Math, asked by CAPTAINJACKSPARROW98, 1 day ago

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Answered by Anonymous
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Answered by mathdude500
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Question :-

\rm \: 1, \:  \omega, \:  {\omega}^{2}  \: are \: the \: roots \: of \: cube \: root \: of \: unity, \: then \\

\rm \: \triangle  \:  =  \: \begin{gathered}\sf \left | \begin{array}{ccc}1& {\omega }^{n} & {\omega }^{2n} \\{\omega }^{n}&{\omega }^{2n}&1\\{\omega }^{2n}&1& {\omega }^{n}\end{array}\right | \end{gathered} \\

is equal to

\rm \:  \:  \:  \:  \: (A) \:  \: 0 \\

\rm \:  \:  \:  \:  \: (B) \:  \: 1 \\

\rm \:  \:  \:  \:  \: (C) \:  \: \omega \\

\rm \:  \:  \:  \:  \: (D) \:  \:  {\omega}^{2}  \\

\large\underline{\sf{Solution-}}

Given determinant is

\rm \: \triangle  \:  =  \: \begin{gathered}\sf \left | \begin{array}{ccc}1& {\omega }^{n} & {\omega }^{2n} \\{\omega }^{n}&{\omega }^{2n}&1\\{\omega }^{2n}&1& {\omega }^{n}\end{array}\right | \end{gathered} \\

Now, we know that,

\boxed{\sf{  \:1 + \omega  +  {\omega }^{2}  = 0 \: }} \\

and

\boxed{\sf{  \: {\omega }^{3} = 1 \: }} \\

Consider

\rm \:  \: 1 + {\omega }^{n} + {\omega }^{2n} \\

Now, 2 cases arises :-

Case :- 1 When n is multiple of 3

Let n = 3m, where m is a natural number.

\rm \:  \: {\omega }^{n} \:  =  \: {\omega }^{3m} \:  = 1 \:  \\

and

\rm \:  \: {\omega }^{2n} \:  =  \: {\omega }^{6m} \:  = 1 \:  \\

Now, Consider

\rm \:  \: 1 + {\omega }^{n} + {\omega }^{2n} \\

can be rewritten as

\rm \: =  \: 1 + {\omega }^{3m} + {\omega }^{6m} \\

\rm \: =  \: 1 + 1 + 1 \\

\rm \: =  \: 3 \\

Case :- 2 When n is not a multiple of 3

Let n = 3m + 1, where m is a natural number.

So,

\rm \:  \: 1 + {\omega }^{n} + {\omega }^{2n} \\

can be rewritten as

\rm \: =  \: 1 + {\omega }^{3m + 1} + {\omega }^{6m + 2} \\

\rm \: =  \: 1 + {\omega }^{3m}\omega  + {\omega }^{6m} {\omega }^{2}  \\

\rm \: =  \: 1 +\omega  +  {\omega }^{2}   \\

\rm \: =  \: 0 \\

So, we have

\begin{gathered}\begin{gathered}\bf\: 1 + {\omega }^{n} + {\omega }^{n} = \begin{cases} &\sf{0 \: if \: n \: is \: not \: multiple \: of \: 3} \\ \\  &\sf{3 \: if \: n \: is \: multiple \: of \: 3} \end{cases}\end{gathered}\end{gathered} \\

Now, Consider

\rm \: \triangle  \:  =  \: \begin{gathered}\sf \left | \begin{array}{ccc}1& {\omega }^{n} & {\omega }^{2n} \\{\omega }^{n}&{\omega }^{2n}&1\\{\omega }^{2n}&1& {\omega }^{n}\end{array}\right | \end{gathered} \\

Case :- 1 When n a multiple of 3, the determinant can be rewritten as

\rm \: \triangle  \:  =  \: \begin{gathered}\sf \left | \begin{array}{ccc}1&1&1\\1&1&1\\1&1& 1\end{array}\right | \end{gathered} \\

Since, any two rows or columns are identical, then determinant value is 0.

\rm\implies \:\triangle  \:  =  \: 0 \\

Now, Consider Case :- 2 When n is not a multiple of 3

\rm \: \triangle  \:  =  \: \begin{gathered}\sf \left | \begin{array}{ccc}1& {\omega }^{n} & {\omega }^{2n} \\{\omega }^{n}&{\omega }^{2n}&1\\{\omega }^{2n}&1& {\omega }^{n}\end{array}\right | \end{gathered} \\

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \boxed{\sf{  \: \: OP \:  \: R_1 \:  \to \: R_1 + R_2 + R_3 \:  \: }} \\

\rm \: \triangle  \:  =  \: \begin{gathered}\sf \left | \begin{array}{ccc}1 + {\omega }^{n} + {\omega }^{2n}&1 +  {\omega }^{n} + {\omega }^{2n} & 1 + {\omega }^{n} + {\omega }^{2n} \\{\omega }^{n}&{\omega }^{2n}&1\\{\omega }^{2n}&1& {\omega }^{n}\end{array}\right | \end{gathered} \\

\rm \: \triangle  \:  =  \: \begin{gathered}\sf \left | \begin{array}{ccc}0& 0 & 0 \\{\omega }^{n}&{\omega }^{2n}&1\\{\omega }^{2n}&1& {\omega }^{n}\end{array}\right | \end{gathered} \\

If any one row or column is completely 0, then determinant value is 0.

\rm\implies \:\triangle  \:  =  \: 0 \\

Hence, from this we concluded that,

\rm \: \triangle  \:  =  \: \begin{gathered}\sf \left | \begin{array}{ccc}1& {\omega }^{n} & {\omega }^{2n} \\{\omega }^{n}&{\omega }^{2n}&1\\{\omega }^{2n}&1& {\omega }^{n}\end{array}\right | \end{gathered} = 0 \\

So, option (A) is correct.

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Additional Information :-

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.

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