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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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  • $\begingroup$ Where does $\pmatrix{-1&2&3&4&0 \\ 2&1&-3&4&1 }$ come from? Do you mean $\pmatrix{-1&2&1&4 \\ 2&1&-1&1 }\cdot \vec x= \pmatrix{0\\ 1}$? $\endgroup$
    – draks ...
    Oct 4, 2012 at 21:11
  • $\begingroup$ Yes, I really wanted an augmented matrix $\endgroup$ Oct 4, 2012 at 21:14
  • $\begingroup$ Why do you add coloumns instead of rows? $\endgroup$
    – draks ...
    Oct 4, 2012 at 21:26
  • $\begingroup$ 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. $\endgroup$
    – Cocopuffs
    Oct 4, 2012 at 21:53

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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  • $\begingroup$ We both copied the system into the matrix incorrectly... Or you were just copying me... $\endgroup$ Oct 4, 2012 at 21:40
  • $\begingroup$ I copied yours, but see above for what it should be. $\endgroup$
    – adam W
    Oct 4, 2012 at 21:42
  • $\begingroup$ Yes, I copied it down wrong from paper. However, I copied down the equations correctly. Why do they not work? $\endgroup$ Oct 4, 2012 at 21:48
  • $\begingroup$ It is not solved correctly yet, see above; do not scale only part of an equation. $\endgroup$
    – adam W
    Oct 4, 2012 at 21:52
  • $\begingroup$ Thanks, that was the issue! Could you also explain how I can check this solution? $\endgroup$ Oct 4, 2012 at 21:54
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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: $$x = \frac{1}{5} \pmatrix{3 \\ -1 \\ 5 \\ 0} x_3 + \frac{1}{5} \pmatrix{2 \\ -9 \\ 0 \\ 5} x_4 + \frac{1}{5} \pmatrix{2 \\ 1 \\ 0 \\ 0}$$

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