Let's say for the eigenvector equation, $ (\lambda I - A)X = 0 $, some eigenvalue of $A$, $ \lambda_1 $, is found and $ \lambda_1 I - A $ is reduced to solve for its respective eigenvector $ X_1 $. If it is reduced to the identity matrix, $ I $, what can you say about the eigenvector $ X_1 $? Does the eigenvector $X_1$ exist? Is $A$ diagonalizable?
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By definition, $x$ is an eigenvector of $A$ for the value $\lambda_1$ if $Ax = \lambda_1 x$, or by rearranging, $(\lambda_1 I - A)x=0$. Also by definition, $\lambda_1$ is an eigenvalue if and only if it has a non-zero eigenvector.
So if $\lambda_1 I-A$ is row-reducible to the identity matrix, then the equation $(\lambda_1 I - A)x=0$ has only the trivial solution $x=0$. But then $\lambda_1$ has no eigenvectors except 0, so $\lambda_1$ is not actually an eigenvalue at all.
In other words, $\lambda_1$ is an eigenvalue of $A$ if and only if $(\lambda_1 I - A)x$ is not row-reducible to the identity.
One other note, your choice of wording implies you might think that each eigenvalue has exactly one eigenvector, which is definitely not the case. The set of eigenvectors of an eigenvalue forms a subspace, so if there is one eigenvector then there are an infinity of them.