If $sen{A}<1$, then $I-A$ is invertible, and $$ex (I-A)^{-1}=I+A+A^2+cdots, eex$$ aa convergent power series. This is called the Neumann series.
Solution. Since $sen{A}<1$, $$ex sum_{n=0}^infty sen{A}^n=frac{1}{1-sen{A}}<infty. eex$$ Due to the completeness of the matrix space, $dps{sum_{n=0}^infty A_n}$ converges. Since $$ex (I-A)(I+cdots+A^{n-1})=I-A^n, eex$$ we may take limit to get $$ex (I-A)cdot sum_{n=0}^infty A^n=I. eex$$