91Ó°ÊÓ

To express a system in theaugmented matrixform, 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:

\[\begin{array}{c}{x_1}\,\,\,\,\,\,\,\,\,\,\, + 3{x_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\\{x_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3{x_4} = 3\\ - 2{x_2} + \,3{x_3}\,\,\, + 2{x_4} = 1\\3{x_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 7{x_4} = - 5\end{array}\]

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

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&{ - 2}&3&2&1\\3&0&0&7&{ - 5}\end{array}} \right]\]

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 fourth equation, perform an elementary row operationon the matrix\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&{ - 2}&3&2&1\\3&0&0&7&{ - 5}\end{array}} \right]\] as shown below.

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

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&{ - 2}&3&2&1\\{3 - 3\left( 1 \right)}&{0 - 3\left( 0 \right)}&{0 - 3\left( 3 \right)}&{7 - 3\left( 0 \right)}&{ - 5 - 3\left( 2 \right)}\end{array}} \right]\]

After the row operation, the matrix becomes

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&{ - 2}&3&2&1\\0&0&{ - 9}&7&{ - 11}\end{array}} \right]\]

Use the \[{x_2}\] term in the second equation to eliminate the \[ - 2{x_2}\] term from the third equation. Perform an elementary row operationon the matrix\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&{ - 2}&3&2&1\\0&0&{ - 9}&7&{ - 11}\end{array}} \right]\] as shown below.

Add 2 times the second row to the third row; i.e., \({R_3} \to {R_3} + 2{R_2}\).

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\{0 + 2\left( 0 \right)}&{ - 2 + 2\left( 1 \right)}&{3 + 2\left( 0 \right)}&{2 + 2\left( { - 3} \right)}&{1 + 2\left( 3 \right)}\\0&0&{ - 9}&7&{ - 11}\end{array}} \right]\]

After the row operation, the matrix becomes

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&0&3&{ - 4}&7\\0&0&{ - 9}&7&{ - 11}\end{array}} \right]\]

Use the \[3{x_3}\] term in the third equation to eliminate the \[ - 9{x_3}\] term from the fourth equation. Perform an elementary row operationon the matrix\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&0&3&{ - 4}&7\\0&0&{ - 9}&7&{ - 11}\end{array}} \right]\] as shown below.

Add 3 times the third row to the fourth row; i.e., \({R_4} \to {R_4} + 3{R_3}\).

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&0&3&{ - 4}&7\\{0 + 3\left( 0 \right)}&{0 + 3\left( 0 \right)}&{ - 9 + 3\left( 3 \right)}&{7 + 3\left( { - 4} \right)}&{ - 11 + 3\left( 7 \right)}\end{array}} \right]\]

After the row operation, the matrix becomes

\[\left[ {\begin{array}{*{20}{c}}{\rm{1}}&0&3&0&2\\0&1&0&{ - 3}&3\\0&0&3&{ - 4}&7\\0&0&0&{ - 5}&{10}\end{array}} \right]\]

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

\[\begin{array}{c}{x_1}\,\,\,\,\,\,\,\,\,\,\, + 3{x_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\\{x_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3{x_4} = 3\\\,3{x_3}\,\,\, - 4{x_4} = 7\\ - 5{x_4} = 10\end{array}\]

A unique value of \[{x_4}\] can be obtained from the fourth equation.

If \[{x_4}\] is substituted by its unique value in the second and third equations, the unique values of \[{x_2}\] and \[{x_3}\] can be calculated. Thus, substituting \[{x_3}\] by its value in the first equation, you will get a unique value of\[{x_1}\].

Since all the values can be uniquely determined, a unique solution exists for the given system.

Hence, the given system is consistent.