determinant of sum of matrices

Question

having A,B,C square matrices of size $n\gt2$, with C = A+B.

If we know that:
$det(A) = 0$, and
B is a diagonal matrix with $det(B)\ne 0 $

Then how can one find $det(C)$?

It should be enough for me to prove $det(C)\ne 0$

Answer 1

No, you can't find $\det C$. Suppose that$$A=\begin{pmatrix}1&1\\1&1\end{pmatrix}\text{ and that }B=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$Then $\det A=0$, $\det B\neq0$, and $\det(A+B)=3$. But $\det(0+B)=1\neq3$.

Answer 2

Counterexample:

Take $A=\begin{pmatrix} 0& 0\\ 0&{-1} \end{pmatrix}$ and $B=\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}.$

$\det A=0, $ $B$ diagonal, $\det B=1\neq 0.$ But $\det C=0 .$