Chapter 2 - Matrices and Systems of Linear Equations - 2.6 The Inverse of a Square Matrix - Problems - Page 178: 36

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A is an $n \times n$ matrix with $A^4 = 0$ We have $(I_n - A)(I_n - A)^{-1}$ since $(I_n - A)^{-1}=I_n+A+A^2+A^3$ so $(I_n - A)(I_n+A+A^2+A^3)$ $=I_n+A+A^2+A^3-A-A^2-A^3-A^4$ $=I_n-A^4$ $=I_n-0$ $=I_n$ Hence, $I_n- A$ is invertible with $(I_n - A)^{-1}=I_n+A+A^2+A^3$