I'm currently studying about how to solve the eigenvalue problem of systems of linear equations and had a question regarding an exercise problem.
I'm aware that in order to determine the eigenvalues $\lambda$, we set the determinant of the characteristic matrix to $0$ and solve:
$$\text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0$$
I'm also aware that when finding the determinant of a matrix that's larger than $2 \times 2$ then we use the equation:
$$ \begin{align} \text{det}(\mathbf{A}) & = \sum_{j = i}^n a_{jk} C_{jk} \\ & = \sum_{j = 1}^n a_{jk}\ (-1)^{j + k}\ M_{jk} \end{align} $$
and we can choose any row $j$ or column $k$ and still have the same result.
The specific exercise problem that I have is to find the eigenvalues for the following matrix:
$$ \mathbf{A} = \begin{bmatrix} -2 & \phantom{-}2 & -3 \\ \phantom{-} 2 & \phantom{-}1 & -6 \\ -1 & -2 & \phantom{-}0 \end{bmatrix} $$
The characteristic matrix is:
$$ \mathbf{A} - \lambda \mathbf{I} = \begin{bmatrix} -2 -\lambda & \phantom{-}2 & -3 \\ \phantom{-}2 & \phantom{-}1 - \lambda & -6 \\ -1 & -2 & -\lambda \end{bmatrix} $$
The correct characteristic equation given by the textbook is:
$$-\lambda^3 - \lambda^2 + 21\lambda + 45 = 0$$
I'll try to demonstrate that I get differing results based on which row/column I choose to calculate the determinant.
1st Column: $k = 1$
$$ \begin{align} \text{det}(\mathbf{A}) & = \sum_{j = 1}^3 a_{j1}\ (-1)^{j + 1}M_{j1} \\ & = a_{11}\ (-1)^{1 + 1} M_{11} + a_{21}\ (-1)^{2 + 1}\ M_{21} + a_{31}\ (-1)^{3 + 1}\ M_{31} \\ & = (-2 - \lambda) \begin{vmatrix} 1 - \lambda & -6 \\ -2 & -\lambda \end{vmatrix} - 2 \begin{vmatrix} \phantom{-}2 & -3 \\ -2 & -\lambda \end{vmatrix} - \begin{vmatrix} 2 & -3 \\ 1 - \lambda & -6 \end{vmatrix} \\ & = (-2 - \lambda)(-\lambda(1 - \lambda) - 12) - 2(-2\lambda - 6) - (-12 + 3(1 - \lambda)) \\ & = -\lambda^3 -\lambda^2 + 21\lambda + 45 \end{align} $$
Now I'll choose the third row. This was actually my initial choice, because it seemed to lead to simpler calculations, but actually keeps giving me the wrong answer.
3rd Row: $j = 3$
$$ \begin{align} \text{det}(\mathbf{A}) & = \sum_{k = 1}^3 a_{3k}\ (-1)^{3 + k}M_{3k} \\ & = a_{31}\ (-1)^{3 + 1} M_{31} + a_{32}\ (-1)^{3 + 2}\ M_{32} + a_{33}\ (-1)^{3 + 3}\ M_{33} \\ & = - \begin{vmatrix} 2 & -3 \\ 1 - \lambda & -6 \end{vmatrix} + 2 \begin{vmatrix} -2 - \lambda & -3 \\ \phantom{-}2 & -6 \end{vmatrix} - \lambda \begin{vmatrix} -2 - \lambda & 2 \\ \phantom{-}2 & 1 - \lambda \end{vmatrix} \\ & = (-12 + 3(1 - \lambda)) + 2(6(\lambda + 2) + 6) - \lambda((\lambda + 2)(\lambda + 1) - 4) \\ & = -\lambda^3 -3\lambda^2 + 17\lambda + 45 \end{align} $$
I'm convinced that there is an error somewhere, but after spending much time I'm having trouble finding it. Is my logic correct? If so, where might the error lie? Thanks in advance!