Question

The value of $\left| {\begin{array}{*{20}{l}}
  1&\omega &{2{\omega ^2}} \\
  2&{2{\omega ^2}}&{4{\omega ^3}} \\
  3&{3{\omega ^3}}&{6{\omega ^4}}
\end{array}} \right|$ is equal to (Where $\omega $ is imaginary cube root of unity)
(a)0
(b)1
(c)2
(d)3

Answer 1

Hint: Here we need to find the value of the determinant of the given matrix. The given matrix is of the order 3. So we will use the rules to find the determinant. If we will get the determinant in terms of the imaginary cube then we will further simplify it to get the required answer.

Complete step-by-step answer:
We will first calculate the value of the given determinant i.e.
$D = \left| {\begin{array}{*{20}{l}}
  1&\omega &{2{\omega ^2}} \\
  2&{2{\omega ^2}}&{4{\omega ^3}} \\
  3&{3{\omega ^3}}&{6{\omega ^4}}
\end{array}} \right|$
So we can write it as;
$ \Rightarrow D = 1\left( {2{\omega ^2} \times 6{\omega ^4} - 3{\omega ^3} \times 4{\omega ^3}} \right) - 2\left( {\omega \times 6{\omega ^4} - 3{\omega ^3} \times 2{\omega ^2}} \right) + 3\left( {\omega \times 4{\omega ^3} - 2{\omega ^2} \times 2{\omega ^2}} \right)$
On multiplying the terms inside all the brackets, we get
$ \Rightarrow D = 1\left( {12{\omega ^6} - 12{\omega ^6}} \right) - 2\left( {6{\omega ^5} - 6{\omega ^5}} \right) + 3\left( {4{\omega ^4} - 4{\omega ^4}} \right)$
On adding and subtracting the like terms present inside the brackets, we get
$ \Rightarrow D = 1 \times 0 - 2 \times 0 + 3 \times 0$
On multiplying all the terms, we get
$ \Rightarrow D = 0 - 0 + 0$
On further simplification, we get
$ \Rightarrow D = 0$
Therefore, the required value of the determinant is equal to 0.
Hence, the correct option is option (a).
Note: Here we have obtained the value of the determinant of the matrix which is of order 3. The most important property of a determinant is that if two or more rows are similar then the value of the determinant is equal to zero. Also, we need to keep in mind that if we interchange the rows and columns the value of determinant remains the same. If all the elements of any one of the rows or columns are zero then the value of the determinant is equal to zero. If we multiply all the elements of a row by a constant then the resultant determinant will be constant times the value of the original determinant.