RREF using mod 2 operations

Question

Can someone please help me calculate the reduced row echelon form of the following matrix: $$ \begin{bmatrix} 1&1&1&0 \\ 1&1&0&1 \\ 0&0&1&1 \end{bmatrix} \in M_{3,4}(F_2)$$ Where $F_2$ denotes the field of scalars $\{0,1\}$ with operations doen using mod $2$ arithmetic.

I am having problems because no matter what I do, I get no leading entry in column 2. For instance, adding $R_1$ to $R_2$ would make $R_2= \{0,0,1,1\}$.

Can there be no leading entry in the second column of second row? From what I have learned, each column must have a leading entry except for in the bottom row.

Answer 1

After adding the first row to the second one, you get, as you wrote:$$\begin{bmatrix}1&1&1&0\\0&0&1&1\\0&0&1&1\end{bmatrix}.$$Then, after adding the second row to the first and to the third ones, you get:$$\begin{bmatrix}1&1&0&1\\0&0&1&1\\0&0&0&0\end{bmatrix}.$$And this matrix is in RREF.

Answer 2

There's no difference in the algorithm: \begin{align} \begin{bmatrix} 1&1&1&0 \\ 1&1&0&1 \\ 0&0&1&1 \end{bmatrix} &\to \begin{bmatrix} 1&1&1&0 \\ 0&0&1&1 \\ 0&0&1&1 \end{bmatrix} && R_2\gets R_2+R_1 \\[2ex]&\to \begin{bmatrix} 1&1&1&0 \\ 0&0&1&1 \\ 0&0&0&0 \end{bmatrix} && R_3\gets R_3+R_2 \\[2ex]&\to \begin{bmatrix} 1&1&0&1 \\ 0&0&1&1 \\ 0&0&0&0 \end{bmatrix} && R_1\gets R_1+R_2 \end{align} Modulo $2$ one has never to reduce the pivot.

There is no pivot in the second column because it's equal to the first column, so it is a linear combination of the preceding pivot columns; in the RREF, pivot columns are those that are not a linear combination of the preceding (pivot) columns; a nonpivot column is a linear combination of the preceding pivot columns, and the coefficients yield precisely the needed coefficients; indeed $$ C_2=1C_1,\qquad C_4=1C_2+1C_3 $$