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.

Thus, the augmented matrix for the system of equations \({x_1} + 2{x_2} + {x_3} = 4\), \({x_2} - {x_3} = 1\),and\({x_1} + 3{x_2} = 0\) can be represented as follows:

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

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

Use the \({x_1}\) term in the first equation to eliminate the \({x_1}\) term from the third equation. Perform an elementary row operationon the matrix\[\left[ {\begin{array}{*{20}{c}}1&2&1&4\\0&1&{ - 1}&1\\1&3&0&0\end{array}} \right]\] as shown below:

Add \( - 1\) times the first row to the third row; i.e., \({R_3} \to {R_3} - {R_1}\).

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

After performing the row operation, the matrix becomes

\[\left[ {\begin{array}{*{20}{c}}1&2&1&4\\0&1&{ - 1}&1\\0&1&{ - 1}&{ - 4}\end{array}} \right]\]

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

Use the \({x_1}\) term in the second equation to eliminate the \({x_2}\) and \( - {x_3}\) terms from the third equation. Perform an elementary row operation on the matrix\[\left[ {\begin{array}{*{20}{c}}1&2&1&4\\0&1&{ - 1}&1\\0&1&{ - 1}&{ - 4}\end{array}} \right]\] as shown below:

Add \( - 1\) times the second row to the third row; i.e., \({R_3} \to {R_3} - {R_2}\).

\[\left[ {\begin{array}{*{20}{c}}1&2&1&4\\0&1&{ - 1}&1\\{0 - 0}&{1 - 1}&{ - 1 - \left( { - 1} \right)}&{ - 4 - 1}\end{array}} \right]\]

After performing the row operation, the matrix becomes

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

To check if the system of equations is consistent, convert the augmented matrix into the system of equations again.

\(\begin{array}{c}{x_1} + 2{x_2} + {x_3} = 4\\{x_2} - {x_3} = 1\\0 = - 5\end{array}\)

The equation \(0 = - 5\) can also be written as \(0{x_1} + 0{x_2} + 0{x_3} = - 5\). No values of \({x_1}\), \({x_2}\),and \({x_3}\) can satisfy the equation \(0{x_1} + 0{x_2} + 0{x_3} = - 5\). It proves that the system is inconsistent, or there is no solution.

Hence, the three planes have no common point.