Answer
Characteristic polynomial: $\lambda^2-2\lambda-1$
$\lambda=1\pm\sqrt{2}$
Work Step by Step
1.) To solve for the characteristic equation, expand $det(A-\lambda I)$.
$(3-\lambda)(-1-\lambda)-(-2)(1)$
$=\lambda^2-2\lambda-1$
2.) Set the characteristic equation equal to zero to solve for the eigenvalues
$\lambda^2-2\lambda-1=0$
Note: Use the quadratic equation to solve for the eigenvalues
$\lambda = (-b\pm \sqrt{ b^2-4ac})/{2a}$
$\lambda= (2\pm\sqrt{(-2)^2-4(1)(-1)})/2(1)$
$\lambda=(2\pm\sqrt{8})/2$
$\lambda=1\pm\sqrt{2}$
