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x4 + 2x5 - x6 = 2
x1 + 2x2 + x5 - x6 = 0
x1 + 2x2 + 2x3 - x5 + x6 = 2

I know there's 6 variables and only 3 equations, but that shouldn't stop you from getting a solution. It says find all solutions. so if the solution is dependent on a variable that's fine. i just need a statement of all solutions

2007-05-01 09:56:51 · 3 answers · asked by brandon 5 in Science & Mathematics Mathematics

3 answers

Now I can show you how to use matrices to solve a system of equations. First, create an augmented matrix:

[0 0 0 1 2 -1 | 2]
[1 2 0 0 1 -1 | 0]
[1 2 2 0 -1 1 | 2]

First, subtract row 2 from row 3:

[0 0 0 1 2 -1 | 2]
[1 2 0 0 1 -1 | 0]
[0 0 2 0 -2 2 | 2]

Divide the third row by 2:

[0 0 0 1 2 -1 | 2]
[1 2 0 0 1 -1 | 0]
[0 0 1 0 -1 1 | 1]

Swap rows 1 and 2:

[1 2 0 0 1 -1 | 0]
[0 0 0 1 2 -1 | 2]
[0 0 1 0 -1 1 | 1]

Now swap rows 2 and 3:

[1 2 0 0 1 -1 | 0]
[0 0 1 0 -1 1 | 1]
[0 0 0 1 2 -1 | 2]

There are now more elementary row operations to perform, since the pivot columns (the 1's to the far left of each row) have nothing but zeros above them. The answer, then, is

x1 = -2x2 - x5 + x6
x2 = free (meaning it can take any value)
x3 = x5 - x6 +1
x4 = -2x5 + x6 + 2
x5 = free
x6 = free

Hope that, along with the other explanation I gave you, helps.

2007-05-01 20:21:42 · answer #1 · answered by Crystal 3 · 1 0

First of all, rewrite the equations so that the different variables are arranged in columns, using zeros as placeholders wherever necessary like this

0x1 + 0x2 + 0x3 + 1x4 + 2x5 - 1x6 = 2
1x1 + 2x2 + 0x3 + 0x4 + 1x5 - 1x6 = 0
1x1 + 2x2 + 2x3 + 0x4 - 1x5 + 1x6 = 2

Now you can use Gauss-Jordan elimination to solve as far as you can. Even after this your answers will still look something like

x4 = (2x2 - x5)/3

You won't get a numerical answer

2007-05-01 17:07:47 · answer #2 · answered by dogsafire 7 · 0 0

I can't see any tricks, so you will have to figure to express the value of each of three selected variables in terms of the other three variables.

2007-05-01 17:04:57 · answer #3 · answered by cattbarf 7 · 0 0

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