So, I'm not really sure how to calculate the amount of solutions for a system with 3 equations.
All I know is that it has something to do with matrices and the discriminant, but I'm not sure where to go with that.
Here's the system that I'm working on:
$$x+2y+3z=7$$ $$3x+4y-z=4$$ $$3x+2y-11z=-13$$
On
Got it, thanks Amzoti!
$$ \left[\begin{array}{rrr|r} 1 & 2 & 3 & 7 \\ 3 & 4 & -1 & 4 \\ 3 & 2 & -11 & -13 \end{array}\right] $$
$$ \left[\begin{array}{rrr|r} 1 & 2 & 3 & 7 \\ 0 & -2 & -10 & -17 \\ 0 & -4 & -20 & -34 \end{array}\right] $$
$$ \left[\begin{array}{rrr|r} 1 & 2 & 3 & 7 \\ 0 & -2 & -10 & -17 \\ 0 & 0 & 0 & 0 \end{array}\right] $$
$$0=0$$
Infinite amount of solutions
We have the system $$x+2y+3z=7$$ $$3x+4y-z=4$$ $$3x+2y-11z=-13$$
The associated augmented coefficient matrix is given by:
$$\begin{pmatrix} 1 & 2 & 3&|&7\\3 & 4 & -1 &|&4 \\ 3 & 2 & -11 &|& -13\end{pmatrix}$$
Use Gaussian Elimination (row reduction using elementary row operations). If you encounter a row with all zeros, then you'll know there are infinitely many solutions. You'll have an inconsistent system, for which there are NO solutions, if you obtain a row such as $$(0\;0\;0\;|\; a)$$ where $a \neq 0$. Otherwise, the system will have one unique solution.
If you row reduce correctly, you should obtain a reduced row form as follows:
$$\begin{pmatrix} 1 & 0 & -7 &|& -10 \\ 0 & 1 &5 & | & \frac{17}{2} \\ 0 & 0 & 0 &|&0\end{pmatrix}$$