Show that $\exists t_0$ such that $C(t_0)$ is not invertible.

Question

Let $A,B$ be $n\times n$ matrix (real) such that $\det A>0,\det B<0$;

Consider $t\in [0,1]$ and $C(t)=tA+(1-t)B$

Show that $\exists t_0$ such that $C(t_0)$ is not invertible.

Answer 1

$\det C(t)$ is a polynomial of $t$ and thus continuous on $[0,1]$. Since $\det C(0)=\det B<0$, $\det C(1)=\det A>0$, by intermediate value theorem, there exists $t_0\in[0,1]$ such that $\det C(t_0)=0$.