6. Reduce the following matrix to reduced row

To reduce the given matrix C to reduced row echelon form, we'll use a series of elementary row operations. The goal is to transform the matrix into a form where each leading coefficient (the leftmost nonzero entry) of a row is equal to 1, and each leading coefficient is the only nonzero entry in its column.
$C=\left[\begin{array}{cccc}1& -1& \frac{1}{2}& \frac{1}{2}\\ 0& 1& \frac{1}{2}& -\frac{1}{2}\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right]$
To make the leading coefficient of the second row equal to 1, we'll multiply the second row by the reciprocal of its leading coefficient, which is 1.
$R_2\to R_2·\frac{1}{1}=R_2$
$C=\left[\begin{array}{cccc}1& -1& \frac{1}{2}& \frac{1}{2}\\ 0& 1& \frac{1}{2}& -\frac{1}{2}\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right]$
To eliminate the entries below the leading coefficient of the first row, we'll use row operations. We'll multiply the second row by the appropriate factor and add it to the first row.
$R_1\to R_1-R_2·1=R_1$
$C=\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& \frac{1}{2}& -\frac{1}{2}\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right]$
To eliminate the entries below the leading coefficient of the second row, we'll use row operations. We'll multiply the third row by the appropriate factor and add it to the second row.
$R_2\to R_2-R_3·\frac{1}{2}=R_2$
$C=\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& -\frac{1}{2}\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right]$
There are no entries below the leading coefficient of the third row, so we move on to the next row.
To make the leading coefficient of the fourth row equal to 1, we'll multiply the fourth row by the reciprocal of its leading coefficient, which is 0.
$R_4\to R_4·\frac{1}{0}=R_4$
$C=\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& -\frac{1}{2}\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right]$