Is it allowed to swap rows/col in 0 = det(A-λI) so I put my 4x4 A matrix in a lower triangular form?

Question

So, I want to diagonalize this 4x4 Matrix A = \begin{bmatrix}2&0&0&0\\-1&3&1&0\\2&-2&0&0\\2&-1&3&-5\end{bmatrix}

and putting it in a lower triangular form gonna make the whole process straightforward. So, my question is is my very first step in finding the eigenvalues as solutions of 0 = det(A-λI) expressed 0 = det of \begin{bmatrix}2-λ&0&0&0\\-1&3-λ&1&0\\2&-2&0-λ&0\\2&-1&3&-5-λ\end{bmatrix} equivalent to 0 = (-1)^(2-1) det of \begin{bmatrix}2-λ&0&0&0\\2&-2-λ&0&0\\-1&3&1-λ&0\\2&-1&3&-5-λ\end{bmatrix} ----(1)

justified by det(A) = (-1)^(i-1)* det(B), where i the row swapped by i-1 and B in this case is: \begin{bmatrix}2&0&0&0\\2&-2&0&0\\-1&3&1&0\\2&-1&3&-5\end{bmatrix}

I am not sure if (1) legit or not but it gives me correct eigenvalues (the diagonal elements) or it is just a coincidence.

Thank you!

Answer 1

Let $$A = \begin{bmatrix} 1 & 2\\3 & 4 \end{bmatrix}$$ $$p_A(\lambda) = \det(A - \lambda I) = (1-\lambda)(4-\lambda) - 6$$ Let $$B = \begin{bmatrix} 2 & 1\\4 & 3 \end{bmatrix}$$ $$p_B(\lambda) = \det(B - \lambda I) = (2-\lambda)(3-\lambda) - 4 = PA$$ where $P$ is a permutation matrix. Arranging $p_A(\lambda)$ and $p_B(\lambda)$, you will find that $$p_A(\lambda)\neq p_B(\lambda)$$

In general, if $P$ is a permutation matrix then $$\det (A - \lambda I) = \det(PP(A-\lambda I)) = \det(P(A-\lambda I)P) = \det(PAP - \lambda I) \neq \det(PA - \lambda I)$$