This operation is telling us to multiply all the entries in Row 2 of the augmented matrix by ½ so let’s do that.

\[\left[ {\begin{array}{rrr|r}1&{ - 6}&2&0\\2&{ - 8}&{10}&4\\3&{ - 4}&{ - 1}&2\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{\frac{1}{2}{R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}1&{ - 6}&2&0\\1&{ - 4}&5&2\\3&{ - 4}&{ - 1}&2\end{array}} \right]\]

This operation is telling us to interchange Row 1 and Row 3 of the augmented matrix. Here is that work.

\[\left[ {\begin{array}{rrr|r}1&{ - 6}&2&0\\2&{ - 8}&{10}&4\\3&{ - 4}&{ - 1}&2\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} \leftrightarrow {R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}3&{ - 4}&{ - 1}&2\\2&{ - 8}&{10}&4\\1&{ - 6}&2&0\end{array}} \right]\]

For this operation we are going to replace Row 1 with the results of taking the original entries from Row 1 and subtract from them 6 times the entries in Row 3.

\[\left[ {\begin{array}{rrr|r}1&{ - 6}&2&0\\2&{ - 8}&{10}&4\\3&{ - 4}&{ - 1}&2\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} - 6{R_{\,3}} \to {R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}{ - 17}&{18}&8&{ - 12}\\2&{ - 8}&{10}&4\\3&{ - 4}&{ - 1}&2\end{array}} \right]\]

Here are the individual computations for this operation.

\[\begin{align*}& {\mbox{Column 1 }}:\,\,\,1 - 6\left( 3 \right) = - 17\\ & {\mbox{Column 2 }}:\,\,\, - 6 - 6\left( { - 4} \right) = 18\\ & {\mbox{Column 3 }}:\,\,\,2 - 6\left( { - 1} \right) = 8\\ & {\mbox{Column 4 }}:\,\,\,0 - 6\left( 2 \right) = - 12\end{align*}\]