Answer
Characteristic Equation: $\lambda^2-9\lambda+32$
There are no real eigenvalues for the given matrix
Work Step by Step
1.) To find the characteristic, solve for and simplify $det(A-\lambda I)$
$(5-\lambda)(4-\lambda)-(3)(-4)$
$=20-4\lambda-5\lambda+\lambda^2+12$
$=\lambda^2-9\lambda+32$
2.) To solve for the eigenvalues, set the characteristic equation equal to zero and solve for $\lambda$
$\lambda^2-9\lambda+32=0$
Note: Use the quadratic equation to solve for $\lambda$
$\lambda = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$\lambda = \frac{9 \pm \sqrt{81-4(1)(32)}}{2(1)}$
$\lambda = \frac{9 \pm \sqrt{-47}}{2}$
Since the determinant of the quadratic equation is negative (-47), there are no real eigenvalues of the matrix