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Zero determinant

Consider the real numbers x_i, x_j for 1 \leq i , j \leq 3 . Prove that

    \[\mathcal{D} = \begin{vmatrix} \sin (x_1+y_1) & \sin (x_1+y_2) & \sin (x_1+y_3)\\ \sin (x_2+y_1) & \sin (x_2+y_2) & \sin (x_2+y_3)\\ \sin (x_3+y_1) & \sin (x_3+y_2) & \sin (x_3+y_3) \end{vmatrix}=0\]

Solution

Using the identity \sin (a+b)=\sin a\cos b+\sin b\cos a in combination with \det AB = \det A \det B we have:

    \[\mathcal{D}=\begin{vmatrix} \sin x_1 & \cos x_1 & 0\\ \sin x_2 & \cos x_2 & 0\\ \sin x_3 & \cos x_3 & 0 \end{vmatrix} \cdot \begin{vmatrix} \cos y_1 & \cos y_2 & \cos y_3 \\ \sin y_1 & \sin y_2 & \sin y_3 \\ 0 & 0 & 0 \end{vmatrix} =0\]

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