Need a help in finding the inverse of an operator . The question and part of its answer is given as follows:

13. Let $K$ be an operator of a finite rank on a Hilbert space $H$. For $\varphi \in H$,
  $$ K\varphi = \sum_{j=1}^{n} \langle \varphi, \varphi_j\rangle\psi_j. $$
  Suppose $\psi_j \in \operatorname{sp}\{ \varphi_1, \cdots, \varphi_n\}^{\perp}$ for $j = 1, \cdots, n$. Prove that $\mathrm{I} + \alpha K$ is invertible for any $\alpha$ and find its inverse.
Solution. Let $\alpha \in \mathbb{C}$, $K'\varphi = \sum_{j=1}^{n} \langle \varphi, \varphi_j\rangle (-\alpha \psi_j)$. By Theorem 7.1,
  \begin{align*}
\text{$\mathrm{I} + \alpha K$ is invertible}
&\quad \Leftrightarrow \quad
\text{$\mathrm{I}-K'$ is invertible} \\
&\quad \Leftrightarrow \quad
\det(\delta_{ij}-\langle(-\alpha\psi_j),\varphi_i\rangle)_{i,j=1}^{n}\neq 0.
\end{align*}
  But
  $$\det(\delta_{ij}-\langle(-\alpha\psi_j),\varphi_i\rangle)_{i,j=1}^{n}
= \det(\delta_{ij}+\alpha\langle\psi_j,\varphi_i\rangle)_{i,j=1}^{n}
= 1
\neq 0$$
  (because $\psi_j \in \operatorname{sp}\{ \varphi_1, \cdots, \varphi_n\}^{\perp}$ for $j = 1, \cdots, n$). Thus $\mathrm{I}+\alpha K$ is invertible.

A part of the theorem is given as follows:

7.1. Theorem. Suppose $K \in L(H)$ is of finite rank, say
  $$ Kx = \sum_{j=1}^{n} \langle x, \varphi_j\rangle\psi_j. $$
  The operator $I-K$ is invertible if and only if
  $$\det(\delta_{ij}-\langle\psi_j,\varphi_i\rangle)_{i,j=1}^{n} \neq 0. $$
  In this case, for every $y \in H$,
  $$ (I-K)^{-1}y = y - \frac{1}{\det(a_{ij})}\det\begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} & \langle y, \varphi_1\rangle \\
a_{21} & a_{22} & \cdots & a_{2n} & \langle y, \varphi_2\rangle \\
\vdots & \vdots & & \vdots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} & \langle y, \varphi_n\rangle \\
\psi_1 & \psi_2 & \cdots & \psi_n & 0
\end{pmatrix} $$

Then the action of the inverse of the operator on $y$ is $y-I$, but how can this tell me what is the inverse of the operator $K$, could anyone clarify this for me please?
Thanks! 

Your hypotheses on $\{\psi_1,\dotsc,\psi_n\}$ and $\{\varphi_1,\dotsc,\phi_n\}$ make an appeal to that theorem completely and utterly unnecessary. Since $\{\psi_1,\dotsc,\psi_n\} \subset \{\varphi_1,\dotsc,\phi_n\}^\perp$, it immediately follows that $K^2 = 0$. What, then, is $I^2 - \alpha^2 K^2$, and why should this suggest to you an explicit inverse for $I + \alpha K$?
