Answer
\[A^{-1}=\frac{abcd}{bc-ad}\left[\begin{array}{cc}
\frac{1}{d}&\frac{-1}{b}\\
\frac{-1}{c}&\frac{1}{a}\end{array}\right]\]
Work Step by Step
\[A=\left[\begin{array}{cc}
\frac{1}{a}&\frac{1}{b}\\
\frac{1}{c}&\frac{1}{d}\end{array}\right]\]
It is given that $a\neq 0,\;b\neq 0,\;c\neq 0,\;d\neq 0$
$\Rightarrow \left(\frac{1}{a}\right)\left(\frac{1}{d}\right)-\left(\frac{1}{b}\right)\left(\frac{1}{c}\right)\neq 0$
$\Rightarrow A$ is invertible
Using Theorem 3.8
\[A^{-1}=\frac{1}{ \left(\frac{1}{a}\right)\left(\frac{1}{d}\right)-\left(\frac{1}{b}\right)\left(\frac{1}{c}\right) }\left[\begin{array}{cc}
\frac{1}{d}&\frac{-1}{b}\\
\frac{-1}{c}&\frac{1}{a}\end{array}\right]\]
$A^{-1}=\frac{1}{\frac{1}{ad}-\frac{1}{bc}}\left[\begin{array}{cc}
\frac{1}{d}&\frac{-1}{b}\\
\frac{-1}{c}&\frac{1}{a}\end{array}\right]$
$\Rightarrow A^{-1}=\frac{abcd}{bc-ad}\left[\begin{array}{cc}
\frac{1}{d}&\frac{-1}{b}\\
\frac{-1}{c}&\frac{1}{a}\end{array}\right]$
Hence $\Rightarrow A^{-1}=\frac{abcd}{bc-ad}\left[\begin{array}{cc}
\frac{1}{d}&\frac{-1}{b}\\
\frac{-1}{c}&\frac{1}{a}\end{array}\right]$ .
