Question

Question: In Exercise 10, determine the values of the parameter s for which the system has a unique solution, and describe the solution.

10.

$\begin{array}{c}s{x}_{\mathbf{1}}-\mathbf{2}{x}_{\mathbf{2}}=\mathbf{1}\\ 4s{x}_{\mathbf{1}}+\mathbf{4}s{x}_{\mathbf{2}}=\mathbf{2}\end{array}$

Short answer

The solution of the given system is unique for $s\ne 0,-2$. For such a system, the solution is $x_1=\frac{s+1}{s\left(s+2\right)}$, and $x_2=-\frac{1}{2\left(s+2\right)}$.

Step-by-step solution

Write the matrix form

The given system is equivalent to $Ax=b$.

Here, $A=\left(\begin{array}{cc}s& -2\\ 4s& 4s\end{array}\right)$, $x=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)$, and $b=\left(\begin{array}{c}1\\ 2\end{array}\right)$.

Then, $A_1\left(b\right)=\left(\begin{array}{cc}1& -2\\ 2& 4s\end{array}\right)$, and $A_2\left(b\right)=\left(\begin{array}{cc}s& 1\\ 4s& 2\end{array}\right)$.

Determine the value of s

Note that the solution is unique for $detA\ne 0$.

$\begin{array}{c}detA=\left|\begin{array}{cc}s& -2\\ 4s& 4s\end{array}\right|\\ =4s^2+8s\\ detA=4s\left(s+2\right)\end{array}$

When $detA=0$, you get:

$\begin{array}{c}4s\left(s+2\right)=0\\ s\left(s+2\right)=0\\ s=0,-2\end{array}$

Hence, the solution of the given system is unique for $s\ne 0,-2$.

Use Cramer’s rule

For such a system, the solution is obtained by using Cramer’s rule, that is,

$x_i=\frac{det{A}_i\left(b\right)}{detA}$, $i=1,2$.

Hence,

$\begin{array}{c}{x}_1=\frac{det{A}_1\left(b\right)}{detA}\\ =\frac{\left|\begin{array}{cc}1& -2\\ 2& 4s\end{array}\right|}{4s\left(s+2\right)}\\ =\frac{4s+4}{4s\left(s+2\right)}\\ x_1=\frac{s+1}{s\left(s+2\right)}\end{array}$

$\begin{array}{c}{x}_2=\frac{det{A}_2\left(b\right)}{detA}\\ =\frac{\left|\begin{array}{cc}s& 1\\ 4s& 2\end{array}\right|}{4s\left(s+2\right)}\\ =\frac{2s-4s}{4s\left(s+2\right)}\\ =\frac{-2s}{4s\left(s+2\right)}\\ x_2=\frac{-1}{2\left(s+2\right)}\end{array}$