Chapter 4 System of Linear Equations

Section 4.4 More general Systems

4.4.3 Exercises

Exercise 4.4.3
Find the $y$-intercept $b$ and the slope $m$ of a line described by the equation $y=mx+b$ which is defined by two points. The first point at ${x}_{1}=\alpha$ lies on the line described by the equation ${y}_{1}\left(x\right)=-2\left(1+x\right)$. The second point at ${x}_{2}=\beta$ lies on the line described by the equation ${y}_{2}\left(x\right)=x-1$. The following figure illustrates the situation.
1. Find the system of equations for the parameters $b$ and $m$.
The first equation reads $m\alpha +b$$=$
;
the second equation reads $m\beta +b$$=$
.
The constants $\alpha$ and $\beta$ have to remain in the solution; for these alpha and beta can be entered.
2. Solve this system of equations for $b$ and $m$. For which values of $\alpha$ and $\beta$ does the system have a unique solution, no solution, or an infinite number of solutions?
For $\alpha =-2$ and $\beta =2$ one obtains, for example, the solution $m$$=$
and $b$$=$
, the case $\alpha =2$ and $\beta =-2$ results in the solution $m$$=$
and $b$$=$ .
The LS has an infinite number of solutions if $\alpha$$=$
and $\beta$$=$ .
The corresponding solutions can be parameterised by $m=r$ and $b$$=$
, $r\in ℝ$.
3. What is the graphical interpretation of the last two cases, i.e. no solution and an infinite number of solutions?

Exercise 4.4.4
Find the solution set of the following LS depending on the parameter $t\in ℝ$.

$\begin{array}{cccc}\hfill x-y+tz& \hfill =\hfill & t\hfill & \text{equation} \left(1\right) ,\hfill \\ \hfill tx+\left(1-t\right)y+\left(1+{t}^{2}\right)z& \hfill =\hfill & -1+t\hfill & \text{equation} \left(2\right) ,\hfill \\ \hfill \left(1-t\right)x+\left(-2+t\right)y+\left(-1+t-{t}^{2}\right)z& \hfill =\hfill & {t}^{2}\hfill & \text{equation} \left(3\right) .\hfill \end{array}$

The LS only has solutions for the following values of the parameter: $t\in \text{}$
.
Set can be entered in the form $\left\{$a;b;c;$\dots \right\}$. The empty set can be entered as $\left\{\right\}$.
For the smallest value of the parameter $t$ the solution is
$x$$=$
, $y$$=$
, $z=r$, $r\in ℝ$.
For the greatest value of the parameter $t$ the solution is
$x$$=$
, $y$$=$
, $z=r$, $r\in ℝ$.