This exercise below has stumped me and my intuition has always told me that having $n$ unknowns with $n$ linear equations will generally have exactly one solution (when the coefficient matrix is invertible).
Give an example of a system of $n$ linear equations in $n$ unknowns with infinitely many solutions.
The only other case is when the system has no solutions at all (like two parallel lines). However, I have never heard of a case where you have $n$ equations with $n$ unknowns and end up with an infinite number of solutions! I can't even visualize what it would look like. The only time you would have an infinite number of solutions is when you have more unknowns than equations, but this is impossible here.
Best Answer
Given $n\geq 2$ variables consider $n-1$ equations in those variables. These will generally have infinitely many solutions because there are more variables then equations.
Now add an $n$-th equation obtained by adding together the previous $n-1$ equations. This $n$-th equation is redundant because every solution of the first $n-1$ will be also a solution of the last one by construction.
Just to make an example in $3$ variables: $$ \left\{ \begin{array}{lcl} 2x-y+z &= &3 \\ x-3y-2z &= &-1 \\ 3x-4y-z &= &2 \end{array} \right. $$