If 1,\omega ,\omega ^2 are the cube roots of unity, then \Delta = | \beginmatrix 1 & \omega ^n & \omega ^2n \omega ^n & \omega ^2n & 1 \omega ^2n & 1 & \omega ^n \endmatrix | is equal to

Matrices and Determinants PYQ — JEE Main 2003 · Shift 0 · Q80 with Solution

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Correct answer: B

Let $a=\omega^n.$ Then, since $\omega^3=1$ , we also have $\omega^{2n}=a^2.$ So the determinant becomes

1 & a & a^2\\ a & a^2 & 1\\ a^2 & 1 & a \end{vmatrix}.$$ 2. Observe that the rows are $$R_1=(1,a,a^2),\quad R_2=(a,a^2,1),\quad R_3=(a^2,1,a).$$ This is a circulant-type matrix. 3. Now use the fact that $a=\omega^n$ and $\omega^3=1$. Hence $a^3=1$. So $a$ can only be one of $$1,\ \omega,\ \omega^2.$$ We check each case. --- ### Case 1: $a=1$ Then $$\Delta=\begin{vmatrix} 1&1&1\\ 1&1&1\\ 1&1&1 \end{vmatrix}=0.$$ ### Case 2: $a=\omega$ Then $$R_1=(1,\omega,\omega^2),\quad R_2=(\omega,\omega^2,1)=\omega R_1,$$ because $$\omega(1,\omega,\omega^2)=(\omega,\omega^2,\omega^3)=(\omega,\omega^2,1).$$ Similarly, $$R_3=(\omega^2,1,\omega)=\omega^2 R_1.$$ So all three rows are proportional, hence determinant is $0$. ### Case 3: $a=\omega^2$ Then $$R_2=(\omega^2,\omega,1)=\omega^2 R_1, \qquad R_3=(\omega,1,\omega^2)=\omega R_1.$$ Again all rows are proportional, so determinant is $0$. --- 4. Therefore, in every possible case, $$\Delta=0.$$ Hence the correct option is $$\boxed{\text{B }(0)}.$$

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Correct answer: B

Let $a=\omega^n.$ Then, since $\omega^3=1$ , we also have $\omega^{2n}=a^2.$ So the determinant becomes

1 & a & a^2\\ a & a^2 & 1\\ a^2 & 1 & a \end{vmatrix}.$$ 2. Observe that the rows are $$R_1=(1,a,a^2),\quad R_2=(a,a^2,1),\quad R_3=(a^2,1,a).$$ This is a circulant-type matrix. 3. Now use the fact that $a=\omega^n$ and $\omega^3=1$. Hence $a^3=1$. So $a$ can only be one of $$1,\ \omega,\ \omega^2.$$ We check each case. --- ### Case 1: $a=1$ Then $$\Delta=\begin{vmatrix} 1&1&1\\ 1&1&1\\ 1&1&1 \end{vmatrix}=0.$$ ### Case 2: $a=\omega$ Then $$R_1=(1,\omega,\omega^2),\quad R_2=(\omega,\omega^2,1)=\omega R_1,$$ because $$\omega(1,\omega,\omega^2)=(\omega,\omega^2,\omega^3)=(\omega,\omega^2,1).$$ Similarly, $$R_3=(\omega^2,1,\omega)=\omega^2 R_1.$$ So all three rows are proportional, hence determinant is $0$. ### Case 3: $a=\omega^2$ Then $$R_2=(\omega^2,\omega,1)=\omega^2 R_1, \qquad R_3=(\omega,1,\omega^2)=\omega R_1.$$ Again all rows are proportional, so determinant is $0$. --- 4. Therefore, in every possible case, $$\Delta=0.$$ Hence the correct option is $$\boxed{\text{B }(0)}.$$