Find the inverse of matrices in Exercise 29-32, if they exist. Use the algorithm introduced in this section.

\(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{ - {\bf{2}}}&{\bf{1}}\\{\bf{4}}&{ - {\bf{7}}}&{\bf{3}}\\{ - {\bf{2}}}&{\bf{6}}&{ - {\bf{4}}}\end{aligned}} \right)\)

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

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

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

At row two, multiply row one by 4 and subtract it from row two, i.e., \({R_2} \to {R_2} - 4{R_1}\).

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

At row three, multiply row two by 2 and subtract it from row three, i.e., \({R_3} \to {R_3} - 2{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}&1\\0&1&{ - 1}\\0&0&0\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\{ - 4}&1&0\\{10}&{ - 2}&1\end{aligned}} \right)\)

So, the inverse of the matrix is not possible.