Consider each matrix in Exercises 5 and 6 as the augmented matrix of a linear system. State in words the next two elementary row operations that should be performed in the process of solving the system.

6. \(\left( {\begin{aligned}{*{20}{c}}1&{ - 6}&4&0&{ - 1}\\0&2&{ - 7}&0&4\\0&0&1&2&{ - 3}\\0&0&3&1&6\end{aligned}} \right)\)

The given augmented matrix of a linear system is provided below:

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 6}&4&0&{ - 1}\\0&2&{ - 7}&0&4\\0&0&1&2&{ - 3}\\0&0&3&1&6\end{aligned}} \right)\)

Since the above matrix has five columns, you can conclude that the given augmented matrix of a linear system consists of four unknown variables, say,\({x_1},\,\,{x_2},\,\,{x_3},\,\,\)and \({x_4}\). To obtain the solution of the system, you need to convert the given augmented matrix into the upper or lowertriangular matrix.

A basic principle states that row operations do not affect the solution set of a linear system.

To eliminate the\(3{x_3}\)term in the fourth row,perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 6}&4&0&{ - 1}\\0&2&{ - 7}&0&4\\0&0&1&2&{ - 3}\\0&0&3&1&6\end{aligned}} \right)\) as shown below.

Subtract 3 times row 3 from row 4,i.e., \({R_4} \to {R_4} - 3{R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 6}&4&0&{ - 1}\\0&2&{ - 7}&0&4\\0&0&1&2&{ - 3}\\{0 - 3\left( 0 \right)}&{0 - 3\left( 0 \right)}&{3 - 3\left( 1 \right)}&{1 - 3\left( 2 \right)}&{6 - 3\left( { - 3} \right)}\end{aligned}} \right)\)

After performing the row operation, the matrix becomes:

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 6}&4&0&{ - 1}\\0&2&{ - 7}&0&4\\0&0&1&2&{ - 3}\\0&0&0&{ - 5}&{15}\end{aligned}} \right)\)

To obtain the\({x_4}\)term in the fourth row,perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 6}&4&0&{ - 1}\\0&2&{ - 7}&0&4\\0&0&1&2&{ - 3}\\0&0&0&{ - 5}&{15}\end{aligned}} \right)\) as shown below.

Divide row 4 by\( - 5\).

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 6}&4&0&{ - 1}\\0&2&{ - 7}&0&4\\0&0&1&2&{ - 3}\\{ - \frac{0}{5}}&{ - \frac{0}{5}}&{ - \frac{0}{5}}&{ - \left( {\frac{{ - 5}}{5}} \right)}&{ - \frac{{15}}{5}}\end{aligned}} \right)\)

After performing the row operation, the matrix becomes:

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 6}&4&0&{ - 1}\\0&2&{ - 7}&0&4\\0&0&1&2&{ - 3}\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

After performing these two operations, the augmented matrix becomes an upper triangular matrix, and the required solution can be computed by converting the matrix into equations.

Hence, the two required elementary row operations that should be performed in the process of solving the system are as follows:

1. Thefirst operation: 3 times the third row subtracted from the fourth row;i.e., \({R_4} \to {R_4} - 3{R_3}.\)
2. The second operation: The fourth row divided by \( - 5;\) i.e., \({R_4} \to - \frac{1}{5}{R_4}.\)