Question

$△=\left|\begin{array}{ccc}a& a+b& a+b+c\\ 3a& 4a+3b& 5a+4b+3c\\ 6a& 9a+6b& 11a+9b+6c\end{array}\right|$ where $a=i$, $b=\omega$, $c={\omega }^2$, then $△$ is equal to

• A. $i$
• B. $-{\omega }^2$
• C. $\omega$
• D. $-i$

Answer 1

The correct option is A $i$

$△=\left|\begin{array}{ccc}a& a+b& a+b+c\\ 3a& 4a+3b& 5a+4b+3c\\ 6a& 9a+6b& 11a+9b+6c\end{array}\right|$

where $a=i$,

$b=\omega$,

$c={\omega }^2$,

then $△=\left|\begin{array}{ccc}i& i+\omega & i+\omega +{\omega }^2\\ 3i& 4i+3\omega & 5i+4\omega +3{\omega }^2\\ 6i& 9i+6\omega & 11i+9\omega +6{\omega }^2\end{array}\right|$

$\therefore i\left[\right(4i+3\omega \left)\right(11i+9\omega +6{\omega }^2)-(9i+6\omega \left)\right(5i+4\omega +3{\omega }^2\left)\right]$

$-(i+\omega )\left[\right(3i\left)\right(11i+9\omega +6{\omega }^2)-(6i\left)\right(5i+4\omega +3{\omega }^2\left)\right]$

$+(i+\omega +{\omega }^2)\left[\right(3i\left)\right(9i+6\omega )-(6i\left)\right(4i+3\omega \left)\right]$

Thus by solving above $△$ we get $i$

Hence $△=i$