Consider a Markov process $X_t$ where $t$ belongs to natural numbers, which takes values on a finite state space $\mathcal{X}$. We say that $x \in \mathcal{X}$ is an absorbing state if $X_t = x$ implies $X_{\tau} = x$ for all $\tau \ge t$. We say a state is non-absorbing if it is not an absorbing state.
Suppose that $n$ states of the given Markov chain are non-absorbing and $s$ states are absorbing. Then we can partition the state transition matrix as \begin{align*} \begin{bmatrix} A & B \\ 0 & I_s \end{bmatrix} \end{align*} where $I_s$ is the $s \times s$ identity matrix.
How does one prove that $$I_n - A$$ is invertible, where, $I_n$ is the $n \times n$ identity matrix?
Source of my question: Definition 8.17 and Theorem 8.9 on page 83, and the first paragraph on page 84 of these notes.