A matrix question(simple one i suppose)

[georgeh]

I have th following matrix and I am suppose to determine whether the matrix is in row-echelon form(R-E), reduced row-echelon(R-RE) form, both, or neither..

I know the following about R-R-E
* If there is a row of all 0's there at the bottom of the matrix
* The upper left most number has to be a 1
* the following number below it has to be to the right of the leading 1 has to be a 1..
* in R-RE any number above the 1 and below have to be 0..
- For R-E they have to be only below the 1 not above it..
all right,
So here is the matrix:
1 2 0 3 0
0 0 1 1 0
0 0 0 0 1
0 0 0 0 0

The book shows the answer as being both a R-R-E and R-E.
I disagree because the 1 below the 3, if it were in R-R-E, that 3 would have to be a 0. Therefore, I believe it is in R-E form. Am I correct in my reasoning?
thanks in advance..

 

[assyrian_77]

This is what PlanetMath has to say about R-E:

A matrix is said to be in row echelon form if each non-zero row has more leading zeros than the previous row

So, your argument that the 3 has to be a 0 would even kill that and you would have neither.

Further more, PlanetMath says this about R-R-E:

For a matrix to be in reduced row echelon form it has to first satisfy the requirements to be in row echelon form and additionally satisfy the following requirements:

The first non-zero element in any row must be 1.
The first element of value 1 in any row must the only non-zero value in its column.

Now, look at these arguments. Can you tell why your matrix is indeed R-R-E?

 

[HallsofIvy]

* in R-RE any number above the 1 and below have to be 0..

Notice the word the in "the 1". It does not say "each 1 in the row". That clearly refers to the leading 1 in the row, not the second 1.

 

[georgeh]

doh! I see said the blind man to his deaf wife.