3. Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots b_{n}$ be $2 n$ distinct real numbers, $M$ be an $n \times n$ matrix, whose element in the $i$-th row and $j$-th column is $a_{1}+b_{3}$. If the product of the elements in each column is 1, then the product of the elements in each row is ( ).
A. 1
B. -1
C. $(-1)^{n-1}$
D. None of the above
3. $C$

Given $\left.\prod_{i=1}^{n}\left(a_{i}+b_{j}\right)=1 \leqslant 1 \leqslant j \leqslant n\right)$, the equation $\prod_{i=1}^{n}\left(a_{i}+x\right)=1$ has $n$ distinct roots $b_{j}(1 \leqslant j \leqslant n)$. Therefore, $\prod_{i=1}^{n}\left(x-b_{j}\right)=\prod_{i}^{\prime}\left(x+a_{i}\right)-1$. Substituting $x=-u$, we get $(-1)^{n} \cdot \prod_{j=1}^{n}\left(a_{i}+b_{j}\right)=-1 \Rightarrow \prod_{j=1}^{n}\left(a_{i}+b_{j}\right)=(-1)^{n-1}$, which means the product is $(-1)^{n-1}$.