# Finding Eigenvalues of a 3x3 Matrix (7.12-17)

Please check my work in finding eigenvalues for the following problem. I am working out of the textbook Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons.

For reference the following identity is given because some textbooks reverse the formula having $\lambda$ subtract the diagonal elements instead of subtracting $\lambda$ from the diagonal elements:

$$det(A - \lambda I) = \begin{vmatrix} 13-\lambda & 0 & -15 \\ -3 & 4-\lambda & 9 \\ 5 & 0 & -7-\lambda \\ \end{vmatrix} = 0$$ Taking the center column we have: $$(4-\lambda) \begin{vmatrix} 13-\lambda & -15 \\ 5 & -7-\lambda \\ \end{vmatrix} \\ = (4-\lambda)[(13-\lambda)(-7-\lambda) + 5(15)] = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda - 91 + 75) = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda -21) = 0 \\ = -\lambda^3 + 10\lambda^2 - 24\lambda + 21\lambda - 84 = 0 \\ = \lambda^3 - 10\lambda^2 + 3\lambda + 84 = 0 \\$$ Using an online calculator the characteristic equation factors into: $$\lambda^3 - 10\lambda^2 + 3\lambda + 84 = 0 \\ (\lambda - 4)(\lambda^2 - 6\lambda - 21) = 0 \\$$ But answer in text is \begin{align*} \lambda_1 = 8 \qquad \lambda_2 = 4 \qquad \lambda_3 = -2 \\ \end{align*} Question: Although I can get $\lambda = 4$ out of the factored equation there is no way to get the other two eigenvalues. I suspect my algebra. Where did I go wrong?

• Your error is in = $$(4-\lambda)(\lambda^2 - 6\lambda - 91 + 75) = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda \color{red}{-21}) = 0$$. It should be $\color{blue}{-16}$ and then it factors well to give you the required answerr – Shailesh Jan 24 '16 at 0:11
• And then instead of multplying out $(4-\lambda)(\lambda^2-6\lambda -16)$, just factor $(4-\lambda)(\lambda-8)(\lambda +2)$. – Tim Raczkowski Jan 24 '16 at 0:13

Your error is in the step $$(4-\lambda)(\lambda^2 - 6\lambda - 91 + 75) = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda \color{red}{-21}) = 0$$ It should be $\color{blue}{-16}$ and then it factors well to give you the required answer