I am stuck with my homework from math. I should calcutate the determinant of a matrix:
$$\begin{bmatrix}
\sin(x) & \sin(2x) & \cos(x) & \cos(2x)\\
\cos(x) & 2\cos(2x) & -\sin(x)& -2\sin(2x)\\
-\sin(x)& -4\sin(2x)& -\cos(x)& -4\cos(2x)\\
-\cos(x)& -8\cos(2x)& \sin(x) & 8\sin(2x)
\end{bmatrix}$$
I have no clue how to do it. I've tried to convert the matrix to the triangular matrix, but I failed to do so.
Then I tried the Laplacian expansion (I hope it is called so) and I was creating sub-matrices from the first and second column, but I got a really long row of sin and cos and I was unable to make it shorter.
Sorry if I wrote someting wrong, but I am not so good in English. Can you please help me? Is there any way how to solve it?
Thank you very much!!!
Best Answer
Call the matrix $M$. Then by using @r9m's suggestion, we interchange the two middle columns (this switches the sign of the determinant) and apply two row replacements (this doesn't change the determinant) in order to obtain a zero lower left submatrix: \begin{align*} |M| &= -\begin{vmatrix} \sin x & \cos x & \sin 2x & \cos 2x \\ \cos x & -\sin x & 2\cos 2x & -2\sin 2x \\ -\sin x & -\cos x & -4\sin 2x& -4\cos 2x \\ -\cos x & \sin x & -8\cos 2x & 8\sin 2x \end{vmatrix} \\ &= -\begin{vmatrix} \sin x & \cos x & \sin 2x & \cos 2x \\ \cos x & -\sin x & 2\cos 2x & -2\sin 2x \\ 0 & 0 & -3\sin 2x& -3\cos 2x \\ 0 & 0 & -6\cos 2x & 6\sin 2x \end{vmatrix} \\ &= -\begin{vmatrix} \sin x & \cos x \\ \cos x & -\sin x \end{vmatrix} \begin{vmatrix} -3\sin 2x & -3\cos 2x \\ -6\cos 2x & 6\sin 2x \end{vmatrix} \\ &= -(-\sin^2 x - \cos^2 x)(-18\sin^2 2x - 18\cos^2 2x) \\ &= -18(\sin^2 x + \cos^2 x)(\sin^2 2x + \cos^2 2x) \\ &= -18 \end{align*}
This used the fact that if $A,B,D$ are square matrices and $0$ is a square matrix of all zeroes, then: $$ \begin{vmatrix} A & B \\ 0 & D \end{vmatrix} = |AD - B0| = |AD| = |A||D| $$