Singularity and non singularity of matrices If $A$ and $B$ are real matrices of order $n$ s.t. $~\det A>0~$ and $~\det B<0~$ . If $~C(t)=tA+(1-t)B, ~~t\in[0,1]~$ then which is true
$1)~~~~C(t)$ is non singular for all $~t~$ in $~(0,1)~$.
$2)~~~~C(t)$ is singular for finite number of $~t~$ in $~(0,1)~$.
$3)~~~~ C(t)$ is singular at infinite number of $~t~$ in $~(0,1)~$.
 A: $det(C(t))$ is a polynomial in $t$. Hence $det(C(t))=0$ for at most finitely many $t$. This answers 2). Also, this polynomial is a continuous function which is negative when $t=0$ and positive when  $t=1$ so it must vanish for some $t$ inbetween . This shows that 1) is false. 
A: Define $f(t)=\det C(t)$ for $t \in [0,1]$. We have that $f$ is continuous , $f(0) <0$ and $f(1)>0.$
Now invoke the intermediate value theorem.
A: $\det C(t) = \det(tA + (1 - t)B); \tag 1$
$\det C(0) = \det (B) < 0; \tag 2$
$\det C(1) = \det(A) > 0; \tag 3$
now since $\det C(t)$ is a continuous function of $t$ we may invoke the intermediate value theorem to assert that 
$\exists t' \in (0, 1), \; \det C(t') = 0, \tag 4$
which shows that option (1) is false, so we affirm that $C(t)$ is singular for a least one $t \in (0, 1)$.  Now $\det C(t)$ as given in (1) is a polynomial of degree $n$ in $t$; therefore $\det C(t)$ has at most $n$ real zeroes; as we have seen, at least one of them lies in $(0, 1)$; this shows us that option (2) binds, and that option (3) is false.
Therefore the correct selection is option (2).
I think it is worthy of note that this problem is similar in many respects to this question, which was put on hold for the usual reasons.
