Solve the following system for $a$, $b$, and $c$: $$\begin{pmatrix}1 & -1 & 2\\2 & -2 & 2 \\ 3 & -3 & 2\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$$
Using row reduction I got to this point: $$\begin{pmatrix}1 & -1 & 2\\0 & 0 & -2 \\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$$
I know the first row gives the equation $$a-b+2c=0$$ and the second row gives the equation $$-2c=0$$
but how should I interpret the last row (the one full of zeros)? Does a row of zeros imply the corresponding variable is zero, so $c=0$? Or does it imply that $c$ can take any value?
Thank you!
As you correctly interpreted the first two equations, the same logic applied to the last one.
A row of zeroes implies that the variable is free, and can thus take any value.
However, the variable to which you applied it is incorrect. $c=0$ by equation #2, but you have no equation fixing $b$, which can take any value.
As a result, you have $b \in \mathbb{R}$. Therefore, when you solve, you do not have a point of intersection of 3 planes, but a line, since one of the planes can be expressed as a weighted sum of the other two planes. Continuing the process: $$\left [ \begin{array}{ccc} 1 & -1 & 2 & 0 \\ 0 & 0 & -2& 0 \\ 0 & 0 & 0 & 0 \end{array} \right ]$$ $$\left [ \begin{array}{ccc} 1 & -1 & 2 & 0 \\ 0 & 0 & 1& 0 \\ 0 & 0 & 0 & 0 \end{array} \right ]$$ $$\left [ \begin{array}{ccc} 1 & -1 & 0 & 0 \\ 0 & 0 & 1& 0 \\ 0 & 0 & 0 & 0 \end{array} \right ]$$ This is the reduced-row-echelon "rref" form. The leading ones in each row have no other non-zero entries in the given column. This is the simplest form of the equations. Translating back to the original variables, you get: $$a-b=0,c=0,b\in \mathbb{R}$$ Therefore, the solution vector is: $$\left ( \begin{array}{ccc} a\\b\\c \end{array} \right )= \left ( \begin{array}{ccc} b\\b\\0 \end{array} \right )= \left ( \begin{array}{ccc} 1\\1\\0 \end{array} \right ) b$$ Notice how we have $a$ and $c$ in terms of $b$, since there is only one free variable.