- #1
Oudeis Eimi
- 77
- 6
Hi, I'm tutoring someone on lineal algebra (upper high-school level).
We're going through some exercises, but we're stuck with a trivial
looking one.
Show that the following determinant is null:
\left| \begin{array}{ccc}
1 & \cos x & \cos 2x \\
\cos x & \cos 2x & \cos 3x \\
\cos 2x & \cos 3x & \cos 4x \end{array} \right|
$$\cos 2x = \cos^2 x - \sin^2 x$$
$$\cos 3x = \cos^3 x - 3 \cos x \sin^2 x$$
$$\cos 4x = \cos^4 x - 6 \cos^2 x \sin^2 x + \sin^4 x$$
Expansion of a determinant by the elements of a row or column.
I tried a brute force approach by expanding the cosines and computing the determinant directly in terms of $$\sin x$$ and $$\cos x$$, but that's an inelegant approach. There has to be a trivial, neater way to show this, but I seem to not be in my brightest lately, as I can't find it. I suspect I should be using the fact that the matrix is symmetric. Any ideas?
We're going through some exercises, but we're stuck with a trivial
looking one.
Homework Statement
Show that the following determinant is null:
\left| \begin{array}{ccc}
1 & \cos x & \cos 2x \\
\cos x & \cos 2x & \cos 3x \\
\cos 2x & \cos 3x & \cos 4x \end{array} \right|
Homework Equations
$$\cos 2x = \cos^2 x - \sin^2 x$$
$$\cos 3x = \cos^3 x - 3 \cos x \sin^2 x$$
$$\cos 4x = \cos^4 x - 6 \cos^2 x \sin^2 x + \sin^4 x$$
Expansion of a determinant by the elements of a row or column.
The Attempt at a Solution
I tried a brute force approach by expanding the cosines and computing the determinant directly in terms of $$\sin x$$ and $$\cos x$$, but that's an inelegant approach. There has to be a trivial, neater way to show this, but I seem to not be in my brightest lately, as I can't find it. I suspect I should be using the fact that the matrix is symmetric. Any ideas?