91Ó°ÊÓ

The inverse of an\(m \times m\)matrix A can be computed using theaugmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\), where\(I\)is the identity matrix. Matrix Ahas an inverse if \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\) is row equivalent to \(\left( {\begin{aligned}{*{20}{c}}I&{{A^{ - 1}}}\end{aligned}} \right)\).

Consider the matrix\(A = \left( {\begin{aligned}{*{20}{c}}1&2\\4&7\end{aligned}} \right)\).

Write the augmented matrix\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)as shown below:

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

Row reduce the augmented matrix.

Use the\({x_1}\)term in the first equation to eliminate the\(4{x_1}\)term from the second equation. Add\( - 4\)times row two to row two.

\(\left( {\begin{aligned}{*{20}{c}}1&2&1&0\\4&7&0&1\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&2&1&0\\0&{ - 1}&{ - 4}&1\end{aligned}} \right)\)

Use the\( - {x_2}\)term in the second equation to eliminate the\(2{x_2}\)term from the first equation. Add 2 times row two to row one.

\(\left( {\begin{aligned}{*{20}{c}}1&2&1&0\\0&{ - 1}&{ - 4}&1\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&0&{ - 7}&2\\0&{ - 1}&{ - 4}&1\end{aligned}} \right)\)

Multiply row two by\( - 1\).

\(\left( {\begin{aligned}{*{20}{c}}1&0&{ - 7}&2\\0&{ - 1}&{ - 4}&1\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&0&{ - 7}&2\\0&1&4&{ - 1}\end{aligned}} \right)\)

By comparing with\(\left( {\begin{aligned}{*{20}{c}}I&{{A^{ - 1}}}\end{aligned}} \right)\), you get\({A^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}{ - 7}&2\\4&{ - 1}\end{aligned}} \right)\).

Thus, theinverse of the matrix is \(\left( {\begin{aligned}{*{20}{c}}{ - 7}&2\\4&{ - 1}\end{aligned}} \right)\).