Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{3}}&{\bf{8}}\\{\bf{2}}&{\bf{4}}&{{\bf{11}}}\\{\bf{1}}&{\bf{2}}&{\bf{5}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{3}}}&{\bf{5}}\\{\bf{1}}&{\bf{5}}\\{\bf{3}}&{\bf{4}}\end{aligned}} \right)\). Compute \({A^{ - {\bf{1}}}}B\) without computing \({A^{ - {\bf{1}}}}\). (Hint: \({A^{ - {\bf{1}}}}B\) is the solution of the equation \(AX = B\).)

Note that \({A^{ - 1}}B\) is the solution of \(Ax = B\); that is, \(x = {A^{ - 1}}B\).

Here, the augmented matrix is:

\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&3&8&{ - 3}&5\\2&4&{11}&1&5\\1&2&5&3&4\end{aligned}} \right)\)

At row 2, multiply row 1 by 2 and subtract it from row 2 i.e., \({R_2} \to {R_2} - 2{R_1}\), and at row 3, subtract row 1 from row 3, i.e., \({R_3} \to {R_3} - {R_1}\).

\( \sim \left( {\begin{aligned}{*{20}{c}}1&3&8&{ - 3}&5\\0&{ - 2}&{ - 5}&7&{ - 5}\\0&{ - 1}&{ - 3}&6&{ - 1}\end{aligned}} \right)\)

Interchange row 2 and row 3, i.e., \({R_2} \leftrightarrow {R_3}\).

\( \sim \left( {\begin{aligned}{*{20}{c}}1&3&8&{ - 3}&5\\0&{ - 1}&{ - 3}&6&{ - 1}\\0&{ - 2}&{ - 5}&7&{ - 5}\end{aligned}} \right)\)

At row 2, divide row 2 by \( - 1\).

\( \sim \left( {\begin{aligned}{*{20}{c}}1&3&8&{ - 3}&5\\0&1&3&{ - 6}&1\\0&{ - 2}&{ - 5}&7&{ - 5}\end{aligned}} \right)\)

At row 3, multiply row 2 by 2 and add it to row 3, i.e., \({R_3} \to {R_3} + 2{R_2}\).

\( \sim \left( {\begin{aligned}{*{20}{c}}1&3&8&{ - 3}&5\\0&1&3&{ - 6}&1\\0&0&1&{ - 5}&{ - 3}\end{aligned}} \right)\)

At row 2, multiply row 3 by 3 and subtract it from row 2, i.e., \({R_2} \to {R_2} - 3{R_3}\), and at row 1, multiply row 3 by 8 and subtract it row 1, i.e., \({R_1} \to {R_1} - 8{R_3}\).

\( \sim \left( {\begin{aligned}{*{20}{c}}1&3&0&{37}&{29}\\0&1&0&9&{10}\\0&0&1&{ - 5}&{ - 3}\end{aligned}} \right)\)

At row 1, multiply row 2 by 3 and subtract it from row 1, i.e., \({R_1} \to {R_1} - 3{R_2}\).

\( \sim \left( {\begin{aligned}{*{20}{c}}1&0&0&{10}&{ - 1}\\0&1&0&9&{10}\\0&0&1&{ - 5}&{ - 3}\end{aligned}} \right)\)

This implies that \(x = \left( {\begin{aligned}{*{20}{c}}{10}&{ - 1}\\9&{10}\\{ - 5}&{ - 3}\end{aligned}} \right)\).

Note that \(x = {A^{ - 1}}B\). Hence, \({A^{ - 1}}B = \left( {\begin{aligned}{*{20}{c}}{10}&{ - 1}\\9&{10}\\{ - 5}&{ - 3}\end{aligned}} \right)\).