Question

If the a, b and c are cube root of unity,
\[\left| \begin{matrix}
   {{e}^{a}} & {{e}^{2a}} & {{e}^{3a}} \\
   {{e}^{b}} & {{e}^{2b}} & {{e}^{3b}} \\
   {{e}^{c}} & {{e}^{2c}} & {{e}^{3c}} \\
\end{matrix} \right|-\left| \begin{matrix}
   {{e}^{a}} & {{e}^{2a}} & 1 \\
   {{e}^{b}} & {{e}^{2b}} & 1 \\
   {{e}^{c}} & {{e}^{2c}} & 1 \\
\end{matrix} \right|\]
1. \[0\]
2. \[e\]
3. \[e^2\]
4. \[e^3\]

Answer 1

Hint: Here in the given question is a, b and c are cube roots of unity. That means\[a=1\], \[b=1\]and\[\,c=1\]. In first step we have to take the common \[{{e}^{a}}{{e}^{b}}{{e}^{c}}\] on\[\left| \begin{matrix}
   {{e}^{a}} & {{e}^{2a}} & {{e}^{3a}} \\
   {{e}^{b}} & {{e}^{2b}} & {{e}^{3b}} \\
   {{e}^{c}} & {{e}^{2c}} & {{e}^{3c}} \\
\end{matrix} \right|\]and\[\left| \begin{matrix}
   {{e}^{a}} & {{e}^{2a}} & 1 \\
   {{e}^{b}} & {{e}^{2b}} & 1 \\
   {{e}^{c}} & {{e}^{2c}} & 1 \\
\end{matrix} \right|\]= \[\left| \begin{matrix}
   1 & {{e}^{a}} & {{e}^{2a}} \\
   1 & {{e}^{b}} & {{e}^{2b}} \\
   1 & {{e}^{c}} & {{e}^{2c}} \\
\end{matrix} \right|\] By using the property of the determinant, if we interchange the column twice, the determinant will be the same.

Complete step by step answer:
Assign the value \[\Delta \]to this Question (You can assign any value for simplicity)
\[\Delta =\left| \begin{matrix}
   {{e}^{a}} & {{e}^{2a}} & {{e}^{3a}} \\
   {{e}^{b}} & {{e}^{2b}} & {{e}^{3b}} \\
   {{e}^{c}} & {{e}^{2c}} & {{e}^{3c}} \\
\end{matrix} \right|-\left| \begin{matrix}
   {{e}^{a}} & {{e}^{2a}} & 1 \\
   {{e}^{b}} & {{e}^{2b}} & 1 \\
   {{e}^{c}} & {{e}^{2c}} & 1 \\
\end{matrix} \right|----(1)\]
By taking common \[{{e}^{a}}{{e}^{b}}{{e}^{c}}\] on equation\[(1)\],
\[\Delta ={{e}^{a}}{{e}^{b}}{{e}^{c}}\left| \begin{matrix}
   1 & {{e}^{a}} & {{e}^{2a}} \\
   1 & {{e}^{b}} & {{e}^{2b}} \\
   1 & {{e}^{c}} & {{e}^{2c}} \\
\end{matrix} \right|-\left| \begin{matrix}
   {{e}^{a}} & {{e}^{2a}} & 1 \\
   {{e}^{b}} & {{e}^{2b}} & 1 \\
   {{e}^{c}} & {{e}^{2c}} & 1 \\
\end{matrix} \right|\]
After using the property of determinant that is if we interchange the column twice then its determinant remains the same. After using this we get:
\[\Delta ={{e}^{a}}{{e}^{b}}{{e}^{c}}\left| \begin{matrix}
   1 & {{e}^{a}} & {{e}^{2a}} \\
   1 & {{e}^{b}} & {{e}^{2b}} \\
   1 & {{e}^{c}} & {{e}^{2c}} \\
\end{matrix} \right|-\left| \begin{matrix}
   1 & {{e}^{a}} & {{e}^{2a}} \\
   1 & {{e}^{b}} & {{e}^{2b}} \\
   1 & {{e}^{c}} & {{e}^{2c}} \\
\end{matrix} \right|\]
Take the determinate common from the above equation and after simplifying we get:

\[\Delta =\left| \begin{matrix}
   1 & {{e}^{a}} & {{e}^{2a}} \\
   1 & {{e}^{b}} & {{e}^{2b}} \\
   1 & {{e}^{c}} & {{e}^{2c}} \\
\end{matrix} \right|\left( {{e}^{a}}{{e}^{b}}{{e}^{c}}-1 \right)----(2)\]
But the determinant of \[\left| \begin{matrix}
   1 & {{e}^{a}} & {{e}^{2a}} \\
   1 & {{e}^{b}} & {{e}^{2b}} \\
   1 & {{e}^{c}} & {{e}^{2c}} \\
\end{matrix} \right|=1({{e}^{b}}{{e}^{2c}}-{{e}^{c}}{{e}^{2b}})-{{e}^{a}}((1\times {{e}^{2c}})-(1\times {{e}^{2b}}))+{{e}^{2a}}((1\times {{e}^{c}})-(1\times {{e}^{b}}))\]
\[\therefore \left| \begin{matrix}
   1 & {{e}^{a}} & {{e}^{2a}} \\
   1 & {{e}^{b}} & {{e}^{2b}} \\
   1 & {{e}^{c}} & {{e}^{2c}} \\
\end{matrix} \right|=1({{e}^{b}}{{e}^{2c}}-{{e}^{c}}{{e}^{2b}})-{{e}^{a}}({{e}^{2c}}-{{e}^{2b}})+{{e}^{2a}}({{e}^{c}}-{{e}^{b}})\]
Substitute the value of determinant in equation\[(2)\]:
\[\Delta =(({{e}^{b}}{{e}^{2c}}-{{e}^{c}}{{e}^{2b}})-{{e}^{a}}({{e}^{2c}}-{{e}^{2b}})+{{e}^{2a}}({{e}^{c}}-{{e}^{b}}))\times ({{e}^{a}}{{e}^{b}}{{e}^{c}}-1)----(3)\]
By using the property of cube root of unity you will get the value of
\[a=1\]
\[\,b=1\]
 \[c=1\]
After substituting all the values of a, b and c in equation \[(3)\] you will get:
\[\Delta =(({{e}^{1}}{{e}^{2\times 1}}-{{e}^{1}}{{e}^{2\times 1}})-{{e}^{1}}({{e}^{2\times 1}}-{{e}^{2\times 1}})+{{e}^{2\times 1}}({{e}^{1}}-{{e}^{1}}))\times ({{e}^{1}}{{e}^{1}}{{e}^{1}}-1)\]
\[\Delta =((e\times {{e}^{2}}-e\times {{e}^{2}})-e({{e}^{2}}-{{e}^{2}})+{{e}^{2}}(e-e))\times ({{e}^{3}}-1)\]
\[\Delta =(({{e}^{3}}-{{e}^{3}})-e({{e}^{2}}-{{e}^{2}})+{{e}^{2}}(e-e))\times ({{e}^{3}}-1)\]
\[\Delta =((0)-e(0)+{{e}^{2}}(0))\times ({{e}^{3}}-1)\]
\[\Delta =(0-0+0)\times ({{e}^{3}}-1)\]
\[\Delta =(0)\times ({{e}^{3}}-1)\]
\[\Delta =0\]

So, the correct answer is “Option 1”.

Note: if in the question is given that cube root of unity of a, b and c then the value of a, b and c must be value 1 according to the concept of complex cube root of unity. Remember that if there is interchange of one column then there will be change of determinant. If there is interchange of two columns then the determinant will remain the same. So, by using this concept of the cube root of unity as well as the property of determinants we can solve this type of problem.