5. (6 pts) Find a basis for the following subspace: $$ \left\{ \begin{bmatrix} r - s - 2t - u \\ -r + 2s + 5t + 2u \\ s + 3t + u \\ 3r + s + 5u \end{bmatrix} \in \mathbb{R}^4 : \text{$r, s, t$ and $u$ are scalars} \right\}. $$
So I have this matrix: $$\begin{pmatrix} 1 &-1 &-2 &-1 \\ -1 &2& 5 &2 \\ 0 &1 &3 &1 \\ 3 &1 & 0 & 5 \end{pmatrix} $$ If I compute the the matrix in reduced echelon form, I get: $$\begin{pmatrix} 1&0&1&0\\ 0&1&0&3\\ 0&0&1&-2/3\\ 0 & 0 & 0 & 0 \end{pmatrix}$$
Please note that i want to put the matrix in reduced echelon form , and not row reduced echelon form.
I know that the three rules of reduced echelon form are:
1) The first non zero number of a row is 1 (leading entry).
2) An all zero row is placed at the end of the matrix.
3)A leading entry of a row is placed to the right of a leading entry of the previous row.
If it were to be in row reduced echelon form , then another property would hold, or else that each column that contains a leading entry, all other entries in the same column are zeros.
So is my matrix correctly put in a row echelon form ?