linear algebra - The effect of elementary row operations on characteristic polynomial

I have a $3 \times 3$ matrix $A$ for which I need to determine the characteristic polynomial.

Suppose I had $\det(A)=f$. If I performed row operations on $A$ I could manipulate $f$ accordingly to find the determinant for the new matrix. So for example if I multiplied row $1$ by $k$ and swapped row $2$ and row $3$ the determinant for the new matrix would be given by $-1 \times k \times f$.

My question: Could I simplify $A$ by performing row operations on it,
then use the simplified $A$ in $(\lambda I_{3}-A)$, calculate $\det(\lambda I_{3}-A)$ and finally manipulate $\det(\lambda I_{3} -A)$ according to the row operations that I did on the original $A$. So if I swapped row $1$ and $3$ to get the simplified $A$, I would multiply $\det(\lambda I_{3}-A)$ by $-1$ in the end to get the final determinant.

Would this change the characteristic polynomial? If yes, why?

Answer

No, elementary matrix operations don't preserve the eigenvalues in general. As an example, consider $A = I$, where $I$ is the $3 \times 3$ identity matrix.

$$\det{(\lambda I - A)} = \det{(\lambda I - I)} = (\lambda-1)^3.$$

Now, lets construct $B$ by swapping rows 1 and 3 while multiplying row 2 by 2 (as suggested in your comment), i.e.
$$B = \begin{pmatrix} 0 & 0 & 1\\ 0 & 2 & 0\\ 1 & 0 & 0 \end{pmatrix},$$
and

$$\det{(\lambda I - B)} = (x-2)(x^2 - 1) \neq -2 (\lambda - 1)^3 = -2 \det{(\lambda I - A)}.$$

However, we could construct $B_\lambda$ by starting with the matrix $\lambda I - A$, swapping rows 1 and 3 and multiplying row 2 by 2, i.e.

$$B_\lambda = \begin{pmatrix} 0 & 0 & \lambda-1\\ 0 & 2\lambda-2 & 0\\ \lambda-1 & 0 & 0 \end{pmatrix}.$$
Then
$$\det{B_\lambda} = -2(\lambda-1)^3 = -2 \det{(\lambda I - A)}$$
as expected.