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

10.

\(\begin{array}{c}s{x_{\bf{1}}} - {\bf{2}}{x_{\bf{2}}} = {\bf{1}}\\4s{x_{\bf{1}}} + {\bf{4}}s{x_{\bf{2}}} = {\bf{2}}\end{array}\)

The given system is equivalent to \(Ax = b\).

Here, \(A = \left( {\begin{array}{*{20}{c}}s&{ - 2}\\{4s}&{4s}\end{array}} \right)\), \(x = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\), and \(b = \left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right)\).

Then, \({A_1}\left( b \right) = \left( {\begin{array}{*{20}{c}}1&{ - 2}\\2&{4s}\end{array}} \right)\), and \({A_2}\left( b \right) = \left( {\begin{array}{*{20}{c}}s&1\\{4s}&2\end{array}} \right)\).

Note that the solution is unique for \(\det A \ne 0\).

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}s&{ - 2}\\{4s}&{4s}\end{array}} \right|\\ = 4{s^2} + 8s\\\det A = 4s\left( {s + 2} \right)\end{array}\)

When \(\det A = 0\), 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 \(s \ne 0, - 2\).

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

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

Hence,

\(\begin{array}{c}{x_1} = \frac{{\det {A_1}\left( b \right)}}{{\det A}}\\ = \frac{{\left| {\begin{array}{*{20}{c}}1&{ - 2}\\2&{4s}\end{array}} \right|}}{{4s\left( {s + 2} \right)}}\\ = \frac{{4s + 4}}{{4s\left( {s + 2} \right)}}\\{x_1} = \frac{{s + 1}}{{s\left( {s + 2} \right)}}\end{array}\)

\(\begin{array}{c}{x_2} = \frac{{\det {A_2}\left( b \right)}}{{\det A}}\\ = \frac{{\left| {\begin{array}{*{20}{c}}s&1\\{4s}&2\end{array}} \right|}}{{4s\left( {s + 2} \right)}}\\ = \frac{{2s - 4s}}{{4s\left( {s + 2} \right)}}\\ = \frac{{ - 2s}}{{4s\left( {s + 2} \right)}}\\{x_2} = \frac{{ - 1}}{{2\left( {s + 2} \right)}}\end{array}\)