Linear Homogeneous Algebraic Equations From the following two linear homogeneous algebraic equations:
$$A \sin\left(\frac{kl}{\sqrt2}\right) = B \sin(kl)$$
$$\frac{kA}{\sqrt2}\cos\left(\frac{kl}{\sqrt2}\right) = kB\cos(kl)$$
form matrix of these 2 equations, and setting the determinant equal to zero will lead to:
$$\frac1{\sqrt2}\cos\left(\frac{kl}{\sqrt2}\right)\sin(kl) - \sin\left(\frac{kl}{\sqrt2}\right)\cos(kl) = 0.$$
k is unknown. l is constant. A and B are constants. I'm trying to find a nonzero solution when the determinant of the equation system vanishes.
How do I solve this?
 A: $(A,B)=(0,0)$ is always a solution.
If $k=0$ then both equations vanish, and any $(A,B)$ at all will be a solution.
If $k\ne 0$ but $l=0$, then the first equation vanishes and the second one reduces to $A=\sqrt 2 B$.
There are other special values of $kl$ where the two equations are proportional such that there are nonzero solutions, but you will have to find them numerically. Your determinant procedure looks correct and simplifies to $\frac{\tan(kl)}{\sqrt 2} = \tan(\frac{kl}{\sqrt 2})$, which probably has no nice closed-form solutions other than $kl=0$.
A: As previous comments indicate, it is not clear what you are trying to find. However, It is possible to solve this by getting a realation between A and B as follows:
$A \sin\left(\frac{kl}{\sqrt2}\right) = B \sin(kl)$ (1)
$\frac{kA}{\sqrt2}\cos\left(\frac{kl}{\sqrt2}\right) = kB\cos(kl)$  (2)
which is:
${A}\cos\left(\frac{kl}{\sqrt2}\right) = {\sqrt2} B\cos(kl) , k \ne 0$ (3)
Square and add both sides of (1) and (3):
$A^2=B^2(1+(\cos(kl))^2)$
When I first posted this comment, I got a simpler experession but I was corrected by Pedja, I hope this can help you...
