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Describe all solutions of the system (the last column is the augmented column)

$-x_1 +2x_2 + x_3 + 4x_4 = 0$

$2x_1 + x_2 -x_3 +x_4 = 1$

$\pmatrix{-1&2&1&4&0 \\ 2&1&-1&1&1 } \sim \pmatrix{1&-2&-1&-4&0 \\ 0&5&1&9&1} \sim \pmatrix{1&0 & -\dfrac{3}{5}& -\dfrac{2}{5}& \dfrac{2}{5} \\ 0&5&1&9&1} $

Solving we get the equations:

$x_1 = \dfrac{3}{5}x_3 + \dfrac{2}{5}x_4 + \dfrac{2}{5} \iff x_1 = 3x_3 + 2x_4 + 2$ (Can I scale like this?)

$5x_2 = -x_3 -9x_4 +1 \iff x_2 = x_3 +9x_4 - 1$ (can I scale like this?)

Now we have:

$\langle x_1, x_2, x_3, x_4\rangle = \langle 3x_3 + 2x_4 +2, -x_3 -9x_4 + 1, x_3, x_4\rangle$

$= x_3\langle 3, -1, 1, 0 \rangle + x_4\langle 2, -9, 0, 1\rangle + \langle 2,1,0,0\rangle $

Where did I go wrong?

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    I don't understand your scaling. The right hand equations aren't the same as the left, so they don't necessarily have the same solution space.2012-10-04

2 Answers 2

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Actually it should start with:

$\pmatrix{-1&2&1&4&0 \\ 2&1&-1&1&1}$

In response to your (can I scale like this) question, no. It is an equation and to preserve the equality, all elements should be scaled, not just one side of the equation.

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    Either algebraically with variable $x_3$ and $x_4$ as unknowns (they should cancel out to get 0=0 and 1=1) or with concrete values for the unknowns (try 1 or 5 to hopefully cancel fractions)2012-10-04
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You need to be careful scaling.

You have the equations: $-x_1+\frac{3}{5}x_3 + \frac{2}{5}x_4 = -\frac{2}{5}, \ \ \ 5 x_2 + x_3 + 9x_4 = 1$ which can be written as $x_1 = \frac{3}{5}x_3 + \frac{2}{5}x_4 + \frac{2}{5}, \ \ \ x_2 = -\frac{1}{5} x_3 - \frac{9}{5} x_4 + \frac{1}{5}$ Thus all the solutions are: $x = \frac{1}{5} \pmatrix{3 \\ -1 \\ 5 \\ 0} x_3 + \frac{1}{5} \pmatrix{2 \\ -9 \\ 0 \\ 5} x_4 + \frac{1}{5} \pmatrix{2 \\ 1 \\ 0 \\ 0}$