For the matrix $$ A=\begin{pmatrix}0 &0 &0 &0 &0 & 0&1&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&1&0&0&0&0&0&0\\ 0&0&0&0&1&0&0&0&0&0&0&0&0\\0&0&0&0&0&1&0&0&0&0&0&0&0\\0&0&0&0&0&0&1&0&0&0&0&0&0\\0&0&0&0&0&0&0&1&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&1&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&1\\ a&b&c&d&e&f&g&h&i&j&k&l&m\end{pmatrix} $$
I would like to compute the determinant (resp. characteristic polynomial).
The matrix looks nearly the same as the companion matrix (only the first and third row differ)
$$ B=\begin{pmatrix}0 &1 &0 &0 &0 & 0&0&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&1&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&1&0&0&0&0&0&0&0&0\\0&0&0&0&0&1&0&0&0&0&0&0&0\\0&0&0&0&0&0&1&0&0&0&0&0&0\\0&0&0&0&0&0&0&1&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&1&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&1\\ a&b&c&d&e&f&g&h&i&j&k&l&m\end{pmatrix} $$
for which it is known that the characteristic polynomial is given by $$ t^{13}-mt^{12}-lt^{11}-kt^{10}-jt^9-it^8-ht^7-gt^6-ft^5-et^4-dt^3-ct^2-bt -a $$
My question is if I can deduce the characteristic polynomial of matrix $A$ from that of $B$.
(of course i could simply compute the determinant of A by, e.g. gauss. but i am interested to know whether it suffices to know the determinant of B)
They are similar and different at the same time.
I don't know where you want to go from there, there are plenty of zeros in the matrices, so the characteristic polynomial is easy to calculate for both.
Anyway, if you note the common part:
$G(t)=t^{13}-mt^{12}-lt^{11}-kt^{10}-jt^9-it^8-ht^7-gt^6-ft^5-et^4-dt^3$
Then $\begin{cases}P_A(t)=G(t)-(a+c)t^5-bt^4\\ P_B(t)=G(t)-ct^2-bt-a\end{cases}$
Eventually you could rewrite:
$P_A(a,b,c,d,e,f,g,h,i,j,k,l,m,t)=P_B(0,0,0,d,b+e,a+c+f,g,h,i,j,k,l,m,t)$