Chapter 2 - Matrices and Systems of Linear Equations - 2.6 The Inverse of a Square Matrix - Problems - Page 178: 35

See answer below

To prove that $I_n- A^3$ is invertible with $(I_n = A^3)^{-1}$ we will do: $(I_n - A^3)(I_n - A^3)^{-1}$ since $(I_n - A^3)^{-1}=I_n+A^3+A^6+A^9$ so $(I_n- A^3)(I_n+A^3+A^6+A^9)$ $=I_n+A^3+A^6+A^9-A^3-A^6-A^9-A^{12}$ $=I_n-A^{12}$ $=I_n-0$ $=I_n$ Hence, $I_n- A^3$ is invertible with $(I_n - A^3)^{-1}=I_n+A^3+A^6+A^9$