Requirement that determinant be zero for characteristic polynomial In one book I'm reading that

[...]
  $\newenvironment{rcases}
  {\left.\begin{aligned}}
  {\end{aligned}\right\rbrace}$
$\begin{rcases}
  (a_{11}-\lambda)x_1+a_{12}x_2=0, \\
  a_{21}x_1+(a_{22}-\lambda)x_2=0
\end{rcases}$
for some scalar $\lambda$. Since the equations for $x_1, x_2$ comprise a linear homogeneous system, and since $x_1^2+x_2^2\ne 0$, an eigenvalue $\lambda$ must be such that the determinant of the system vanishes: thus
$$\lambda^2 - (a_{11}+a_{22})\lambda+a_{11}a_{22}-a_{12}a_{21}=0$$

I don't see how the fact that  $x_1^2+x_2^2\ne 0$ implies the necessity for the determinant to be equal zero. Can someone please clarify?

$x_1^2 + x_2^2 \neq 0$ is just another way of saying that the vector $(x_1, x_2)$ is not $(0,0)$. So, $(A -\lambda I) x = 0$ has a nontrivial solution. Thus, $A - \lambda I$ is singular, that is, its determinant equals zero.
