Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

\(\begin{aligned}{c}\det A = 1\left( {12} \right) - 2\left( 5 \right)\\ = 12 - 10\\\det A = 2 \ne 0\end{aligned}\)

This implies that A is invertible.

(a)

\({A^{ - 1}} = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}&{ - 2}\\{ - 5}&1\end{aligned}} \right)\)

The solution for the system \(Ax = {b_1}\) is obtained as shown below:

\(\begin{aligned}{c}x = {A^{ - 1}}{b_1}\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}&{ - 2}\\{ - 5}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 1}\\3\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{ - 12 - 6}\\{5 + 3}\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{ - 18}\\8\end{aligned}} \right)\\x = \left( {\begin{aligned}{*{20}{c}}{ - 9}\\4\end{aligned}} \right)\end{aligned}\)

The solution for the system \(Ax = {b_2}\) is obtained as shown below:

\(\begin{aligned}{c}x = {A^{ - 1}}{b_2}\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}&{ - 2}\\{ - 5}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1\\{ - 5}\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12 + 10}\\{ - 5 - 5}\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{22}\\{ - 10}\end{aligned}} \right)\\x = \left( {\begin{aligned}{*{20}{c}}{11}\\{ - 5}\end{aligned}} \right)\end{aligned}\)

The solution for the system \(Ax = {b_3}\) is obtained as shown below:

\(\begin{aligned}{c}x = {A^{ - 1}}{b_3}\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}&{ - 2}\\{ - 5}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}2\\6\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{24 - 12}\\{ - 10 + 6}\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}\\{ - 4}\end{aligned}} \right)\\x = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\end{aligned}} \right)\end{aligned}\)

The solution for the system \(Ax = {b_4}\) is obtained as shown below:

\(\begin{aligned}{c}x = {A^{ - 1}}{b_4}\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}&{ - 2}\\{ - 5}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}3\\5\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{36 - 10}\\{ - 15 + 5}\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{26}\\{ - 10}\end{aligned}} \right)\\x = \left( {\begin{aligned}{*{20}{c}}{13}\\{ - 5}\end{aligned}} \right)\end{aligned}\)

{\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\)

(b)

\(\left( {\begin{aligned}{*{20}{c}}A&{{b_1}}&{{b_2}}&{{b_2}}&{{b_4}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&2&{ - 1}&1&2&3\\5&{12}&3&{ - 5}&6&5\end{aligned}} \right)\)

At row two, multiply row one by 5 and subtract it from row two, i.e., \({R_2} \to {R_2} - 5{R_1}\).

\( \sim \left( {\begin{aligned}{*{20}{c}}1&2&{ - 1}&1&2&3\\0&2&8&{ - 10}&{ - 4}&{ - 10}\end{aligned}} \right)\)

Divide row two by 2.

\( \sim \left( {\begin{aligned}{*{20}{c}}1&2&{ - 1}&1&2&3\\0&1&4&{ - 5}&{ - 2}&{ - 5}\end{aligned}} \right)\)

At row one, multiply row two by 2 and subtract it from row one, i.e., \({R_1} \to {R_1} - 2{R_2}\).

\( \sim \left( {\begin{aligned}{*{20}{c}}1&0&{ - 9}&{11}&6&{13}\\0&1&4&{ - 5}&{ - 2}&{ - 5}\end{aligned}} \right)\)

This implies that the solutions are \(\left( {\begin{aligned}{*{20}{c}}{ - 9}\\4\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}{11}\\{ - 5}\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\end{aligned}} \right),\) and \(\left( {\begin{aligned}{*{20}{c}}{13}\\{ - 5}\end{aligned}} \right)\), same as in part (a).