Where invertibility breaks

A square matrix fails to be invertible exactly when its determinant is zero.

1. The determinant is $\det (M)=k(k+1)-4\times 3=k^2+k-12$.
2. Setting it to zero gives $k^2+k-12=(k-3)(k+4)=0$.
3. The roots are $k=3$ and $k=-4$, and only the first is positive.
4. So the matrix is not invertible at $k=3$.

Answer: $3$