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To express a system in theaugmented matrix form, extract the coefficients of the variables and the constants and place these entries in the column of the matrix.

The given system of equations is as follows:

\(\begin{aligned}{c}{x_1} - 3{x_2} + 4{x_3} = - 4\\3{x_1} - 7{x_2} + 7{x_3} = - 8\\ - 4{x_1} + 6{x_2} - {x_3} = 7\end{aligned}\)

So, the augmented matrix for the given system can be written as follows:

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 3}&4&{ - 4}\\3&{ - 7}&7&{ - 8}\\{ - 4}&6&{ - 1}&7\end{aligned}} \right)\)

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

To eliminate the \(3{x_1}\) term from the second equation, perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 3}&4&{ - 4}\\3&{ - 7}&7&{ - 8}\\{ - 4}&6&{ - 1}&7\end{aligned}} \right)\) as shown below.

Add \( - 3\) times the first row to the second row; i.e., \({R_2} \to {R_2} - 3{R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 3}&4&{ - 4}\\{3 - 3\left( 1 \right)}&{ - 7 - 3\left( { - 3} \right)}&{7 - 3\left( 4 \right)}&{ - 8 - 3\left( { - 4} \right)}\\{ - 4}&6&{ - 1}&7\end{aligned}} \right)\)

After the row operation, the matrix becomes

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

To eliminate the \( - 4{x_1}\) term from the third equation, perform an elementary row operation on the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 3}&4&{ - 4}\\0&2&{ - 5}&4\\{ - 4}&6&{ - 1}&7\end{aligned}} \right)\) as shown below.

Add 4 times the first row to the third row; i.e., \({R_3} \to {R_3} + 4{R_1}\).

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

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 3}&4&{ - 4}\\0&2&{ - 5}&4\\0&{ - 6}&{15}&{ - 9}\end{aligned}} \right)\)

Use the \(2{x_2}\) term of the second equation to eliminate the \( - 6{x_2}\) term from the third equation. Perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 3}&4&{ - 4}\\0&2&{ - 5}&4\\0&{ - 6}&{15}&{ - 9}\end{aligned}} \right)\) as shown below.

Add 3 times the second row to the third row; i.e., \({R_3} \to {R_3} + 3{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 3}&4&{ - 4}\\0&2&{ - 5}&4\\{0 + 3\left( 0 \right)}&{ - 6 + 3\left( 2 \right)}&{15 + 3\left( { - 5} \right)}&{ - 9 + 3\left( 4 \right)}\end{aligned}} \right)\)

After the row operation, the matrix becomes

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

From the obtained augmented matrix, the system of equations can be written as follows:

\(\begin{aligned}{c}{x_1} - 3{x_2} + 4{x_3} = - 4\\2{x_2} - 5{x_3} = 4\\0 = 3\end{aligned}\)

The last equation \(0 = 3\) is never true. Therefore, the system of equations is inconsistent.

Hence, the solution set is empty, and the given system has no solution.