I was asked in a recent message (Sunday, 2/6/2011) if the answers shown are the only answers. Here is my response.
Let's look at a very simple example: consider the (idiotic) 1-by-1 matrix, [7].

It is already in row echelon form. But [-13] is also a row echelon form of this matrix. Sigh. There are many row echelon forms of a matrix.

So there can be many different answers to the question: "Change the matrix BLAH to row echelon form." I fear the worst when grading that problem.

HOWEVER, the RREF is unique. That is, there is also one and only one RREF of a matrix (for the matrix [7], that unique RREF is [1]).

This is why I was a bit careful in formulating the quiz language. The first question asks "Find a row echelon form of the matrix" and the second question asks "Find the reduced row echelon form of the matrix". There is a difference between "a" and "the". The same is true for the answers. I used Maple's commands to get "a" row echelon form. As it happens (I checked a few of the examples by hand) what I got when I worked through the text's "Forward pass" happens to be what the Maple command printed. BUT any valid answer (and there are many, many of them!) will or would be acceptable.

I hope this is helpful.

Example 1: Try 
[ 0     0    2     2     1]
[                         ]
[ 1    -1    0    -1     2]
[                         ]
[-1     0    0     1     1]
[                         ]
[ 2     0    0     2    -1]
Do the "forward pass" of the textbook's form of Gaussian elimination to get a row echelon form of the matrix, and then complete the "backward pass" to get the row reduced echelon form.
An answer is here.

Example 2: Try 

[-1    1     1    0    -1]
[                        ]
[ 2    0     0    1     2]
[                        ]
[ 2    0    -1    1     2]
[                        ]
[-1    2    -1    0     1]

Do the "forward pass" of the textbook's form of Gaussian elimination to get a row echelon form of the matrix, and then complete the "backward pass" to get the row reduced echelon form.
An answer is here.

Example 3: Try 

[-1     1     2    -1    0]
[                         ]
[-1     0     1     2    1]
[                         ]
[ 1     0    -1     1    1]
[                         ]
[ 1    -1     2     1    1]

Do the "forward pass" of the textbook's form of Gaussian elimination to get a row echelon form of the matrix, and then complete the "backward pass" to get the row reduced echelon form.
An answer is here.

Make your own examples!
I used Maple to create the examples I've given here. I will also use the same method to create the problems which I will give you on Monday. I will give you a "random" matrix for the first part, and a matrix in row echelon form (unrelated!) for the second part. So if you want to practice more, you could create your own examples. Here is the "software", not too elaborate: 

> with(linalg):
> A:=rand(-1..2):
> B:=proc()local T;T:=matrix(4,5,[seq(A(),j=1..20)]);print(T,LUdecomp(T),rref(T));end:
> B();
         [2     0     1    -1    -1]  [2     0     1      -1     -1 ]  [1    0    0    0    0]
         [                         ]  [                             ]  [                     ]
         [1    -1    -1     0     0]  [0    -1    -3/2    1/2    1/2]  [0    1    0    0    0]
         [                         ], [                             ], [                     ]
         [0     1     0     2     2]  [0     0    -3/2    5/2    5/2]  [0    0    1    0    0]
         [                         ]  [                             ]  [                     ]
         [1     0    -1    -1    -1]  [0     0     0      -3     -3 ]  [0    0    0    1    1]
In fact, you could do every possible example. There are 20 entries, and I'll put –1, 0, 1, or 2 in each entry. So there are only 420=1099511627776 (that's 13 digits) examples. Sigh. Most (but not all!) will have rank 4 and therefore nullity 1.

Maintained by greenfie@math.rutgers.edu and last modified 2/1/2011.