Answer

Verified

432.9k+ 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.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Assertion The resistivity of a semiconductor increases class 13 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

Using the following information to help you answer class 12 chemistry CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Change the following sentences into negative and interrogative class 10 english CBSE

What organs are located on the left side of your body class 11 biology CBSE