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.
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.