My instructor claims that it's inefficient and superfluous to compute eigenvectors de novo for each $2$ by $2$ matrix. He suggested a trick instead which resembles the eigenvectors and cases here. He avers that once you find $\lambda$, then immediately conclude from $A - \lambda I = \begin{bmatrix} a-\lambda & b \\ c & d - \lambda \\ \end{bmatrix}$,
$1.$ that an eigenvector is always $\begin{bmatrix} \color{#FF4F00}{\LARGE{-}} b \\ a - \lambda \end{bmatrix} $ or its negative : $\begin{bmatrix} { b \\ \color{#FF4F00}{\LARGE{-}} ( a - \lambda ) } \end{bmatrix} $$2.$ If the first row of $ A - \lambda I = \mathbf{ 0} $, then an eigenvector is always $\begin{bmatrix} \color{#FF4F00}{\LARGE{-}}( d - \lambda) \\ c \end{bmatrix} $ or the negative of this.
$3.$ If $ A - \lambda I = $ 0 matrix, then any vector is an eigenvector. I think this is the reason.
Informally and intuitively, would someone please explain/expound on his assertions? What about for eigenvalues or higher dimensions? No formal proofs or arguments please.
If so, why don't textbooks explain this? Strang adverts to it on P288 but only in 2 sentences.