Algebra - Augmented Matrices

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Section 7.3 : Augmented Matrices

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

\[\left[ {\begin{array}{rrr|r}4&{ - 1}&3&5\\0&2&5&9\\{ - 6}&1&{ - 3}&{10}\end{array}} \right]\]

\(8{R_{\,1}}\)
\({R_{\,2}} \leftrightarrow {R_{\,3}}\)
\({R_{\,2}} + 3{R_{\,1}} \to {R_{\,2}}\)

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a Show Solution

This operation is telling us to multiply all the entries in Row 1 of the augmented matrix by 8 so let’s do that.

\[\left[ {\begin{array}{rrr|r}4&{ - 1}&3&5\\0&2&5&9\\{ - 6}&1&{ - 3}&{10}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{8{R_{\,1}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}{32}&{ - 8}&{24}&{40}\\0&2&5&9\\{ - 6}&1&{ - 3}&{10}\end{array}} \right]\]

b Show Solution

This operation is telling us to interchange Row 2 and Row 3 of the augmented matrix. Here is that work.

\[\left[ {\begin{array}{rrr|r}4&{ - 1}&3&5\\0&2&5&9\\{ - 6}&1&{ - 3}&{10}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} \leftrightarrow {R_{\,3}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}4&{ - 1}&3&5\\{ - 6}&1&{ - 3}&{10}\\0&2&5&9\end{array}} \right]\,\]

c Show Solution

For this operation we are going to replace Row 2 with the results of taking the original entries from Row 2 and add to them 3 times the entries in Row 1.

\[\left[ {\begin{array}{rrr|r}4&{ - 1}&3&5\\0&2&5&9\\{ - 6}&1&{ - 3}&{10}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}{{R_{\,2}} + 3{R_{\,1}} \to {R_{\,2}}}\\ \to \end{array}\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{rrr|r}4&{ - 1}&3&5\\{12}&{ - 1}&{14}&{24}\\{ - 6}&1&{ - 3}&{10}\end{array}} \right]\,\]

Here are the individual computations for this operation.

\[\begin{align*}& {\mbox{Column 1 }}:\,\,\,0 + 3\left( 4 \right) = 12\\ & {\mbox{Column 2 }}:\,\,\,2 + 3\left( { - 1} \right) = - 1\\ & {\mbox{Column 3 }}:\,\,\,5 + 3\left( 3 \right) = 14\\ & {\mbox{Column 4 }}:\,\,\,9 + 3\left( 5 \right) = 24\end{align*}\]