# 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?

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Which of those letters are the unknowns, and which are constants? –  Henning Makholm Oct 30 '11 at 7:18
@Chris: please use mathjax commands for your question. The question is not clear. –  Hassan Muhammad Oct 30 '11 at 7:36
@Henning: It appears that $A$ and $B$ are the unknowns. –  Brian M. Scott Oct 30 '11 at 7:39
These are not algebraic equations. Also there's very little to do with linear algebra here. –  Gerry Myerson Oct 30 '11 at 9:53

$(A,B)=(0,0)$ 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 $kl=0$.

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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...

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$A^2=B^2(1+(\cos(kl))^2)$ –  pedja Oct 30 '11 at 8:30
@pedja: Thanks, you are correct of course. I have fixed that in the post. –  Emmad Kareem Oct 30 '11 at 8:38
Replace <>= by $\backslash\mathtt{ne}$. –  Did Oct 30 '11 at 8:43
@Didier Piau: Thanks, done. I am not very familiar with the syntax yet. –  Emmad Kareem Oct 30 '11 at 8:45
@Emmad, right. The best way to learn is to look at the source of other questions or answers and to see how they encode such and such. –  Did Oct 30 '11 at 8:48