Let $A,B,C$ be $3×3$ matrices such that$$ \det(A)=2,\ \det(B)=-\frac{1}{3},\\ C=\begin{pmatrix} 1&-1&4\\ 0&0&15\\ 0&0&-3\end{pmatrix}.$$ Calculate $\det(AC+BC)$.
Any ideas how can I proceed?
Sorry, I got wrong $\det(AC+AB)$. It was $\det(AC+BC)$ instead. Thanks you all for the reply!
One has $$\det(AC+BC) =\det([A+B] C) =\det(A+B) \det(C)=\det(A+B) \cdot 0=0$$ since the first two columns of $C$ are clearly dependent so that $\det(C) =0$.