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Given the system of equations:

$$x_2+3x_3-x_4+2x_5=0$$

$$2x_1+3x_2+x_3+3x_4=0$$

$$x_1+x_2-x_3+2x_4-x_5=0$$

The solution is a subspace in $\mathbb{R}^5.$ Determine a basis for this subspace.

Starting with matrix $ \begin{pmatrix} 0 & 2 & 1 \\ 1 & 3 & 1 \\ 3 & 1 &-1 \\ -1 & 3 & 2 \\ 2 & 0 &-1 \\ \end{pmatrix} $ I got the reduced matrix $ \begin{pmatrix} 1 & 0 &-\frac12 \\ 0 & 1 & \frac12 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $.

The solution I got is:

$$(x_1,x_2,x_3)=t(1/2,-1/2,1),t\in \mathbb{R}$$

How do I determine a basis with only one vector? Obviously they have to be linearly independent.


Never mind, I messed up. I accidentally transposed the matrix without noticing. I should get more sleep. The new solution I got was:

$$(x_1,x_2,x_3,x_4,x_5)=t_1(4,-3,1,0,0)+t_2(-3,1,0,1,0)+t_3(3,-2,0,0,1)$$

Which is correct.

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    $\begingroup$ Shouldn't your solution be for $(x_1,x_2,x_3,x_4,x_5)$? How many free variables appear in your reduced system? $\endgroup$ – G Tony Jacobs Dec 2 '16 at 4:01
  • $\begingroup$ Only one. Rows 3, 4 and 5 in reduced form are zero vectors. $\endgroup$ – Steve Dec 2 '16 at 4:02
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    $\begingroup$ This is the reduced matrix I get. From that, I only have one free variable: $$ \begin{pmatrix} 1 & 0 & -1/2 \\ 0 & 1 & 1/2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $$ $\endgroup$ – Steve Dec 2 '16 at 4:09
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    $\begingroup$ You are transposed. And put any work that you have done in the original post, and not in the comments. $\endgroup$ – Doug M Dec 2 '16 at 4:35
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    $\begingroup$ As other users said in the above comments, knowing how you obtained the (incorrect) solution is rather essential for the answers - otherwise they cannot say what you did wrong. I believe it can be more-or-less guessed from your comments what you did. Please, check my edit, and if what I wrote is substantially different from what you did, then edit your post again. $\endgroup$ – Martin Sleziak Dec 2 '16 at 10:37
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Sketch:

Step 1: Create the augmented matrix for this system.

Step 2: Row reduce the system into reduced row echelon form.

Step 3: Turn your system back into equations

Step 4: Solve for your pivot variables in terms of your free variables.

Step 5: One at a time, set one free variable equal to $1$ and the remaining free variables equal to $0$. The values of all variables in this case (pivot and free) form your basis vectors.

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  • $\begingroup$ What if my row reduced system only contains one free variable? See above for what it looks like. $\endgroup$ – Steve Dec 2 '16 at 4:18
  • $\begingroup$ That is not the augmented system. The augmented system should have five columns and three rows. $\endgroup$ – Michael Burr Dec 2 '16 at 4:23

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