Question: 6. Use Cramer’s rule to compute the solution of the following system.

\(\begin{array}{c}{x_{\bf{1}}} + {\bf{3}}{x_{\bf{2}}} + \,{x_{\bf{3}}} = {\bf{4}}\\ - {x_{\bf{1}}} + \,\,\,\,\,\,\,\,\,\,{\bf{2}}{x_{\bf{3}}} = {\bf{2}}\\{\bf{3}}{x_{\bf{1}}} + \,{x_{\bf{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\, = {\bf{2}}\end{array}\)

The matrix form of the given system is:

\(\left( {\begin{array}{*{20}{c}}1&3&1\\{ - 1}&0&2\\3&1&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4\\2\\2\end{array}} \right)\)

Thus, \(Ax = b\), where \(A = \left( {\begin{array}{*{20}{c}}1&3&1\\{ - 1}&0&2\\3&1&0\end{array}} \right)\), \(b = \left( {\begin{array}{*{20}{c}}4\\2\\2\end{array}} \right)\), and \(x = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\).

Then, \({A_1}\left( b \right) = \left( {\begin{array}{*{20}{c}}4&3&1\\2&0&2\\2&1&0\end{array}} \right)\),\({A_2}\left( b \right) = \left( {\begin{array}{*{20}{c}}1&4&1\\{ - 1}&2&2\\3&2&0\end{array}} \right)\), and \({A_3}\left( b \right) = \left( {\begin{array}{*{20}{c}}1&3&4\\{ - 1}&0&2\\3&1&2\end{array}} \right)\).

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}1&3&1\\{ - 1}&0&2\\3&1&0\end{array}} \right|\\ = - 3\left| {\begin{array}{*{20}{c}}{ - 1}&2\\3&0\end{array}} \right| + 0 - 1\left| {\begin{array}{*{20}{c}}1&1\\{ - 1}&2\end{array}} \right|\\ = - 3\left( { - 6} \right) - \left( 3 \right)\\\det A = 15\end{array}\)

\(\begin{array}{c}\det {A_1}\left( b \right) = \left| {\begin{array}{*{20}{c}}4&3&1\\2&0&2\\2&1&0\end{array}} \right|\\ = - 3\left| {\begin{array}{*{20}{c}}2&2\\2&0\end{array}} \right| + 0 - 1\left| {\begin{array}{*{20}{c}}4&1\\2&2\end{array}} \right|\\ = - 3\left( { - 4} \right) - 6\\ = 12 - 6\\\det {A_1}\left( b \right) = 6\end{array}\)

\(\begin{array}{c}\det {A_2}\left( b \right) = \left| {\begin{array}{*{20}{c}}1&4&1\\{ - 1}&2&2\\3&2&0\end{array}} \right|\\ = 1\left| {\begin{array}{*{20}{c}}{ - 1}&2\\3&2\end{array}} \right| - 2\left| {\begin{array}{*{20}{c}}1&4\\3&2\end{array}} \right| + 0\\ = - 8 - 2\left( { - 10} \right)\\ = - 8 + 20\\\det {A_2}\left( b \right) = 12\end{array}\)

\(\begin{array}{c}\det {A_3}\left( b \right) = \left| {\begin{array}{*{20}{c}}1&3&4\\{ - 1}&0&2\\3&1&2\end{array}} \right|\\ = - 3\left| {\begin{array}{*{20}{c}}{ - 1}&2\\3&2\end{array}} \right| + 0 - 1\left| {\begin{array}{*{20}{c}}1&4\\{ - 1}&2\end{array}} \right|\\ = - 3\left( { - 8} \right) - 6\\ = 24 - 6\\\det {A_3}\left( b \right) = 18\end{array}\)

By Cramer’s rule, \({x_i} = \frac{{\det {A_i}\left( b \right)}}{{\det A}}\), \(i = 1,2,3\). Hence,

\(\begin{array}{c}{x_1} = \frac{{\det {A_1}\left( b \right)}}{{\det A}}\\ = \frac{6}{{15}}\\{x_1} = \frac{2}{5}\end{array}\)

\(\begin{array}{c}{x_2} = \frac{{\det {A_2}\left( b \right)}}{{\det A}}\\ = \frac{{12}}{{15}}\\{x_2} = \frac{4}{5}\end{array}\)

\(\begin{array}{c}{x_3} = \frac{{\det {A_3}\left( b \right)}}{{\det A}}\\ = \frac{{18}}{{15}}\\{x_3} = \frac{6}{5}\end{array}\)