11.16 For what value of $m$ does the system of equations

$$
\left\{\begin{array}{l}
2 x+(m-1) y=3 \\
(m+1) x+4 y=-3
\end{array}\right.
$$

have an infinite number of solutions? No solutions?
11.16 The system has an infinite number of solutions or no solutions if the coefficients of $x$ and $y$ are proportional. Therefore, $\frac{2}{m+1}=\frac{m-1}{4}$ (here $m+1 \neq 0$; it is easy to verify that when $m=-1$, the system has a unique solution). Further, we have $(m+1)(m-1)=8 ; m^{2}-1=8$, from which $m= \pm 3$. When $m=3$, we get $\frac{2}{4}=\frac{2}{4} \neq \frac{3}{-3}$, i.e., the system has no solutions. When $m=-3$, we get $\frac{2}{-2}=\frac{-4}{4}=\frac{3}{-3}$, i.e., the system has an infinite number of solutions.

Answer: an infinite number of solutions when $m=-3$; no solutions when $m=3$.