Question

Is the matrix below in REF? I think no, because all leading coefficients must be 1 right? Since the leading coefficient of the 2nd row is not equal to one, then it should not be in REF... But how come the teacher says my reasoning is not right and he even told us that "all leading entries must be 1 when a matrix is in RREF, not in REF". So what is this matrix? Is it already in REF or not? I'm confused now.

Answer 1

@Jhannybean @Loser66

Answer 2

@Directrix

Answer 3

why is it reduced echelon form already?

Answer 4

it's not reduced

Answer 5

there's a difference between reduced echelon form, and just echelon form

Answer 6

lol you just said it O.o

Answer 7

"it's in reduced echelon form (REF) "

Answer 8

meant to say row echelon form

Answer 9

just to clarify:

REF = row echelon form
RREF = reduced row echelon form

in REF, the first nonzero term can be something other than 1

Answer 10

now it's gone.. oh okay..

Answer 11

anyway, your matrix satisfies the requirements for row echelon form but not reduced echelon form

the requirements for REF are:

- all nonzero rows must be above the all-zero rows
- all entries below a leading entry must be zero
- each leading entry must be in a column to the right of the leading entry above it

Answer 12

aah now it's clear! Thank you so much!

Answer 13

wait, can you also make points for RREF... pleasse?

Answer 14

like the requirements for RREF too..

Answer 15

for RREF, it must satisfy all the requirements of REF with the addition of

- the leading coeffiecient must be 1, with the exception of zero rows

Answer 16

basically, if you can draw a "staircase" above the non-zero entries, then it's REF