Solving a Linear System with Dimensional Analysis

[TheFerruccio]

Homework Statement

Solve the Ax=b

Homework Equations

A=$\left( \begin{array}{ccc} 0 & \cos (L \lambda ) & \sin (L \lambda ) \\ \sin \left(\frac{L \lambda }{2}\right) & -\cos \left(\frac{L \lambda }{2}\right) & -\sin \left(\frac{L \lambda }{2}\right) \\ \lambda \cos \left(\frac{L \lambda }{2}\right)+4 \sin \left(\frac{L \lambda }{2}\right) & \lambda \sin \left(\frac{L \lambda }{2}\right) & -\lambda \cos \left(\frac{L \lambda }{2}\right) \\ \end{array} \right)$
b=zero vector.
x = what I am trying to solve for.

The Attempt at a Solution

This can be done a variety of ways, including in Mathematica. Do I try to solve for λ first, then plug it into the matrix, and solve for x? I've found that even being very precise with my λ yields x = 0. Is there any other way to approach this problem that is simpler?

 

[vela]

What's the context for this problem? I ask because if it's from a physical situation where L represents a length, then it looks, based on dimensional analysis, like you may have made a mistake in deriving the matrix.

When you say you solved for $\lambda$, what criterion were you using — that A is singular?

 

[TheFerruccio]

Whoa, you're right! I usually check that. I totally forgot. λ in this case is √(P/(E I)) which has dimensions of 1/length. So, I corrected the matrix (algebra error that I subsequently found):

$\left( \begin{array}{ccc} 0 & \cos (L \lambda ) & \sin (L \lambda ) \\ \sin \left(\frac{L \lambda }{2}\right) & -\cos \left(\frac{L \lambda }{2}\right) & -\sin \left(\frac{L \lambda }{2}\right) \\ \lambda \cos \left(\frac{L \lambda }{2}\right)+\frac{4 \sin \left(\frac{L \lambda }{2}\right)}{L} & \lambda \sin \left(\frac{L \lambda }{2}\right) & -\lambda \cos \left(\frac{L \lambda }{2}\right) \\ \end{array} \right)$

Also, yes, I'm solving it with the assumption that I want the matrix to be singular, so I have to solve for λ first, by taking the determinant and solving the characteristic equation.

This yields the first λ = 4.0575/L. Converting from the def. of λ, I get $P=\frac{(4.0575^2) \text{E} \text{I}}{L^2}$.
This problem is an excerpt from a larger problem I have been working on in the engineering section, but I felt that this part would get some visibility here. Thanks for pointing out the dimensional error.

Now, the only thing that remains baffling me is the lack of a "k" anywhere in my matrix. The problem is from here: https://www.physicsforums.com/showthread.php?t=727016. I will post my dilemma in that thread.

A side note: I really hope I am following the forum policy correctly. Engineering problems are tough to pose in this environment, as they bring in a lot of different mathematics concepts into one. I felt compelled to lift the linear algebra problem out of there and into somewhere with more visibility. This problem, as far as the linear algebra of it is concerned, is solved now. Thanks again! Apologies if I accidentally violated policy by not properly encapsulating my problems into single threads.