Let:
$$A=\begin{pmatrix} 3 & 0 & -2 & -3\\ 4 & -8 & 14 & -15\\ 2 & -4 & 7 & -7\\ 0 & 2 & -4 & 3 \end{pmatrix}$$
Find a change of basis matrix $P$ such that $PAP^{-1}$ is in Jordan canonical form.
After finding the characteristic polynomial I can find the Jordan form as it has eigenvalues $1$ with multiplicity $3$ and $2$ with multiplicity $1$. However, what I am struggling with is finding the characteristic polynomial in a fast and effective way rather than finding the $\mathrm{det}(A-\lambda I)$ as it would take a lot of time and error could happen easily. It was described here, but I am wondering if this method is unique or does the exchanging of rows and dividing them depend on the elements of the matrix.
Thanks in advance!