# If $\omega $ is a cube root of unity, then the value of determinant $\left| {\begin{array}{*{20}{c}}

{1 + \omega }&{{\omega ^2}}&\omega \\

{{\omega ^2} + \omega }&{ - \omega }&{{\omega ^2}} \\

{1 + {\omega ^2}}&\omega &{{\omega ^2}}

\end{array}} \right|$ is equal to

A. $ - 1 + \omega $

B. $1 - \omega $

C. 0

D. ${\omega ^2}$

Answer

Verified

360.3k+ views

Hint: In this question, we will use the property of the cube root of unity to solve the given determinant. it is given that $\omega $ is a cube root of unity. The value of $\omega $ = $\dfrac{{ - 1 + \sqrt 3 i}}{2}$. Now we will use the property which is $1 + \omega + {\omega ^2}{\text{ = 0}}$. We will use this property in expanding the determinant.

Complete step-by-step answer:

Now, performing the operation ${{\text{C}}_1} \to {{\text{C}}_1} + {{\text{C}}_2}$ on column \[{{\text{C}}_1}\].

Therefore,

$\vartriangle {\text{ = }}\left| {\begin{array}{*{20}{c}}

{1 + \omega + {\omega ^2}}&{{\omega ^2}}&\omega \\

{{\omega ^2} + \omega - \omega }&{ - \omega }&{{\omega ^2}} \\

{1 + {\omega ^2} + \omega }&\omega &{{\omega ^2}}

\end{array}} \right|$

Using the property $1 + \omega + {\omega ^2}{\text{ = 0}}$, we get

$\vartriangle {\text{ = }}\left| {\begin{array}{*{20}{c}}

0&{{\omega ^2}}&\omega \\

{{\omega ^2}}&{ - \omega }&{{\omega ^2}} \\

0&\omega &{{\omega ^2}}

\end{array}} \right|$

Taking ${\omega ^2}$ common from column \[{{\text{C}}_1}\] and $\omega $ from both the column \[{{\text{C}}_2}\] and \[{{\text{C}}_3}\], we get

$\vartriangle {\text{ = }}{\omega ^4}\left| {\begin{array}{*{20}{c}}

0&\omega &1 \\

1&{ - 1}&\omega \\

0&1&\omega

\end{array}} \right|$

Expanding the determinant through column \[{{\text{C}}_1}\].

$\vartriangle {\text{ = }}{\omega ^4}\{ 0\left| {\begin{array}{*{20}{c}}

{ - 1}&\omega \\

1&\omega

\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}

\omega &1 \\

1&\omega

\end{array}} \right| + 0\left| {\begin{array}{*{20}{c}}

\omega &1 \\

{ - 1}&\omega

\end{array}} \right|\} $

$\vartriangle {\text{ = }}{\omega ^4}\{ 0 - 1({\omega ^2} - 1) + 0\} $

$\vartriangle {\text{ = - }}{\omega ^4}({\omega ^2} - 1)$

Now the value of $\omega $ = $\dfrac{{ - 1 + \sqrt 3 i}}{2}$ . So, we can see that the value of ${\omega ^3}$ = 1.

Therefore,

$\vartriangle {\text{ = - }}\omega ({\omega ^2} - 1) = {\text{ - (}}{\omega ^3} - \omega )$

$\vartriangle {\text{ = - (1 - }}\omega {\text{) = - 1 + }}\omega $

So, the answer is ${\text{ - 1 + }}\omega $ i.e. option (A).

Note: While solving such problems which have a cube root of unity, always apply the properties of the cube root to easily solve the given problem. If instead properties value of cube root is used it will also lead you to the correct answer but the process is very lengthy. Also, simplify the determinant by using properties of determinant before expanding it.

Complete step-by-step answer:

Now, performing the operation ${{\text{C}}_1} \to {{\text{C}}_1} + {{\text{C}}_2}$ on column \[{{\text{C}}_1}\].

Therefore,

$\vartriangle {\text{ = }}\left| {\begin{array}{*{20}{c}}

{1 + \omega + {\omega ^2}}&{{\omega ^2}}&\omega \\

{{\omega ^2} + \omega - \omega }&{ - \omega }&{{\omega ^2}} \\

{1 + {\omega ^2} + \omega }&\omega &{{\omega ^2}}

\end{array}} \right|$

Using the property $1 + \omega + {\omega ^2}{\text{ = 0}}$, we get

$\vartriangle {\text{ = }}\left| {\begin{array}{*{20}{c}}

0&{{\omega ^2}}&\omega \\

{{\omega ^2}}&{ - \omega }&{{\omega ^2}} \\

0&\omega &{{\omega ^2}}

\end{array}} \right|$

Taking ${\omega ^2}$ common from column \[{{\text{C}}_1}\] and $\omega $ from both the column \[{{\text{C}}_2}\] and \[{{\text{C}}_3}\], we get

$\vartriangle {\text{ = }}{\omega ^4}\left| {\begin{array}{*{20}{c}}

0&\omega &1 \\

1&{ - 1}&\omega \\

0&1&\omega

\end{array}} \right|$

Expanding the determinant through column \[{{\text{C}}_1}\].

$\vartriangle {\text{ = }}{\omega ^4}\{ 0\left| {\begin{array}{*{20}{c}}

{ - 1}&\omega \\

1&\omega

\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}

\omega &1 \\

1&\omega

\end{array}} \right| + 0\left| {\begin{array}{*{20}{c}}

\omega &1 \\

{ - 1}&\omega

\end{array}} \right|\} $

$\vartriangle {\text{ = }}{\omega ^4}\{ 0 - 1({\omega ^2} - 1) + 0\} $

$\vartriangle {\text{ = - }}{\omega ^4}({\omega ^2} - 1)$

Now the value of $\omega $ = $\dfrac{{ - 1 + \sqrt 3 i}}{2}$ . So, we can see that the value of ${\omega ^3}$ = 1.

Therefore,

$\vartriangle {\text{ = - }}\omega ({\omega ^2} - 1) = {\text{ - (}}{\omega ^3} - \omega )$

$\vartriangle {\text{ = - (1 - }}\omega {\text{) = - 1 + }}\omega $

So, the answer is ${\text{ - 1 + }}\omega $ i.e. option (A).

Note: While solving such problems which have a cube root of unity, always apply the properties of the cube root to easily solve the given problem. If instead properties value of cube root is used it will also lead you to the correct answer but the process is very lengthy. Also, simplify the determinant by using properties of determinant before expanding it.

Last updated date: 22nd Sep 2023

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