Question

If $ \alpha ,\,\beta ,\,\gamma $ are the cube roots of unity, then the value of the determinant $ \left| \begin{matrix} {{e}^{\alpha }} & {{e}^{2\alpha }} & ({{e}^{3\alpha }}-1) \\ {{e}^{\beta }} & {{e}^{2\beta }} & ({{e}^{2\beta }}-1) \\ {{e}^{\gamma }} & {{e}^{2\gamma }} & ({{e}^{3\gamma }}-1) \\ \end{matrix} \right| $ is equal to

• A. $ -2 $
• B. $ -1 $
• C. 0
• D. 1
• E. 2

Answer 1

Answer: (C)

Let $\triangle=\left|\begin{array}{ccc} e ^{ a } & e ^{2 a } & e ^{3 a } \\ e ^{ b } & e ^{2 b } & e ^{3 b } \\ e ^{ c } & e ^{2 c } & e ^{3 c }\end{array}\right|$
$=\left|\begin{array}{lll} e ^{ a } & e ^{2 a } & 1 \\ e ^{ b } & e ^{2 b } & 1 \\ e ^{ c } & e ^{2 c } & 1 \end{array}\right|$
$= e ^{ a } \cdot e ^{ b } \cdot e ^{ c }\left|\begin{array}{ccc}1 & e ^{ a } & e ^{2 a } \\ 1 & e ^{ b } & e ^{2 b } \\ 1 & e ^{ c } & e ^{2 c }\end{array}\right|+\left|\begin{array}{ccc} e ^{ b } & 1 & e ^{2 a } \\ e ^{ b } & 1 & e ^{2 b } \\ e ^{ c } & 1 & e ^{2 b }\end{array}\right|$
$= e ^{(a+b+c)}=\left|\begin{array}{ccc}1 & e^{a} & e^{2 a} \\ 1 & e^{b} & e^{2 b} \\ 1 & e^{c} & e^{2 c}\end{array}\right|-\left|\begin{array}{ccc}e^{b} & 1 & e^{2 a} \\ e^{b} & 1 & e^{2 b} \\ e^{c} & 1 & e^{2 b}\end{array}\right|$
$=\left(e^{a+b+c}-1\right)\left|\begin{array}{lll}1 & e^{a} & e^{2 a} \\ 1 & e^{b} & e^{2 b} \\ 1 & e^{c} & e^{2 c}\end{array}\right|$
$= 0 (\because a + b + c = 0 )$