Question

If a, b and c are cube roots of unity, $\begin{array}{rcl}\left[\begin{array}{ccc}{e}^a& e^{2a}& e^{3a}\\ e^b& e^{2b}& e^{3b}\\ e^c& e^{2c}& e^{3c}\end{array}\right]-\left[\begin{array}{ccc}{e}^a& e^{2a}& 1\\ e^b& e^{2b}& 1\\ e^c& e^{2c}& 1\end{array}\right]& =& \end{array}$

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

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Finding the value:

Let,

$\begin{array}{rcl}△& =& \left[\begin{array}{ccc}{e}^a& e^{2a}& e^{3a}\\ e^b& e^{2b}& e^{3b}\\ e^c& e^{2c}& e^{3c}\end{array}\right]-\left[\begin{array}{ccc}{e}^a& e^{2a}& 1\\ e^b& e^{2b}& 1\\ e^c& e^{2c}& 1\end{array}\right]\\ & =& e^a{e}^b{e}^c\left[\begin{array}{ccc}1& e^a& e^{3a}\\ 1& e^b& e^{3b}\\ 1& e^c& e^{3c}\end{array}\right]-\left[\begin{array}{ccc}1& e^a& e^{2a}\\ 1& e^b& e^{2b}\\ 1& e^c& e^{2c}\end{array}\right]\\ & =& e^{a+b+c}\left[\begin{array}{ccc}1& e^a& e^{3a}\\ 1& e^b& e^{3b}\\ 1& e^c& e^{3c}\end{array}\right]-\left[\begin{array}{ccc}1& e^a& e^{2a}\\ 1& e^b& e^{2b}\\ 1& e^c& e^{2c}\end{array}\right]\\ & =& \left(e^{a+b+c}-1\right)\left[\begin{array}{ccc}1& e^a& e^{3a}\\ 1& e^b& e^{3b}\\ 1& e^c& e^{3c}\end{array}\right]\end{array}$

$\mathbf{\therefore }\mathbf{△}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$ $\left[\because a+b+c=0\right]$

Hence, Option ‘A’ is Correct.