Is the following matrix in reduced row echelon form?

Question

\begin{pmatrix}1 & 3 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 1 & 5 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}

Answer 1

For a matrix to be in reduced row echelon form, it must be in row echelon form.

For a matrix to be in row echelon form, the leading coefficient of a nonzero row

must be always strictly to the right of the leading coefficient of the row above it.

Therefore, your matrix is not in reduced row echelon form.

Answer 2

Is is not even in row echelon form.

You need to swap rows two and three to get $$\begin{pmatrix}1 & 3 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 5 & 1 & 0 \\ 0 & 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix},$$ which is in row echelon form.

To obtain the reduced row echelon form, the new third row needs to be divided by $4$ and subtracted from the second and first rows $5$ and $1$ times, respectively, to obtain $$\begin{pmatrix}1 & 3 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}.$$ This is the reduced row echelon form of your matrix.