Step 1
We have to prove that if $\lambda$ is an eigenvalue of an invertible matrix A, then ${\lambda ^{ - 1}}$ is an eigenvalue of ${A^{ - 1}}$

An eigenvector of an $n \times n$ matrix A is a nonzero vector x such that $A{\bf{x}} = \lambda {\bf{x}}$ for some scalar $\lambda $. A scalar $\lambda $ is called an eigenvalue of A if there is a nontrivial solution $\bf{x}$ of $A{\bf{x}} = \lambda {\bf{x}}$; such an $\bf{x}$ is called an eigenvector corresponding to $\lambda $.



Step 2: Write the equation for eigenvector
\[\begin{array}{l}A{\bf{x}} = \lambda {\bf{x}}\\{A^{ - 1}}A{\bf{x}} = \lambda {A^{ - 1}}{\bf{x}}\,\,::\,\,\left\{ {{\rm{Multiply}}\,\,{\rm{by}}\,\,{A^{ - 1}}} \right\}\\I{\bf{x}} = \lambda {A^{ - 1}}{\bf{x}}\,\,\,::\,\,\,\left\{ {{A^{ - 1}}A\, = I} \right\}\\{\bf{x}} = \lambda {A^{ - 1}}{\bf{x}}\,\,\,::\,\,\,\left\{ {I{\bf{x}}\, = {\bf{x}}} \right\}\\{\lambda ^{ - 1}}{\bf{x}} = {A^{ - 1}}{\bf{x}}\,\,\,::\,\,\,\left\{ {{\rm{move}}\,\,\,{\rm{\lambda }}\,\,{\rm{to}}\,\,{\rm{left}}\,\,\,{\rm{side}}} \right\}\\{A^{ - 1}}{\bf{x}} = {\lambda ^{ - 1}}{\bf{x}}\,::\,\,\,\,\left\{ {\,{\rm{Switch}}\,\,\,{\rm{sides}}} \right\}\end{array}\]

ANSWER
As we can see that the last equation satisfied the condition that ${\lambda ^{ - 1}}$ is an eigenvalue for ${A^{ - 1}}$