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].Example 1: TryIt 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.
[ 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.
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.