Exercises 25 and 26 prove Theorem 4 for \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\).

26. Show that if \(ad - bc \ne {\bf{0}}\), the formula for \({A^{ - 1}}\) works.

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 theidentity matrix. Matrix Ahas an inverse only 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\(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\).

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

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}a&b&1&0\\c&d&0&1\end{aligned}} \right)\)

Row reduce the augmented matrix as shown below:

Multiply row one by\(\frac{1}{a}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&{b/a}&{1/a}&0\\c&d&0&1\end{aligned}} \right)\)

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

\(\left( {\begin{aligned}{*{20}{c}}1&{b/a}&{1/a}&0\\0&{d - \frac{{bc}}{a}}&{ - c/a}&1\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&{b/a}&{1/a}&0\\0&{\frac{{ad - bc}}{a}}&{ - c/a}&1\end{aligned}} \right)\)

Use the\(\frac{{b{x_2}}}{a}\)term in the first equation to eliminate the\(\left( {\frac{{ad - bc}}{a}} \right){x_2}\)term from the second equation. Add\( - \left( {\frac{b}{{ad - bc}}} \right)\)times row two to row one.

\(\left( {\begin{aligned}{*{20}{c}}1&{b/a}&{1/a}&0\\0&{\frac{{ad - bc}}{a}}&{ - c/a}&1\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&0&{\frac{d}{{ad - bc}}}&{ - \frac{b}{{ad - bc}}}\\0&{\frac{{ad - bc}}{a}}&{ - c/a}&1\end{aligned}} \right)\)

Multiply row two by\(\left( {\frac{a}{{ad - bc}}} \right)\).

\(\left( {\begin{aligned}{*{20}{c}}1&0&{\frac{d}{{ad - bc}}}&{ - \frac{b}{{ad - bc}}}\\0&1&{\frac{{ - c}}{{ad - bc}}}&{\frac{a}{{ad - bc}}}\end{aligned}} \right)\)

By comparing with\(\left( {\begin{aligned}{*{20}{c}}I&{{A^{ - 1}}}\end{aligned}} \right)\), you get\({A^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}{\frac{d}{{ad - bc}}}&{ - \frac{b}{{ad - bc}}}\\{ - \frac{c}{{ad - bc}}}&{\frac{a}{{ad - bc}}}\end{aligned}} \right)\), or\({A^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{aligned}{*{20}{c}}d&{ - b}\\{ - c}&a\end{aligned}} \right)\).

Thus, theinverseof matrix is \({A^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{aligned}{*{20}{c}}d&{ - b}\\{ - c}&a\end{aligned}} \right)\) if \(ad - bc \ne 0\).