Linear combination of invertible matrices which is singular Let $A$ and $B$ be two n×n matrices such that $\det(A)>0$ and $\det(B)<0$. Then show that there is a real number $t$ which is between $0$ and $1$, such that $tA+(1-t)B$ is singular.
I have no idea how to solve this. It looks similar to convexity but how can we define convexity of set of matrices?

If
$A, B \in M_n(\Bbb R), \tag 1$
that is, $A$ and $B$ are $n \times n$ real matrices, then evidently the function
$t \mapsto tA + (1 - t)B \tag 2$
is continuous from $[0, 1]$ to $M_n(\Bbb R)$; and since
$\det:M_n(\Bbb R) \to \Bbb R \tag 3$
is also continuous, the composition
$f(t): t \mapsto tA + (1 - t)B \mapsto \det(tA + (1 - t)B) \tag 4$
a function from $[0, 1] \to \Bbb R$, is itself in fact continuous; furthermore we have
$f(0) = \det(B) < 0, \tag 5$
and
$f(1) = \det(A) > 0; \tag 6$
it now follows from the intermediate value theorem that 
$\exists t_0 \in (0, 1), \; f(t_0) = 0, \tag 7$
that is
$\det(t_0A + (1 - t_0)B) = f(t_0) = 0. \tag 8$
