }^3=1$ , $1+\omega +{\omega }^2=0$
so, $\Delta =\left|\begin{array}{ccc}a& a+b& a+b+c\\ 3a& 4a+3b& 5a+4b+3c\\ 6a& 9a+6b& 11a+9b+6c\end{array}\right|$
By column transformation,
$c_3\to c_3-c_2$ and $c_2\to c_2-c_1$
$\Delta =\left|\begin{array}{ccc}a& b& c\\ 3a& 3b+a& 3c+4b+a\\ 6a& 6b+3a& 6c+3b+2a\end{array}\right|$
By row transformation
$r_2\to r_2-3r_1$ and $r_3\to r_3-6r_1$
$\Delta =\left|\begin{array}{ccc}a& b& c\\ 0& a& b+a\\ 0& 3a& 3b+2a\end{array}\right|$
so, $\Delta =a(3ab+2a^2-3ab-3a^2)=-a^3$
$=-1$