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Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!

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Loki:
Okay, so last time I did linear algebra was a year ago, so my memory is rusty.

I have a set of some five matrices in Q^(2x2). I am supposed to check whether this set a spanning set of Q^(2x2).
What I would do is transform a matrix ((a,b),(c,d)) with a,b as the first row and c,d as the second row into a column vector (a,b,c,d)^t, then use those column vectors to create a 4x5 matrix, which I would Gauss to check whether it has a rank of 4.

Iff it has a rank of 4, then my set is a spanning set.

Is what I am doing correct?

Schmee:
...I'm pretty sure that's right. I just finished a semester on linalg, and your method seems sound. I haven't heard of Gaussing a matrix, though - we probably used a different term. You use row operations to reduce it to a row echelon form, right?

Loki:
Yes. I was using a colloquial used here for performing a Gaussian elimination (aka reducing it to row echelon form).

Thanks for doublechecking!

Schmee:
Okay, so I've hit upon a fairly nasty inhomogenous second-order DE, and I'm stuck. Any help?

We have y'' - y' - y = cosh(x), where y(0) = 0 and y'(0) = 0.
I can get the homogenous part to be y = Ae-x + Be2x, but then I get stuck - I can't think of a substitution for y to solve the second part.
y = c cosh(x) + d sinh(x) doesn't work, so the answer can't be in terms of sinh and cosh.
I've tried separating cosh(x) into 1/2(ex + e-x), and treating it as two separate functions, but again I can't find a solution.

Ideas?

Skewbrow:
Didn't check, but you should be able to handle that ex/2 part of hyperbolic cosine with the usual ansatz Aex. On the other hand the e-x/2 part requires an ansatz of the form Axe-x because e-x is a solution of the homogeneous DE.

But wait just a minute! Recheck the solution to the homogeneous part. The solutions of r2-r-1=0 are r1=(1+sqrt(5))/2 and r2=(1-sqrt(5))/2. Or perhaps you meant that your DE is y''-y'-2y=cosh(x) ?

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