Wikipedia:Reference desk/Archives/Mathematics/2007 February 15

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February 15

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matrices

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Hey there are these weird matrices problems that need to be solved using a fast method. Things like:

1  2  3 | 4
5  6  7 | 8
9 10 11 | 12

This is like a matrix and if you know how to answer thsi please tell me, and write the method that would be dependable and fast, well to humans (i know quite a lot of ways to solve this in comp programming but requires lots and lots of doing math) fast.

Gaussian elimination is your friend.
If you're solving it by computer, it's called an "augmented matrix", and there are a number of methods for solving it quickly. If you're solving it by hand, it's called a "demented matrix", and there's no fast way to solve it. --Carnildo 20:47, 15 February 2007 (UTC)[reply]
So please tell me how people solved "demented matrices" before electronic computers were invented? What methods do they used in the 19th century? 202.168.50.40 02:27, 16 February 2007 (UTC)[reply]
They didn't. Augmented matrices are a technique for solving systems of linear equations designed to be done by a computer or programmable calculator. --Carnildo 19:27, 16 February 2007 (UTC)[reply]
Essentially the question is how to solve a system of linear equations by manual calculation, reliably and quickly. The augmented matrix
 
represents the matrix equation
 
which, in turn, corresponds to
 
The standard solution techiques in numerical linear algebra factor the matrix. The most popular general method is LU decomposition with partial pivoting (and perhaps other safeguards). The singular value decomposition is more robust and flexible, but also much more expensive.
When a problem is presented as an augmented matrix, chances are you are expected to solve it using Gaussian elimination. This is a naive, more expensive, and less robust historical predecessor of LU decomposition. However, the theoretical underpinnings are the same. We employ two tools to change the form without changing the solution. First, we can multiply any row by a scalar value. Second, we can add or subtract a scalar multiple of any row to any other. To be robust, we must include a third option, exchanging two rows (pivoting).
In most cases, a series of such operations will produce
 
at which point we have revealed the solution, x = a, y = b, z = c. If we are unlucky, the last "1" (at least) will be a "0", preventing solution; but if we are really lucky, we simultaneously get c = 0, so we have a solution after all (in fact, any value of z will do).
With today's widespread availability of open-source computer algebra systems, there is rarely any practical need to solve such problems manually. The main reason to do so is to understand the fundamental ideas (which I recommend); in which case, being fast misses the point. --KSmrqT 04:09, 16 February 2007 (UTC)[reply]
The command to do this to any matrix on your Texas Instruments graphing calculator is rref( found under a MATRIX submenu. I suggest doing your best to understand what KSmrq said even if you are in middle or high school and asking back for clarification. [Mαc Δαvιs] X (How's my driving?) ❖ 08:23, 19 February 2007 (UTC)[reply]

Regression Analysis

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(question moved from Misc RD)

How would you best explain regression analysis in mathematical and practical terms? Thanks. --61.6.206.104 15:48, 15 February 2007 (UTC)[reply]

Did you read our entry on regression analysis?  --LambiamTalk 20:39, 17 February 2007 (UTC)[reply]