Describe and compare the solution sets of \({x_1} + 9{x_2} - 4{x_3} = 0\), and \({x_1} + 9{x_2} - 4{x_3} = - 2\).

The given equation is \({x_1} + 9{x_2} - 4{x_3} = 0\).

It can be re-written as \({x_1} = - 9{x_2} + 4{x_3}\).

Here, consider that \({x_1}\) is a basic variable, and \({x_2}\) and \({x_3}\) are free variables. By using the free variable, the basic variables can be solved.

The general solutions are shown below:

\(\left\{ \begin{array}{l}{x_1} = - 9{x_2} + 4{x_3}\\{x_2}{\rm{ is free}}\\{x_3}{\rm{ is free}}\end{array} \right.\)

Let, \({x_2} = r\) and \({x_3} = s\).

Substitute \({x_2} = r\) and \({x_3} = s\) in the equation \({x_1} = - 9{x_2} + 4{x_3}\) to obtain the general solution.

\({x_1} = - 9r + 4s\)

Obtain the vector in the parametric form by using \({x_1} = - 9r + 4s\), \({x_2} = r\), and \({x_3} = s\).

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 9r + 4s}\\r\\s\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 9r}\\r\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{4s}\\0\\s\end{array}} \right)\\ = r\left( {\begin{array}{*{20}{c}}{ - 9}\\1\\0\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}4\\0\\1\end{array}} \right)\end{array}\)

Or it can be written as shown below:

\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}{ - 9}\\1\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}4\\0\\1\end{array}} \right)\)

Thus, the solution in the parametric vector form is \(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}{ - 9}\\1\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}4\\0\\1\end{array}} \right)\).

The equation\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = {x_2}\left( {\begin{array}{*{20}{c}}{ - 9}\\1\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}4\\0\\1\end{array}} \right)\)shows that the solution of the equation \({x_1} = - 9{x_2} + 4{x_3}\) is a linear combination of vectors \(\left( {\begin{array}{*{20}{c}}{ - 9}\\1\\0\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}4\\0\\1\end{array}} \right)\).

Thus, the solution set shows that it is a plane that passes through the origin.

The given equation is \({x_1} + 9{x_2} - 4{x_3} = - 2\).

It can be re-written as \({x_1} = - 2 - 9{x_2} + 4{x_3}\).

Here, consider that \({x_1}\) is a basic variable, and \({x_2}\) and \({x_3}\) are free variables. By using the free variable, the basic variables can be solved.

The general solutions are shown below:

\(\left\{ \begin{array}{l}{x_1} = - 2 - 9{x_2} + 4{x_3}\\{x_2}{\rm{ is free}}\\{x_3}{\rm{ is free}}\end{array} \right.\)

Let, \({x_2} = r\), and \({x_3} = s\).

Substitute \({x_2} = r\) and \({x_3} = s\) in the equation \({x_1} = - 2 - 9{x_2} + 4{x_3}\) to obtain the general solution.

\({x_1} = - 2 - 9r + 4s\)

Obtain the vector in the parametric form by using \({x_1} = - 2 - 9r + 4s\), \({x_2} = r\), and \({x_3} = s\).

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2 - 9r + 4s}\\r\\s\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 2}\\0\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 9r}\\r\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{4s}\\0\\s\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 2}\\0\\0\end{array}} \right) + r\left( {\begin{array}{*{20}{c}}{ - 9}\\1\\0\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}4\\0\\1\end{array}} \right)\end{array}\)

Or it can be written as shown below:

\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2}\\0\\0\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 9}\\1\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}4\\0\\1\end{array}} \right)\)

Thus, the solution in the parametric vector form is \(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2}\\0\\0\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 9}\\1\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}4\\0\\1\end{array}} \right)\).

The equation\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2}\\0\\0\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 9}\\1\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}4\\0\\1\end{array}} \right)\)shows that the solution of the equation \({x_1} + 9{x_2} - 4{x_3} = - 2\) is a linear combination of vectors \(\left( {\begin{array}{*{20}{c}}{ - 9}\\1\\0\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}4\\0\\1\end{array}} \right)\), that passes through \(\left( {\begin{array}{*{20}{c}}{ - 2}\\0\\0\end{array}} \right)\).

Thus, the solution set shows that it is a plane that does not pass through the origin.

The comparison of the solution sets of the equations \({x_1} + 9{x_2} - 4{x_3} = 0\) and \({x_1} + 9{x_2} - 4{x_3} = - 2\) shows that the equation \({x_1} + 9{x_2} - 4{x_3} = 0\) passes through the origin, while the equation \({x_1} + 9{x_2} - 4{x_3} = - 2\) does not.