If a_r = \cos 2r\pi \over 9 + i\sin 2r\pi \over 9 , r = 1, 2, 3, ....., i = \sqrt - 1 , then the determinant | \beginmatrix a_1 & a_2 & a_3 a_4 & a_5 & a_6 a_7 & a_8 & a_9 \endmatrix | is equal to :

Matrices and Determinants PYQ — JEE Main 2021 · 31 Aug · Shift 1 · Q36 with Solution

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Correct answer: C

a_r=\cos\frac{2r\pi}{9}+i\sin\frac{2r\pi}{9}=e^{\frac{2\pi i r}{9}}.

So if we put

\omega=e^{\frac{2\pi i}{9}},

then

a_r=\omega^r.

Hence the matrix is

\begin{vmatrix} \omega & \omega^2 & \omega^3\\ \omega^4 & \omega^5 & \omega^6\\ \omega^7 & \omega^8 & \omega^9 \end{vmatrix}.

Since $\omega^9=1$ , this becomes

\begin{vmatrix} \omega & \omega^2 & \omega^3\\ \omega^4 & \omega^5 & \omega^6\\ \omega^7 & \omega^8 & 1 \end{vmatrix}.

Factor out powers of $\omega$ from each row:

From row 1, factor out $\omega$
From row 2, factor out $\omega^4$
From row 3, factor out $\omega^7$

Then

D=\omega^{1+4+7} \begin{vmatrix} 1 & \omega & \omega^2\\ 1 & \omega & \omega^2\\ 1 & \omega & \omega^2 \end{vmatrix}.

But this is not correct because the third row after factoring should be

(1,\omega,\omega^2)

only if the entries are $(\omega^7,\omega^8,\omega^9)$ , which they are. So indeed all three rows become identical.

Therefore,

D=\omega^{12}\begin{vmatrix} 1 & \omega & \omega^2\\ 1 & \omega & \omega^2\\ 1 & \omega & \omega^2 \end{vmatrix}=0.

Let us also verify directly by row proportionality:

(a_4,a_5,a_6)=\omega^3(a_1,a_2,a_3),

(a_7,a_8,a_9)=\omega^6(a_1,a_2,a_3).

So all rows are scalar multiples of the first row. Hence determinant is zero.

Now check options.

We use $a_r=\omega^r$ and $\omega^9=1$ .

Option A:

a_2a_6-a_4a_8=\omega^{2+6}-\omega^{4+8}=\omega^8-\omega^{12}=\omega^8-\omega^3\neq 0

in general.

Option B:

a_9=\omega^9=1\neq 0.

Option C:

a_1a_9-a_3a_7=\omega^{1+9}-\omega^{3+7}=\omega^{10}-\omega^{10}=0.

So this matches the determinant.

Option D:

a_5=\omega^5\neq 0.

Thus the determinant equals

a_1a_9-a_3a_7.

Therefore the correct option is:

\boxed{\text{C}}

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View written solutionFree

Correct answer: C

a_r=\cos\frac{2r\pi}{9}+i\sin\frac{2r\pi}{9}=e^{\frac{2\pi i r}{9}}.

So if we put

\omega=e^{\frac{2\pi i}{9}},

then

a_r=\omega^r.

Hence the matrix is

\begin{vmatrix} \omega & \omega^2 & \omega^3\\ \omega^4 & \omega^5 & \omega^6\\ \omega^7 & \omega^8 & \omega^9 \end{vmatrix}.

Since $\omega^9=1$ , this becomes

\begin{vmatrix} \omega & \omega^2 & \omega^3\\ \omega^4 & \omega^5 & \omega^6\\ \omega^7 & \omega^8 & 1 \end{vmatrix}.

Factor out powers of $\omega$ from each row:

From row 1, factor out $\omega$
From row 2, factor out $\omega^4$
From row 3, factor out $\omega^7$

Then

D=\omega^{1+4+7} \begin{vmatrix} 1 & \omega & \omega^2\\ 1 & \omega & \omega^2\\ 1 & \omega & \omega^2 \end{vmatrix}.

But this is not correct because the third row after factoring should be

(1,\omega,\omega^2)

only if the entries are $(\omega^7,\omega^8,\omega^9)$ , which they are. So indeed all three rows become identical.

Therefore,

D=\omega^{12}\begin{vmatrix} 1 & \omega & \omega^2\\ 1 & \omega & \omega^2\\ 1 & \omega & \omega^2 \end{vmatrix}=0.

Let us also verify directly by row proportionality:

(a_4,a_5,a_6)=\omega^3(a_1,a_2,a_3),

(a_7,a_8,a_9)=\omega^6(a_1,a_2,a_3).

So all rows are scalar multiples of the first row. Hence determinant is zero.

Now check options.

We use $a_r=\omega^r$ and $\omega^9=1$ .

Option A:

a_2a_6-a_4a_8=\omega^{2+6}-\omega^{4+8}=\omega^8-\omega^{12}=\omega^8-\omega^3\neq 0

in general.

Option B:

a_9=\omega^9=1\neq 0.

Option C:

a_1a_9-a_3a_7=\omega^{1+9}-\omega^{3+7}=\omega^{10}-\omega^{10}=0.

So this matches the determinant.

Option D:

a_5=\omega^5\neq 0.

Thus the determinant equals

a_1a_9-a_3a_7.

Therefore the correct option is:

\boxed{\text{C}}

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