Inspired by the thread "Mathematical Logic Fails Me", or whatever the actual name was. Also by the multitude of Maths threads we've had in the past (0.9 recurring (i.e. 9's forever) DOES equal 1, foolish people!).
As some of you may have read in my other posts or on my profile, I'm currently a second year student at the University of Leeds, studying Mathematics BSc. As a result of the varied modules I do I see a lot of crazy equations and problems, and so I thought I'd see how mathematical the Escapist is. What I want is for everyone to pose a problem, no matter how simple or complex it is, and try to solve other people's problems. If you want to, add your own problem's solution in a spoiler box so people can check their answer against yours. And no flaming please, if someone can prove their reasoning properly then they are obviously correct (the basis of maths in general), so if you don't agree with someone's solution then don't comment unless you can find a better one.
I'll start you off with a (relatively) easy one:
Find the solution to the following set of linear equations by forming a matrix and reducing it to reduced row echelon form (i.e. diagonal is all 1's, all other numbers are zero save the constant terms at the end) using Gaussian Elimination.
x + 2y - z = 1
x + 3y - 2z = -1
2x + y = 2
As some of you may have read in my other posts or on my profile, I'm currently a second year student at the University of Leeds, studying Mathematics BSc. As a result of the varied modules I do I see a lot of crazy equations and problems, and so I thought I'd see how mathematical the Escapist is. What I want is for everyone to pose a problem, no matter how simple or complex it is, and try to solve other people's problems. If you want to, add your own problem's solution in a spoiler box so people can check their answer against yours. And no flaming please, if someone can prove their reasoning properly then they are obviously correct (the basis of maths in general), so if you don't agree with someone's solution then don't comment unless you can find a better one.
I'll start you off with a (relatively) easy one:
Find the solution to the following set of linear equations by forming a matrix and reducing it to reduced row echelon form (i.e. diagonal is all 1's, all other numbers are zero save the constant terms at the end) using Gaussian Elimination.
x + 2y - z = 1
x + 3y - 2z = -1
2x + y = 2
Take this as a matrix:
[1 2 -1 ... 1]
[1 3 -2 ... -1]
[2 1 0 ... 2]
Then reduce as follows:
[1 2 -1 ... 1]
[0 1 -1 ... -2]
[0 -3 2 ... 0]
[1 2 -1 ... 1]
[0 1 -1 ... -2]
[0 0 -1 ... -6]
[1 0 1 ... 5]
[0 1 -1 ... -2]
[0 0 1 ... 6]
[1 0 0 ... -1]
[0 1 0 ... 4]
[0 0 1 ... 6]
Hence we can see that x=(-1), y=4, z=6.
To check, plug these into the original linear equations:
x + 2y - z = -1+8-6 = 1 which is correct.
x + 3y - 2z = -1+12-12 = -1 which is correct.
2x + y = -2+4 = 2 which is correct.
Hence all equations are correct with these values, so the solution set is valid:
x = -1
y = 4
z = 6
[1 2 -1 ... 1]
[1 3 -2 ... -1]
[2 1 0 ... 2]
Then reduce as follows:
[1 2 -1 ... 1]
[0 1 -1 ... -2]
[0 -3 2 ... 0]
[1 2 -1 ... 1]
[0 1 -1 ... -2]
[0 0 -1 ... -6]
[1 0 1 ... 5]
[0 1 -1 ... -2]
[0 0 1 ... 6]
[1 0 0 ... -1]
[0 1 0 ... 4]
[0 0 1 ... 6]
Hence we can see that x=(-1), y=4, z=6.
To check, plug these into the original linear equations:
x + 2y - z = -1+8-6 = 1 which is correct.
x + 3y - 2z = -1+12-12 = -1 which is correct.
2x + y = -2+4 = 2 which is correct.
Hence all equations are correct with these values, so the solution set is valid:
x = -1
y = 4
z = 6
ROFIT!