Q. $\mathrm{\Delta} = \left| \begin{matrix} a & a + b & a + b + c \\ 3a & 4a + 3b & 5a + 4b + 3c \\ 6a & 9a + 6b & 11a + 9b + 6c \end{matrix} \right|$, where $a = i,\ b = \omega,\ c = \omega^{2},$ then $\mathrm{\Delta}$ is equal to
A) $i$
B) $- \omega^{2}$
C) $\omega$
D) $- i$
Explanation
(a)
We have,
$\mathrm{\Delta} = \left| \begin{matrix}
a & a + b & a + b + c \\
3a & 4a + 3b & 5a + 4b + 3c \\
6a & 9a + 6b & 11a + 9b + 6c
\end{matrix} \right|$
Applying $R_{2} \rightarrow R_{2} - 3R_{1},\ R_{3} \rightarrow R_{3} - {2R}_{2}$
$= \left| \begin{matrix}
a & a & a + b + c \\
0 & a & 2a + b \\
0 & a & a + b
\end{matrix} \right|$
$= a\lbrack a^{2} + ab - 2a^{2} - ab\rbrack$
$= - a^{3} = i$ $(\because a = i,$ given)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.