Question

Use the inverse found in Exercise 1 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)

Short answer

The solutions are \({x_1} = 7\) and \({x_2} = - 9\).

Step-by-step solution

Write the matrix form

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

Here, \(A = \left( {\begin{aligned}{*{20}{c}}8&6\\5&4\end{aligned}} \right),{\rm{ }}x = \left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right),\) and \(b = \left( {\begin{aligned}{*{20}{c}}2\\{ - 1}\end{aligned}} \right)\).

Write the inverse obtained in Exercise 1

From Exercise 1, \({\left( {\begin{aligned}{*{20}{c}}8&6\\5&4\end{aligned}} \right)^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}2&{ - 3}\\{ - 2.5}&4\end{aligned}} \right)\).

Express the solution

\(\begin{aligned}{c}x = {A^{ - 1}}b\\ = \left( {\begin{aligned}{*{20}{c}}2&{ - 3}\\{ - 2.5}&4\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}2\\{ - 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4 + 3}\\{ - 5 - 4}\end{aligned}} \right)\\x = \left( {\begin{aligned}{*{20}{c}}7\\{ - 9}\end{aligned}} \right)\end{aligned}\)

Thus, \({x_1} = 7\) and \({x_2} = - 9\).