Finding eigenvalues of a $3\times3$ matrix with Laplace expansion

Question

Currently working on problem for a linear algebra class, but having a difficult time grasping eigenvalues. Here are the steps I'm doing:

$$A=\begin{bmatrix}-5 & 1 & 0 \\ 0 & -4 & 3 \\ 5 & 1 & 0\end{bmatrix}$$ $$\det(A-\lambda I_3)\rightarrow\begin{bmatrix}-5-\lambda & 1 & 0 \\ 0 & -4-\lambda & 3 \\ 5 & 1 & -\lambda\end{bmatrix}$$

Note: I unsure about the $-\lambda$ in the $0$ spot. I've been using this as a reference: https://math.dartmouth.edu/archive/m22s05/public_html/math_22_HW_8.pdf

To simplify, I chose to use Laplace expansion since we can begin from the $a_{2,1}$ index of the matrix.

$$0\times \det\begin{bmatrix}1 & 0 \\ 1 & -\lambda\end{bmatrix}+(-4-\lambda)\times \det\begin{bmatrix}-5-\lambda & 0 \\ 5 & -\lambda\end{bmatrix}+3\times \det\begin{bmatrix}-5-\lambda & 1 \\ 5 & 1\end{bmatrix}$$

Simplifying we get $$(-4-\lambda)\times \det\begin{bmatrix}-5-\lambda & 0 \\ 5 & -\lambda\end{bmatrix}+3\times \det\begin{bmatrix}-5-\lambda & 1 \\ 5 & 1\end{bmatrix}$$

Number crunching, we get:

$$(-4-\lambda)(-5-\lambda)(-\lambda)+3(-5-\lambda)-5=-(\lambda^3+9\lambda^2+23\lambda+20)$$

Yet, I'm getting the incorrect answer. Am I performing the Laplace expansion correctly? Is there a hole in my process somewhere? Is it correct to replace a $0$ on the diagonal with $-\lambda$? If so, why?

Answer 1

I see two errors.

$-0\times \det\begin{bmatrix}1 & 0 \\ 1 & -\lambda\end{bmatrix}+(-4-\lambda)\times \det\begin{bmatrix}-5-\lambda & 0 \\ 5 & -\lambda\end{bmatrix}-3\times \det\begin{bmatrix}-5-\lambda & 1 \\ 5 & 1\end{bmatrix}$

and

$(-4-\lambda)(-5-\lambda)(-\lambda)-3(-5-\lambda-5)$

Answer 2

When you perform Laplacian expansion, remember to switch sign from entry to entry. I think it should be -3 rather than +3

$$0\times \det\begin{bmatrix}1 & 0 \\ 1 & -\lambda\end{bmatrix}+(-4-\lambda)\times \det\begin{bmatrix}-5-\lambda & 0 \\ 5 & -\lambda\end{bmatrix}-3\times \det\begin{bmatrix}-5-\lambda & 1 \\ 5 & 1\end{bmatrix}$$