Let \(A = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{2}}}&{ - {\bf{7}}}&{ - {\bf{9}}}\\{\bf{2}}&{\bf{5}}&{\bf{6}}\\{\bf{1}}&{\bf{3}}&{\bf{4}}\end{aligned}} \right)\). Find the third column of \({A^{ - {\bf{1}}}}\) without computing the other columns.

The expression \(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right)\) can be calculated as shown below:

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - 2}&{ - 7}&{ - 9}\\2&5&6\\1&3&4\end{aligned}\,\,\begin{aligned}{*{20}{c}}0\\0\\1\end{aligned}} \right)\)

Interchange rows one and three, i.e., \({R_1} \leftrightarrow {R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&4\\2&5&6\\{ - 2}&{ - 7}&{ - 9}\end{aligned}\,\,\begin{aligned}{*{20}{c}}1\\0\\0\end{aligned}} \right)\)

For row three, multiply row one by 2 and add it to row three, i.e., \({R_3} \to {R_3} + 2{R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&4\\2&5&6\\0&{ - 1}&{ - 1}\end{aligned}\,\,\begin{aligned}{*{20}{c}}1\\0\\2\end{aligned}} \right)\)

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

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&4\\0&{ - 1}&{ - 2}\\0&{ - 1}&{ - 1}\end{aligned}\,\,\begin{aligned}{*{20}{c}}1\\{ - 2}\\2\end{aligned}} \right)\)

At row three, subtract row two from row three, i.e., \({R_3} \to {R_3} - {R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&4\\0&{ - 1}&{ - 2}\\0&0&1\end{aligned}\,\,\begin{aligned}{*{20}{c}}1\\{ - 2}\\4\end{aligned}} \right)\)

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

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&4\\0&{ - 1}&0\\0&0&1\end{aligned}\,\,\begin{aligned}{*{20}{c}}1\\6\\4\end{aligned}} \right)\)

At row one, multiply row three by 4 and subtract it from row one, i.e., \({R_1} \to {R_1} - 4{R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&0\\0&{ - 1}&0\\0&0&1\end{aligned}\,\,\begin{aligned}{*{20}{c}}{ - 15}\\6\\4\end{aligned}} \right)\)

At row one, multiply row two by 3 and add it to row one, i.e., \({R_1} \to {R_1} + 3{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{ - 1}&0\\0&0&1\end{aligned}\,\,\begin{aligned}{*{20}{c}}3\\6\\4\end{aligned}} \right)\)

At row two, multiply row two by \( - 1\), i.e., \({R_2} \to - {R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}\,\,\begin{aligned}{*{20}{c}}3\\{ - 6}\\4\end{aligned}} \right)\)

So, the third column of the inverse matrix is \(\left( {\begin{aligned}{*{20}{c}}3\\{ - 6}\\4\end{aligned}} \right)\).