We claim that the statement is false.
As a counterexample, consider the matrices
\[A=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 0\\
0& 1
\end{bmatrix}.\]
Then
\[A+B=\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}\]
and we have
\[\det(A+B)=\begin{vmatrix}
1 & 0\\
0& 1
\end{vmatrix}=1.\]

On the other hand, the determinants of $A$ and $B$ are
\[\det(A)=0 \text{ and } \det(B)=0,\]
and hence
\[\det(A)+\det(B)=0\neq 1=\det(A+B).\]

Therefore, the statement is false and in general we have
\[\det(A+B)\neq \det(A)+\det(B).\]

Remark.

When we computed the $2\times 2$ matrices, we used the formula
\[\begin{vmatrix}
a & b\\
c& d
\end{vmatrix}=ad-bc.\]

This problem showed that the determinant does not preserve the addition.
However, the determinant is multiplicative.
In general, the following is true:
\[\det(AB)=\det(A)\det(B).\]

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