Chapter 4 System of Linear Equations

Section 4.4 More general Systems

4.4.2 Systems with a Free Parameter

We start with an example that is very easy but demonstrates the essential point concerning free parameters in systems of linear equations.
Example 4.4.1
Find the solution set of the following system of linear equations

$\begin{array}{ccccc}\hfill \text{equation} \left(1\right):& \hfill \hfill & \hfill x-2y& \hfill =\hfill & 3 ,\hfill \\ \hfill \text{equation} \left(2\right):& \hfill \hfill & \hfill -2x+4y& \hfill =\hfill & \alpha \hfill \end{array}$

depending on the parameter $\alpha$.
Multiplying equation $\left(1\right)$ by the factor $2$ and adding equation $\left(2\right)$ results in

$2·\left(x-2y\right)+\left(-2x+4y\right)=2·3+\alpha ⇔2x-4y-2x+4y=6+\alpha ⇔0=6+\alpha .$

Now, two cases have to be distinguished:
Case A ($\alpha \ne -6$): If the given free parameter $\alpha$ is not equal to $-6$, the last equation is a contradiction. In this case the system of linear equations has no solution, i.e. $L=\text{∅}$.
Case B ($\alpha =-6$): If the given free parameter $\alpha$ is equal to $-6$, the last equation is always satisfied ($0=0$). In fact, the two initial equations in this case are multiples of each other such that only one of them indeed carries information. Accordingly, the solution set is infinite: $L=\left\{\left(x=3+2t;y=t\right) : t\in ℝ\right\}$.

The example shows that the solution set may strongly depend on the value of the free parameter.
Such a free parameter can occur not only on one of the right-hand sides of the system of linear equations, but also on the left-hand sides, multiple times or in a function both on the left-hand sides and on the right-hand sides. Also several parameters can occur in a system at the same time.
Let us now consider a slightly more complex example.
Example 4.4.2
Find the solution set of the following system of linear equations

$\begin{array}{ccc}\multicolumn{1}{c}{x+y+\alpha z}& =\hfill & 1 ,\hfill \\ \multicolumn{1}{c}{x+\alpha y+z}& =\hfill & 1 ,\hfill \\ \multicolumn{1}{c}{\alpha x+y+z}& =\hfill & 1\hfill \end{array}$

depending on the value of the parameter $\alpha$.
For this, e.g. the first equation is solved for the variable $x$

$x=1-y-\alpha z\mathrm{ }: \text{equation} \left(1\text{'}\right) ,$

and the result is substituted into the second and third equation:

$\begin{array}{ccccccccc}\hfill \left(1-y-\alpha z\right)+\alpha y+z& \hfill =\hfill & 1\hfill & \hfill ⇔\hfill & \hfill -\left(1-\alpha \right)y+\left(1-\alpha \right)z& \hfill =\hfill & 0\hfill & \hfill \hfill & : \text{equation} \left(2\text{'}\right) ,\hfill \\ \hfill \alpha \left(1-y-\alpha z\right)+y+z& \hfill =\hfill & 1\hfill & \hfill ⇔\hfill & \hfill \left(1-\alpha \right)y+\left(1-{\alpha }^{2}\right)z& \hfill =\hfill & 1-\alpha \hfill & \hfill \hfill & : \text{equation} \left(3\text{'}\right) .\hfill \end{array}$

This results in a system of two linear equations in two variables $y$ and $z$. It can immediately be seen that for the value $\alpha =1$ something happens. Hence, a case analysis is required.
Case 1 ($\alpha =1$): In this case the two equations $\left(2\text{'}\right)$ and $\left(3\text{'}\right)$ are satisfied identically ($0=0$) and provide no further information. The only relation between the variables $x,y$, and $z$ is equation $\left(1\text{'}\right)$ or equation $\left(1\right)$, respectively, that reads for $\alpha =1$ as follows:

$x+y+z=1\mathrm{ }: \text{equation} \left(\stackrel{^}{1}\right) .$

Hence, the solution set has an infinite number of elements. The set can be described using two free parameters, e.g.

$L=\left\{\left(s;t;1-s-t\right) : s,t\in ℝ\right\} .$

Geometrically, the solution set is exactly the plane described by equation $\left(\stackrel{^}{1}\right)$.
Case 2: ($\alpha \ne 1$): In this case both equation $\left(2\text{'}\right)$ and equation $\left(3\text{'}\right)$ can be divided by $\left(1-\alpha \right)$. Using the third binomial formula ($\left(1-{\alpha }^{2}\right)=\left(1-\alpha \right)\left(1+\alpha \right)$) results in

$\begin{array}{ccccc}\hfill -y+z& \hfill =\hfill & 0\hfill & \hfill \hfill & : \text{equation} \left(2\text{'}\text{'}\right) ,\hfill \\ \hfill y+\left(1+\alpha \right)z& \hfill =\hfill & 1\hfill & \hfill \hfill & : \text{equation} \left(3\text{'}\text{'}\right) .\hfill \end{array}$

According to equation $\left(2\text{'}\text{'}\right)$ one has $y=z$. This is substituted into equation $\left(3\text{'}\text{'}\right)$:

$z+\left(1+\alpha \right)z=1⇔\left(2+\alpha \right)z=1\mathrm{ }: \text{equation} \left(☆\right) .$

Again, one has to take care and an additional case analysis is required since $\alpha =-2$ and $\alpha \ne -2$ have different consequences:
Case 2a ($\alpha =-2$): In this (sub)case equation $\left(☆\right)$ reads $0=1$. This is a contradiction and the initial system of equations has no solution, i.e. $L=\text{∅}$.
Case 2b: $\alpha \ne -2$: In this (sub)case the last equation can be solved easily for $z$:

$z=\frac{1}{2+\alpha } .$

So, $y$ ($y=z$) and $x$ ($x=1-y-\alpha z$) are determined. The initial system of linear equations has a unique solution, namely $L=\left\{\left(x=\frac{1}{2+\alpha };y=\frac{1}{2+\alpha };z=\frac{1}{2+\alpha }\right)\right\}$.

The previous example indicates the relevance of a clear and precise case analysis. Depending on the value of $\alpha$ ($\alpha =1$, $\alpha =-2$, or $\alpha \in ℝ\setminus \left\{-2;1\right\}$) the solution set is completely different! In the first case it is an infinite set, in the second the empty set, and in the third the solution set consists of exactly one element!
By the way, the exceptionality of the case $\alpha =1$ could have seen directly from the initial system of linear equations: For $\alpha =1$, the same equation occurs three times, namely $x+y+z=1$, i.e. two of the three equations in the initial system do not contribute any information and thus, they are unnecessary. For $\alpha =1$, only the equation $x+y+z=1$ relates the three variables.