Let $det(A)>0$ and $det(B)<0$, where $A,B$ are $n\times n $ matrices. Let $C(t)=tA+(1-t)B\space \forall t\in [0,1]$. Then is $C(t)$ invertible only for finitely many $t \in [0,1]$?
Its easy to see that $C(t)$ is invertible at $0,1$. Also, determinant is a polynomial of degree $n$. But, since $ t $ takes real values, I dont see how only finite values of $t$ would make $C(t)$ invertible. Any hints?
As stated in the comments by i707107, the answer relies upon the fact that the determinant polynomial $p(t)=|C(t)|$ is a continuous function with $p(0)<0<p(1)$ because of the given conditions. Hence, by the intermediate value theorem, there exists a $t_0\in (0,1)$ such that $p(t_0)=0$. However, being a polynomial, it is zero at only atmost $n$ points in the interval $(0,1)$.