Answer to Question #216171 in Linear Algebra for Chris

Question

Two of the eigenvalues of a 3*3 matrix A are 2 and 3 and the determinant is 48. Find the third eigenvalue and the trace (A)

Answer 1

Determinant is the product of eigenvalues

$$\det A=\lambda_1\lambda_2\lambda_3=48$$

Given $\lambda_1=2, \lambda_2=3.$ Then

$$\lambda_3=\dfrac{\det A}{\lambda_1\lambda_2}=\dfrac{48}{2(3)}=8$$

$$tr(A)=\lambda_1+\lambda_2+\lambda_3=2+3+8=13$$