Answer
Characteristic Equation: $\lambda^2-11\lambda+40$
There are no real eigenvalues for the given matrix
Work Step by Step
1.) To solve for the characteristic equation, solve the $det(A-\lambda I)$
$(3-\lambda)(8-\lambda)-(-4)(4)$
$=(24-11\lambda+\lambda^2)+16$
$=\lambda^2-11\lambda+40$
2.) To solve for the eigenvalues, set the characteristic equation equal to zero and solve for $\lambda$
$\lambda^2-11\lambda+40=0$
Note: Use the quadratic equation to solve for $\lambda$
$\lambda=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\lambda=\frac{11 \pm \sqrt{121-4(1)(40)}}{2(1)}$
$\lambda=\frac{11 \pm \sqrt{-39}}{2}$
Since the discriminant of the quadratic formula is negative (-39), there are no real eigenvalues.
