First off, I would like to apologize if this is the wrong forum I am posting in. I looked at all the forums and I believe that the Calculus forum is the most appropriate for this question, but if I made a wrong decision then I'm sorry!

The following equation system is a subspace in $\displaystyle \mathbb{R}^5$:

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

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

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

Questions:

How many dimensions is the subspace?

What is a basis for this subspace?

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Okay so first off, I want to figure out how many dimensions the subspace is.

I put the equation system into a totalmatrix:

$\displaystyle \begin{bmatrix}0 & 1 & 3 & -1 & 2 & 0\\ 2 & 3 & 1 & 3 & 0 & 0\\ 1 & 1 & -1 & 2 & -1 & 0\end{bmatrix}$

I then find its parametric solution:

$\displaystyle \begin{bmatrix}x_1\\ x_2\\ -2x_1-3x_2-3x_4\\ x_4\\ 3x_1+4x_2+5x_4\end{bmatrix}$

But here comes my problem...

According to my textbook, the answer is that it has 3 dimensions!

And in my parametric solution I only have 2 vectors (row 3 and 5) and two vectors only span over a 2 dimension!

So I have no idea what it is I have done wrong. Why am I not getting 3 dimensions like my textbook?