Answer
452.4k+ views
Hint:
In this problem, \[\omega \] is the cube root of unity. And \[1 + \omega + {\omega ^2} = 0\] i.e., \[ - 1 - {\omega ^2} = \omega \] is the property of cube roots of unity. So, use this concept to reach the solution of the problem.
Complete step-by-step answer:
Given determinant is \[\left| {\begin{array}{*{20}{c}}
1&1&1 \\
1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\
1&{{\omega ^2}}&{{\omega ^4}}
\end{array}} \right|\]
By using the property \[ - 1 - {\omega ^2} = \omega \], the determinant can be converted into \[\left| {\begin{array}{*{20}{c}}
1&1&1 \\
1&\omega &{{\omega ^2}} \\
1&{{\omega ^2}}&{{\omega ^4}}
\end{array}} \right|\]
By solving the determinant, we have
\[
\Rightarrow 1\left[ {\left( \omega \right)\left( {{\omega ^4}} \right) - \left( {{\omega ^2}} \right)\left( {{\omega ^2}} \right)} \right] - 1\left[ {1\left( {{\omega ^4}} \right) - 1\left( {{\omega ^2}} \right)} \right] + 1\left[ {1\left( {{\omega ^2}} \right) - 1\left( \omega \right)} \right] \\
\Rightarrow 1\left[ {{\omega ^5} - {\omega ^4}} \right] - 1\left[ {{\omega ^4} - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\
\]
This can be rewrite as
\[ \Rightarrow 1\left[ {{\omega ^3}{\omega ^2} - {\omega ^3}\omega } \right] - 1\left[ {{\omega ^3}\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right]\]
We know that \[{\omega ^3} = 1\], by using this formula we have
\[
\Rightarrow 1\left[ {\left( 1 \right){\omega ^2} - \left( 1 \right)\omega } \right] - 1\left[ {\left( 1 \right)\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\
\Rightarrow 1\left[ {{\omega ^2} - \omega } \right] - 1\left[ {\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\
\Rightarrow {\omega ^2} - \omega - \omega + {\omega ^2} + {\omega ^2} - \omega \\
\]
Cancelling the common terms, we have
\[
\Rightarrow 3{\omega ^2} - \omega \\
\Rightarrow 3\omega \left( {\omega - 1} \right) \\
\]
Therefore, the determinant of \[\left| {\begin{array}{*{20}{c}}
1&1&1 \\
1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\
1&{{\omega ^2}}&{{\omega ^4}}
\end{array}} \right|\] is \[3\omega \left( {\omega - 1} \right)\]
Thus, the correct option is B. \[3\omega \left( {\omega - 1} \right)\]
Note: In the given question ‘\[\omega \]’is the imaginary root. Always use the properties of cube roots of unity whenever you find ‘\[\omega \]’ in the question if possible so that you can solve the problem more easily.
In this problem, \[\omega \] is the cube root of unity. And \[1 + \omega + {\omega ^2} = 0\] i.e., \[ - 1 - {\omega ^2} = \omega \] is the property of cube roots of unity. So, use this concept to reach the solution of the problem.
Complete step-by-step answer:
Given determinant is \[\left| {\begin{array}{*{20}{c}}
1&1&1 \\
1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\
1&{{\omega ^2}}&{{\omega ^4}}
\end{array}} \right|\]
By using the property \[ - 1 - {\omega ^2} = \omega \], the determinant can be converted into \[\left| {\begin{array}{*{20}{c}}
1&1&1 \\
1&\omega &{{\omega ^2}} \\
1&{{\omega ^2}}&{{\omega ^4}}
\end{array}} \right|\]
By solving the determinant, we have
\[
\Rightarrow 1\left[ {\left( \omega \right)\left( {{\omega ^4}} \right) - \left( {{\omega ^2}} \right)\left( {{\omega ^2}} \right)} \right] - 1\left[ {1\left( {{\omega ^4}} \right) - 1\left( {{\omega ^2}} \right)} \right] + 1\left[ {1\left( {{\omega ^2}} \right) - 1\left( \omega \right)} \right] \\
\Rightarrow 1\left[ {{\omega ^5} - {\omega ^4}} \right] - 1\left[ {{\omega ^4} - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\
\]
This can be rewrite as
\[ \Rightarrow 1\left[ {{\omega ^3}{\omega ^2} - {\omega ^3}\omega } \right] - 1\left[ {{\omega ^3}\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right]\]
We know that \[{\omega ^3} = 1\], by using this formula we have
\[
\Rightarrow 1\left[ {\left( 1 \right){\omega ^2} - \left( 1 \right)\omega } \right] - 1\left[ {\left( 1 \right)\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\
\Rightarrow 1\left[ {{\omega ^2} - \omega } \right] - 1\left[ {\omega - {\omega ^2}} \right] + 1\left[ {{\omega ^2} - \omega } \right] \\
\Rightarrow {\omega ^2} - \omega - \omega + {\omega ^2} + {\omega ^2} - \omega \\
\]
Cancelling the common terms, we have
\[
\Rightarrow 3{\omega ^2} - \omega \\
\Rightarrow 3\omega \left( {\omega - 1} \right) \\
\]
Therefore, the determinant of \[\left| {\begin{array}{*{20}{c}}
1&1&1 \\
1&{ - 1 - {\omega ^2}}&{{\omega ^2}} \\
1&{{\omega ^2}}&{{\omega ^4}}
\end{array}} \right|\] is \[3\omega \left( {\omega - 1} \right)\]
Thus, the correct option is B. \[3\omega \left( {\omega - 1} \right)\]
Note: In the given question ‘\[\omega \]’is the imaginary root. Always use the properties of cube roots of unity whenever you find ‘\[\omega \]’ in the question if possible so that you can solve the problem more easily.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)