4.4.2 Systems with a Free Parameter

Find the solution set of the following system of linear equations

${\displaystyle \begin{array}{ccccc}\hfill \text{equation}\hspace{0.5em}(1):& \hfill \hfill & \hfill x-2y& \hfill =\hfill & 3\hspace{0.5em},\hfill \\ \hfill \hspace{0.5em}\text{equation}\hspace{0.5em}(2):& \hfill \hfill & \hfill -2x+4y& \hfill =\hfill & \alpha \hfill \end{array}}$

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

${\displaystyle 2·(x-2y)+(-2x+4y)=2·3+\alpha \iff 2x-4y-2x+4y=6+\alpha \iff 0=6+\alpha \hspace{0.5em}.}$


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: $L=\{(x=3+2t;y=t)\hspace{0.5em}:\hspace{0.5em}t\in ℝ\}$.

Find the solution set of the following system of linear equations

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{x+y+\alpha z}& =\hfill & 1\hspace{0.5em},\hfill \\ \multicolumn{1}{c}{x+\alpha y+z}& =\hfill & 1\hspace{0.5em},\hfill \\ \multicolumn{1}{c}{\alpha x+y+z}& =\hfill & 1\hfill \end{array}}$
depending on the value of the parameter $\alpha$.
For this, e.g. the first equation is solved for the variable $x$

${\displaystyle x=1-y-\alpha z\mathrm{\hspace{0.5em}\hspace{0.5em}}:\hspace{0.5em}\text{equation}\hspace{0.5em}(1\text{'})\hspace{0.5em},}$

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

${\displaystyle \begin{array}{ccccccccc}\hfill (1-y-\alpha z)+\alpha y+z& \hfill =\hfill & 1\hfill & \hfill \iff \hfill & \hfill -(1-\alpha )y+(1-\alpha )z& \hfill =\hfill & 0\hfill & \hfill \hfill & :\hspace{0.5em}\text{equation}\hspace{0.5em}(2\text{'})\hspace{0.5em},\hfill \\ \hfill \alpha (1-y-\alpha z)+y+z& \hfill =\hfill & 1\hfill & \hfill \iff \hfill & \hfill (1-\alpha )y+(1-{\alpha }^2)z& \hfill =\hfill & 1-\alpha \hfill & \hfill \hfill & :\hspace{0.5em}\text{equation}\hspace{0.5em}(3\text{'})\hspace{0.5em}.\hfill \end{array}}$

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 $(2\text{'})$ and $(3\text{'})$ are satisfied identically ($0=0$) and provide no further information. The only relation between the variables $x,y$, and $z$ is equation $(1\text{'})$ or equation $(1)$, respectively, that reads for $\alpha =1$ as follows:

${\displaystyle x+y+z=1\mathrm{\hspace{0.5em}\hspace{0.5em}}:\hspace{0.5em}\text{equation}\hspace{0.5em}(\hat{1})\hspace{0.5em}.}$

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

${\displaystyle L=\{(s;t;1-s-t)\hspace{0.5em}:\hspace{0.5em}s,t\in ℝ\}\hspace{0.5em}.}$

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

${\displaystyle \begin{array}{ccccc}\hfill -y+z& \hfill =\hfill & 0\hfill & \hfill \hfill & :\hspace{0.5em}\text{equation}\hspace{0.5em}(2\text{'}\text{'})\hspace{0.5em},\hfill \\ \hfill y+(1+\alpha )z& \hfill =\hfill & 1\hfill & \hfill \hfill & :\hspace{0.5em}\text{equation}\hspace{0.5em}(3\text{'}\text{'})\hspace{0.5em}.\hfill \end{array}}$

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

${\displaystyle z+(1+\alpha )z=1\iff (2+\alpha )z=1\mathrm{\hspace{0.5em}\hspace{0.5em}}:\hspace{0.5em}\text{equation}\hspace{0.5em}(☆)\hspace{0.5em}.}$

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 $(☆)$ 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 $z$:

${\displaystyle z=\frac{1}{2+\alpha }\hspace{0.5em}.}$

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