Step 1
Given set of Equations:
\[\left\{ \begin{array}{l}{x_2} + 4{x_3} = - 5\\{x_1} + 3{x_2} + 5{x_3} = - 2\\3{x_1} + 7{x_2} + 7{x_3} = 6\end{array} \right.\]We have to Solve the system of equations given above by using elementary row operations on the equations or on the augmented matrix. We will prefer the augmented matrix method.

Step 2
Convert the given system of Equations into augmented matrix form . \[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}0&1&4\\1&3&5\\3&7&7\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 5}\\{ - 2}\\6\end{array}} \right]\\\\A = \left[ {\begin{array}{*{20}{c}}0&1&4&{ - 5}\\1&3&5&{ - 2}\\3&7&7&6\end{array}} \right]\end{array}\]

Step 3: Interchange $R_2$ with $R_1$
To obtain pivot element in first row
\[A = \left[ {\begin{array}{*{20}{c}}1&3&5&{ - 2}\\0&1&4&{ - 5}\\3&7&7&6\end{array}} \right]\]

Step 4: ${R_3} = {R_3} -3{R_1} $
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&3&5&{ - 2}\\0&1&4&{ - 5}\\3&7&7&6\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&{ - 2}\\0&1&4&{ - 5}\\{3 - 3\left( 1 \right)}&{7 - 3\left( 3 \right)}&{7 - 3\left( 5 \right)}&{6 - 3\left( { - 2} \right)}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&{ - 2}\\0&1&4&{ - 5}\\0&{ - 2}&{ - 8}&{12}\end{array}} \right]\end{array}\]

Step 4: ${R_3} = {R_3} + 2{R_2} $
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&3&5&{ - 2}\\0&1&4&{ - 5}\\0&{ - 2}&{ - 8}&{12}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&{ - 2}\\0&1&4&{ - 5}\\0&{ - 2 + 2\left( 1 \right)}&{ - 8 + 2\left( 4 \right)}&{12 + 2\left( { - 5} \right)}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&{ - 2}\\0&1&4&{ - 5}\\0&0&0&2\end{array}} \right]\end{array}\]

Step 5: Write the reduced form in system of equations
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&3&5&{ - 2}\\0&1&4&{ - 5}\\0&0&0&2\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}1&3&5\\0&1&4\\0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 5}\\2\end{array}} \right]\\\\{x_1} + 3{x_2} + 5{x_3} = - 2\\{x_2} + 4{x_4} = - 5\\0 = 2\end{array}\]

Step 6
From the last row, we have\[0 = 2\]Since, this is false, the system has no solution making it an inconsistent system.

Step 4: ANSWERS
The system has no solution/Inconsistent.