In Exercises 7–10, the augmented matrix of a linear system has been reduced by row operations to the form shown. In each case, continue the appropriate row operations and describe the solution set of the original system.

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

The augmented matrix of a linear system is given as

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

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

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

Add 4 times the fourth row to the second row; i.e., \({R_2} \to {R_2} + 4{R_4}\).

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

After the row operation, the matrix becomes

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

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

Subtract 3 times the fourth row from the first row; i.e., \({R_1} \to {R_1} - 3{R_4}\).

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

After the row operation, the matrix becomes

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

To eliminate the \( - 2{x_2}\) term in the first equation, perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&0&7\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\) as shown below.

Add 2 times the second row to the first row; i.e., \({R_1} \to {R_1} + 2{R_2}\).

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

After the row operation, the matrix becomes

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

Convert the augmented matrix into equations to get \({x_1} = - 3,\,\,{x_2} = - 5,\,\,{x_3} = 6,\,\,\) and \({x_4} = - 3\).

Hence, the required solution is \(\left( {\begin{aligned}{*{20}{c}}{ - 3}\\{ - 5}\\6\\{ - 3}\end{aligned}} \right)\).