Algebra - Augmented Matrices

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Section 7.3 : Augmented Matrices

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3. For the following augmented matrix perform the indicated elementary row operations.

\[\left[ {\begin{array}{rrr|r}{10}&{ - 1}&{ - 5}&1\\4&0&7&{ - 1}\\0&7&{ - 2}&3\end{array}} \right]\]

\( - 9{R_{\,3}}\)
\({R_{\,1}} \leftrightarrow {R_{\,2}}\)
\({R_{\,3}} - {R_{\,1}} \to {R_{\,3}}\)

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a Show Solution

This operation is telling us to multiply all the entries in Row 3 of the augmented matrix by -9 so let’s do that.

\[\left[ {\begin{array}{rrr|r}{10}&{ - 1}&{ - 5}&1\\4&0&7&{ - 1}\\0&7&{ - 2}&3\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{ - 9{R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}{10}&{ - 1}&{ - 5}&1\\4&0&7&{ - 1}\\0&{ - 63}&{18}&{ - 27}\end{array}} \right]\]

b Show Solution

This operation is telling us to interchange Row 1 and Row 2 of the augmented matrix. Here is that work.

\[\left[ {\begin{array}{rrr|r}{10}&{ - 1}&{ - 5}&1\\4&0&7&{ - 1}\\0&7&{ - 2}&3\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,1}} \leftrightarrow {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}4&0&7&{ - 1}\\{10}&{ - 1}&{ - 5}&1\\0&7&{ - 2}&3\end{array}} \right]\]

c Show Solution

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

\[\left[ {\begin{array}{rrr|r}{10}&{ - 1}&{ - 5}&1\\4&0&7&{ - 1}\\0&7&{ - 2}&3\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,3}} - {R_{\,1}} \to {R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}{10}&{ - 1}&{ - 5}&1\\4&0&7&{ - 1}\\{ - 10}&8&3&2\end{array}} \right]\]

Here are the individual computations for this operation.

\[\begin{align*}& {\mbox{Column 1 }}:\,\,\,0 - \left( {10} \right) = - 10\\ & {\mbox{Column 2 }}:\,\,\,7 - \left( { - 1} \right) = 8\\ & {\mbox{Column 3 }}:\,\,\, - 2 - \left( { - 5} \right) = 3\\ & {\mbox{Column 4 }}:\,\,\,3 - \left( 1 \right) = 2\end{align*}\]