Question 2.1.7: Determine the solution set of the system of 2 linear equatio...

First, notice that neither equation contains x_{2}. This may seem peculiar, but it happens in some applications that one of the variables of interest does not appear in any of the linear equations. If it truly is one of the variables of the problem, ignoring it is incorrect. We rewrite the equations to make it explicit:

x_{1}+0x_{2}+2x_{3}+x_{4}=14\\x_{1}+0x_{2}+3x_{3}+3x_{4}=19

As in Example 2.1.6, we want our leading variable in the first equation to be to the left of the leading variable in the second equation, and we want the leading variable to be eliminated from the second equation. Thus, we use a type (3) step to eliminate x_{1} from the second equation.
Add (−1) times the first equation to the second equation:

x_{1}+0x_{2}+2x_{3}+x_{4}=14\\x_{3}+2x_{4}=5

Observe that x_{2} is not shown in the second equation because the leading variable must have a non-zero coefficient. Moreover, we have already finished our elimination procedure as we have our desired form. The solution can now be completed by back-substitution.

Note that the equations do not completely determine both x_{3} and x_{4}: one of them can be chosen arbitrarily, and the equations can still be satisfied. For consistency, we always choose the variables that do not appear as a leading variable in any equation to be the ones that will be chosen arbitrarily. We will call these free variables.

Thus, in the revised system, we see that neither x_{2} nor x_{4} appears as a leading variable in any equation. Therefore, x_{2} and x_{4} are the free variables and may be chosen arbitrarily (for example, x_{4}=t\in \mathbb{R} \ \ \ and \ \ \ x_{2}=s\in \mathbb{R} ). Then the second equation can be solved for the leading variable x_{3}:

x_{3}=5-2x_{4}=5-2t

Now, solve the first equation for its leading variable x_{1}:

x_{1}=14-2x_{3}-x_{4}=14-2(5-2t)-t=4+3t

Thus, the solution set of the system is

\left [ \begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix} \right ] =\left [ \begin{matrix} 4+3t \\ s \\ 5-2t \\ t \end{matrix} \right ] , \ \ \ \ \ s,t\in \mathbb{R}

In this case, there are infinitely many solutions because for each value of s and for each value of t that we choose, we get a different solution. We say that this equation is the general solution of the system, and we call s and t the parameters of the general solution. For many purposes, it is useful to recognize that this solution can be split into a constant part, a part in t, and a part in s:

\left [ \begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix} \right ] =\left [ \begin{matrix} 4 \\ 0 \\ 5 \\ 0 \end{matrix} \right ] +s\left [ \begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix} \right ] +t\left [ \begin{matrix} 3 \\ 0 \\ -2 \\ 1 \end{matrix} \right ] , \ \ \ \ \ \ s,t\in \mathbb{R}

This will be the standard format for displaying general solutions. It is acceptable to leave x_{2} in the place of s and x_{4} in the place of t, but then you must say x_{2},x_{4}\in \mathbb{R} . Observe that one immediate advantage of this form is that we can instantly see the geometric interpretation of the solution. The intersection of the two hyperplanes x_{1}+2x_{3}+x_{4}=14 \ \ \ and \ \ \ x_{1}+3x_{3}+3x_{4}=19 \ \ \ in \ \ \ \mathbb{R} ^{4} in \mathbb{R} ^{4} is the plane in \mathbb{R} ^{4} that passes through P(4, 0, 5, 0) with vector equation

\vec{x}=\left [ \begin{matrix} 4 \\ 0 \\ 5 \\ 0 \end{matrix} \right ] +s\left [ \begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix} \right ] +t\left [ \begin{matrix} 3 \\ 0 \\ -2 \\ 1 \end{matrix} \right ] , \ \ \ \ \ s,t\in \mathbb{R}