Answer
See solution
Work Step by Step
If A is similar to B,
$A=PBP^{-1}$
If A is invertible, all of its eigenvalues are nonzero. This means all the diagonal values of B are nonzero, so each column in B is a pivot column, which means B is invertible.
$B=P^{-1}AP$
$B^{-1}=(P^{-1}AP)^{-1}=P^{-1}A^{-1}P$, which shows us that $B^{-1}$ is similar to $A^{-1}$
