Question

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

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

Short answer

\(\left( {\begin{aligned}{*{20}{c}}8&3&1\\{10}&4&1\\{\frac{7}{2}}&{\frac{3}{2}}&{\frac{1}{2}}\end{aligned}} \right)\)

Step-by-step solution

Find the expression \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

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

Apply the row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

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

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&{ - 2}\\{ - 3}&1&4\\0&{ - 3}&8\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\{ - 2}&0&1\end{aligned}} \right)\)

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

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

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&{ - 2}\\0&1&{ - 2}\\0&{ - 3}&8\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\3&1&0\\{ - 2}&0&1\end{aligned}} \right)\)

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

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

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

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

At row two, add rows two and three, i.e., \({R_2} \to {R_2} + {R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&{ - 2}\\0&1&0\\0&0&2\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\{10}&4&1\\7&3&1\end{aligned}} \right)\)

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

At row one, add rows three and one, i.e., \({R_1} \to {R_1} + {R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&2\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}8&3&1\\{10}&4&1\\7&3&1\end{aligned}} \right)\)

Divide row three by 2, i.e., \({R_3} \to \frac{{{R_3}}}{2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}8&3&1\\{10}&4&1\\{\frac{7}{2}}&{\frac{3}{2}}&{\frac{1}{2}}\end{aligned}} \right)\)

So, the inverse of the matrix is \(\left( {\begin{aligned}{*{20}{c}}8&3&1\\{10}&4&1\\{\frac{7}{2}}&{\frac{3}{2}}&{\frac{1}{2}}\end{aligned}} \right)\).