Best way to find Reduced Row Echelon Form (rref) of a matrix? I'm sitting here doing rref problems and many of them seem so tedious. Any tricks out there to achieve rref with less effort or am I stuck with rewriting the matrix for every 2/3 operations?
I know TI calculators can do it, but I'm gonna have to do this on my midterm, so I must learn how to do this the most efficient way possible. Thanks.
 A: I don't know any general techniques but here are a couple of specific techniques I can think of off the top of my head:

$1)$ If you see a row which is a linear combination of the other rows then you can immediately replace it with a row of zeroes.  Example:
$$\begin{bmatrix} 3 & 1 & 2 \\ 7 & 4 & 2 \\ -1 & -2 & 2 \end{bmatrix}$$
If you happen to notice that $R_3$ (my notation for the third row) is just $R_3 = 2R_1 - R_2$, then you can immediately simplify this to
$$\begin{bmatrix} 3 & 1 & 2 \\ 7 & 4 & 2 \\ 0 & 0 & 0 \end{bmatrix}$$

$2)$ If you're working on the $n$th column and you notice that dividing any of the rows by the first number will result in fractions (which slow you down), try to look for a number which is a factor of another number minus 1.  Example:
$$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 3 & 4 & 5 \\ 0 & 7 & 4 & 2 \\ 0 & 8 & 5 & 11\end{bmatrix}$$
Here if we divided $R_2$ by $3$, for instance, to get a $1$ in a pivot position, then we'd be getting fractions.  Instead you could notice that $3$ is a factor of $7-1=6$.  So perform the operation $R_3 \to R_3 - 2R_2$ to get
$$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 3 & 4 & 5 \\ 0 & 1 & -4 & -8 \\ 0 & 8 & 5 & 11\end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & -4 & -8 \\ 0 & 3 & 4 & 5 \\ 0 & 8 & 5 & 11\end{bmatrix}$$
Now you've got a $1$ in your $2$nd pivot position without any fractions, yet.

$3)$ Work on an entire column at once.  Starting at where I left off in the last example, we'd then have
$$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & -4 & -8 \\ 0 & 3 & 4 & 5 \\ 0 & 8 & 5 & 11\end{bmatrix}_{\begin{array} 3R_1 \to R_1 - 2R_2 \\ R_3 \to R_3 - 3R_2 \\ R_4 \to R_4 - 8R_2\end{array}} \sim \begin{bmatrix} 1 & 0 & 11 & 20 \\ 0 & 1 & -4 & 8 \\ 0 & 0 & 16 & 29 \\ 0 & 0 & 37 & 75\end{bmatrix}$$
if I didn't make any mistakes.  This saves time writing which can be crucial during an exam.

$2.5)$ (This is really just the same technique as in $2$)  Looking at the above matrix it looks like we're going to have to have some nasty fractions at this point, right?  If we still wanted to avoid fractions, check this out:  First subtract $2R_3$ from $R_4$ to get a $5$ in the third column of $R_4$, then subtract $3(R_4)_{\text{new}}$ from $R_3$ to get a $1$ in the pivot position. And of course, don't write down the entire matrix twice to do these two steps.  Just do the little bit of arithmetic on the side (or in your calculator or in your head) and only write down the matrix once you've gotten the $1$ in the row you want it in. -- Keep in mind, though, that if your TA can't follow your logic, you may be counted off even if you are correct.

$4)$ This one should be a no brainer, but sometimes you forget about basic stuff in the middle of an exam.  If you happen to know that the matrix equation $Ax=b$ has a unique solution $x$ or that the determinant of your square matrix is non-zero, then you can immediately say that the RREF of that matrix is the identity matrix -- no Gauss-Jordan required.
A: I dont know it this answers your concerns, but here is my advice. The main problem is arithmetical errors, and there are two things which increase the likelyhood of such errors, 1)Fractions, 2)Copying errors.  
So use any $1$'s which occur in the matrix to make that column zero, even if it is not the first column. Second write only one new matrix for this, in which other entries in this column are zero, in other words do not rewrite the matrix for each row operation, this reduces work and the likelyhood of copying errors.
Often following these two rules will make the task much easier. 
