Question

If $a,b,c$ are cube roots of unity, then $\left|\begin{array}{ccc}{e}^a& e^{2a}& e^{3a}-1\\ e^b& e^{2b}& e^{3b}-1\\ e^c& e^{2c}& e^{3c}-1\end{array}\right|$ is equal to

• A. $0$
• B. $e$
• C. $e^2$
• D. $e^3$

Answer 1

Answer: (A)

$0$

Let $△=\left|\begin{array}{ccc}{e}^a& e^{2a}& e^{3a}\\ e^b& e^{2b}& e^{3b}\\ e^c& e^{2c}& e^{3c}\end{array}\right|$

$=\left|\begin{array}{ccc}{e}^a& e^{2a}& 1\\ e^b& e^{2b}& 1\\ e^c& e^{2c}& 1\end{array}\right|$
$=e^a\cdot e^b\cdot e^c\left|\begin{array}{ccc}1& e^a& e^{2a}\\ 1& e^b& e^{2b}\\ 1& e^c& e^{2c}\end{array}\right|+\left|\begin{array}{ccc}{e}^b& 1& e^{2a}\\ e^b& 1& e^{2b}\\ e^c& 1& e^{2b}\end{array}\right|$
$=e^{(a+b+c)}=\left|\begin{array}{ccc}1& e^a& e^{2a}\\ 1& e^b& e^{2b}\\ 1& e^c& e^{2c}\end{array}\right|-\left|\begin{array}{ccc}{e}^b& 1& e^{2a}\\ e^b& 1& e^{2b}\\ e^c& 1& e^{2b}\end{array}\right|$
$=(e^{a+b+c}-1)\left|\begin{array}{ccc}1& e^a& e^{2a}\\ 1& e^b& e^{2b}\\ 1& e^c& e^{2c}\end{array}\right|$
$=0(\because a+b+c=0)$.