[.retired.]

what is linear algebra?

[dth656]

its a field of math pertaining to vectors, vector spaces, matrices used to solve sets of linear equations, and the like:

http://en.wikipedia.org/wiki/Linear_algebra

i dont know about other schools, but at berkeley, it was typically a 2nd year lower division required course for engineers and hard science majors. i remember it had applications in thermodynamics (calculating temperature gradients in a metal cooling fin as a function of time, for example), and also linear programming (solving a set of equations to determine to the optimal solution to a real life problem).

Originally Posted by Mike@PYSpeed.com

what is linear algebra?

[drift90]

Originally Posted by Blake

i can't believe you've done this

AH man i was gunna do that lol beat me too it!

[Seminole]

Originally Posted by dth656

i remember it had applications in thermodynamics (calculating temperature gradients in a metal cooling fin as a function of time, for example), and also linear programming (solving a set of equations to determine to the optimal solution to a real life problem).

[darkphantom]

lawl prof said she'd do it in class next week...*sigh*

[10M3MAN]

Originally Posted by darkphantom

lawl..

[doubled24]

Originally Posted by darkphantom

Alright, so its been a while since I've taken this stuff, but wanted some quick help if anyone remembered off the top of their head.

Solving augmented matrices with the complex numbers (i)

trying to figure out if these two matrices A and B are equivalent.

A:
(2+i)x1 - x2 = 1
x1+ix2 =i

this can be expressed as:
[2+i -1 | 1]
[1 i | i ]


B:
[1+2i -i | 2+i ]
[2+2i -2 | 0]


thanks in advance. easy I know, but I forgot how to do the row elem operations...

I Wouldn't try to reduce this to echelon form, I'd just go ahead and solve both the equations for x1 and x2 and compare results. For matrix A, if multiply the entire 2nd row by i and then add the rows together you can eliminate x2 and find x1 = 1, then simply substitute to solve x2. The algebra on the next one is a little hairier, but still straight forward.... I multiplied the top row by i and then added twice the top row to the bottom row, simplified and solved. I got different solutions for x1 but its awfully early in the morning & I havn't checked my work, so dont take my word for it.

Its easier to work with reduced echelon form for larger matrices, but introduction of complex variables make algebra an absolute crock of shit... thats what we have computers for. Key operations to row reduction are : 1) any row can be multiplied by a constant. 2) any 2 rows can be added together to reform either one of the rows. If you can work the leading term in the first row into an integer (preferably like 1 or something), you can subtract multiplies of the first row to eliminate all of the leading terms in the rows belows.