Find the eigenvalues of an unsymmetrical matrix

Question

Eigenvalues of $\begin{bmatrix}4 & -5 & 1 \\ 1 & 0 & -1\ \\ 0 & 1 & -1\end{bmatrix}$

I usually set $0$ equal to $\det(A- \lambda I)$ to find the eigenvalues, but the book says they are $2, 1$ and $0$, which isn't what I got. Am I missing a step?

Answer 1

I will take the determinant by cofactor expansion about the first column:

$$p(\lambda) = \det\begin{pmatrix} 4 - \lambda & -5 & 1 \\ 1 & -\lambda & -1 \\ 0 & 1 & -1 - \lambda\end{pmatrix}$$

$$= (4 - \lambda)((-\lambda)(-1 - \lambda) + 1) - (-5(-1 - \lambda) - 1)$$

$$= (4 - \lambda)(\lambda^2 + \lambda + 1) - 5\lambda - 4$$

$$= -\lambda^3 + 3\lambda^2 - 2\lambda$$

$$= -\lambda(\lambda^2 - 3\lambda + 2)$$

$$= -\lambda(\lambda - 2)(\lambda - 1)$$

Now if the characteristic polynomial is set equal to zero, we get $\lambda = 0,1,2$.