As I'm learning linear algebra, this questions confuses me so much.
which of the following matrices are in row echelon form? For each matrix not in row echelon form, explain why it is not
$$ a) \left( \begin{matrix} 1 & -1 & 4 \\ 0 & 2 & -7 \\ 0 & 0 & 3 \\ \end{matrix} \right) $$
$$ b) \left( \begin{matrix} 1 & 4 & -4 & -5 \\ 0 & 0 & 4 & -8 \\ 0 & 0 & 0 & 0 \\ \end{matrix} \right) $$
$$ c) \left( \begin{matrix} 1 & -2 & 1 & -2 & 1 \\ 0 & 0 & 1 & -8 & -3 \\ 0 & 1 & 0 & 0 & 4 \\ \end{matrix} \right) $$
The reason why i'm getting super confused is because all these 3 matrices can be transformed into row echelon form. For instance if you divide row 2 in matrix $a$ and divide 3, that gets you row echelon form. And for $c$) if you change row 2, 3 then you get row echelon form. However, this question specifically asked if these 3 matrices are in row echelon form. So I don't know if transforming is allowed. What do you guys think about this question?
Because i know in order to be row echelon form implies that:
- The leading (first) entry in each row must be 1
- The leading entry on each subsequent row must be on a new column to the right
- All rows where all entries are zero are below rows where NOT all entries are zero
It should be noted that if a matrix is not in row echelon form, then it is not in row echelon form (even if it can be transformed to row echelon form!)
Definitions of REF can vary (see the Wikipedia one: https://en.wikipedia.org/wiki/Row_echelon_form), so by your definition none of these matrices are in REF, since none of them have leading entries all equal to 1.
However, if this condition on the leading entries were relaxed (that is, leading entries do not have to be equal to 1), then (a) and (b) are in REF, and (c) is not (because it violates condition 2...can you see why?).
Hope this helps! :)