When solving via gauss-jordan When I solve via Gauss-Jordan, taking a $3 \times3$ matrix as an example...
Should I always try and get a $1$ in the upper left corner and $0$'s in the rest of the column followed by getting a $1$ in the middle $2^{\text{nd}}$ column, $2^{\text{nd}}$ row followed by $0$'s in the $2^{\text{nd}}$ column and then finally get a $1$ in the bottom-right corner followed by $0$'s in the $3^{\text{rd}}$ column?
Is this THE system to solve these or does it just vary? Could I be trying to get $1\ 0\ 0$ in the first row for example and work rows like that?
 A: Looking at the $3\times3$ case, you only need to get it in the form
$$
        \begin{pmatrix}
        a & b & d \\
        0 & c & e \\
        0 & 0 & f \\
        \end{pmatrix}
$$
This is called an upper triangluar matrix.  When you are solving the system $A\mathbf{x}=\mathbf{b}$, this allows you to solve the system by substitution.  If $\mathbf{b}=(1 1 1)^T$, for example, then $fz=1$ which implies that $z=\frac1{f}$.  Now you can back substitute in row 2 and solve for y, and then back substitute again and solve for x.  Solving in this way allows us the substitution method for solving systems of equations.  It would be easier to solve though if our matrix was in the form
$$
        \begin{pmatrix}
        1 & a & b \\
        0 & 1 & c \\
        0 & 0 & 1 \\
        \end{pmatrix}
$$
since now there is no dividing by the coefficient to get $x$, say.  Finally, you can understand why it is so efficient to put the system into RREF, or reduced row echelon form which is
$$
        \begin{pmatrix}
        1 & 0 & 0 \\
        0 & 1 & 0 \\
        0 & 0 & 1 \\
        \end{pmatrix}
$$
or something similar to this depending on the dependence of the vectors involved since then it is easy to see that, in our $(1 1 1)^T$ example that $x=1, y=1,$ and $z=1$.  I hope this helps.
