Question: In Exercise 7, determine the values of the parameter s for which the system has a unique solution, and describe the solution.

7.

\(\begin{array}{c}{\bf{6}}s{x_{\bf{1}}} + {\bf{4}}{x_{\bf{2}}} = {\bf{5}}\\{\bf{9}}{x_{\bf{1}}} + {\bf{2}}s{x_{\bf{2}}} = - {\bf{2}}\end{array}\)

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

Here, \(A = \left( {\begin{array}{*{20}{c}}{6s}&4\\9&{2s}\end{array}} \right)\), \(x = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\), and \(b = \left( {\begin{array}{*{20}{c}}5\\{ - 2}\end{array}} \right)\).

Then, \({A_1}\left( b \right) = \left( {\begin{array}{*{20}{c}}5&4\\{ - 2}&{2s}\end{array}} \right)\), and \({A_2}\left( b \right) = \left( {\begin{array}{*{20}{c}}{6s}&5\\9&{ - 2}\end{array}} \right)\).

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

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}{6s}&4\\9&{2s}\end{array}} \right|\\ = 12{s^2} - 36\\\det A = 12\left( {{s^2} - 3} \right)\end{array}\)

When \(\det A = 0\), you get:

\(\begin{array}{c}12\left( {{s^2} - 3} \right) = 0\\{s^2} = 3\\s = \pm \sqrt 3 \end{array}\)

Hence, the solution of the given system is unique for \(s \ne \pm \sqrt 3 \).

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,

\(\begin{array}{c}{x_1} = \frac{{\det {A_1}\left( b \right)}}{{\det A}}\\ = \frac{{\left| {\begin{array}{*{20}{c}}5&4\\{ - 2}&{2s}\end{array}} \right|}}{{12\left( {{s^2} - 3} \right)}}\\ = \frac{{10s + 8}}{{12\left( {{s^2} - 3} \right)}}\\{x_1} = \frac{{5s + 4}}{{6\left( {{s^2} - 3} \right)}}\end{array}\)

\(\begin{array}{c}{x_2} = \frac{{\det {A_2}\left( b \right)}}{{\det A}}\\ = \frac{{\left| {\begin{array}{*{20}{c}}{6s}&5\\9&{ - 2}\end{array}} \right|}}{{12\left( {{s^2} - 3} \right)}}\\ = \frac{{ - 12s - 45}}{{12\left( {{s^2} - 3} \right)}}\\{x_2} = \frac{{ - 4s - 15}}{{4\left( {{s^2} - 3} \right)}}\end{array}\)