Answer
See below
Work Step by Step
a) Let $S$ be a matrix such that $S^{-1}AS = diag(\lambda_1, \lambda_2,..., \lambda_n)$
then $\det (S^{-1}AS )=\lambda_1, \lambda_2,..., \lambda_n \\
\det (S^{-1}AS )\ne 0$
Hence, $A$ is invertible.
b) We have $S^{-1}AS = diag(\lambda_1, \lambda_2,..., \lambda_n)$
then $(S^{-1}AS)^{-1}= diag(\lambda_1, \lambda_2,..., \lambda_n)^{-1}\\
\rightarrow (S^{-1}AS)^{-1}= diag(\frac{1}{\lambda_1}, \frac{1}{\lambda_2},..., \frac{1}{\lambda_n})\\
\rightarrow S^{-1}A^{-1}S= diag(\frac{1}{\lambda_1}, \frac{1}{\lambda_2},..., \frac{1}{\lambda_n})$
