How to show that there exist $x_1,x_2,x_3\in \mathbb{C}$ such that the matrix $x_1A+x_2B+x_3C$is invertible? Suppose $A,B,C\in M_2(\mathbb{C})$and they are linearly independent.Try to show that there exist $x_1,x_2,x_3\in \mathbb{C}$ such that the matrix $x_1A+x_2B+x_3C$is invertible.
I know the span $L(A,B,C)$ is a subspace  of  $M_2(\mathbb{C})$,and all invertilbe matrix is dense in $M_2(\mathbb{C})$  .It seems the conclusion is obvious,but I can't give a rigorous proof.
 A: Let's assume the contrary, that a $\Bbb C$-linear subspace of $M_2(\Bbb C)$
consists entirely of invertible matrices. If we write a generic matrix
$$M=\pmatrix{x&y\\z&t}$$
then this subspace is given by a homogeneous linear equation $f(x,y,z,t)=0$.
By assumption that each solution of $f(x,y,z,t)=0$ satisfies $g(x,y,z,t)=0$
where $g(x,y,z,t)=xt-yz$.
Therefore $V(f)\subseteq V(g)$, where in general $V(h)$ denotes the solution
set of the equation $h(x,y,z,t)=0$ in $\Bbb C^4$.
For $A\subseteq \Bbb C^4$ Write $I(A)$ for the set of polynomials vanishing
on $A$. Then $A\subseteq B$ implies $I(A)\supseteq I(B)$ and so
$I(V(f))\subseteq I(V(g))$.
By Hilbert's Nullstellensatz (this depends on on $\Bbb C$ being algebraically closed),
$I(V(f))=\sqrt{(f)}$. This is the radical of $f$, the set of all polynomials
$h$ for which $h^k$ lies in the principal ideal $(f)$ for some $k\ge1$.
In this case as $f$ is an irreducible polynomial $I(V(f))=(f)$.
For similar reasons, $I(V(g))=(g)$ and we find $(f)\supseteq (g)$.
That means that $g$ is a factor of $f$, but it isn't, and we get our contradiction.
A: Assume that we have a hyperplane inside the subset of singular matrices. It follows that there exist constants $a$, $b$, $c$, $d$, so that 
$$ x (a x + b y + c z + d)- y z =0$$ for all values of $x$, $y$, $z$.
In the above, take $x=1$. We get 
$$b y + c z + (a+d) = y z$$
or 
$$(y -c)(z-b)= a+d+b c$$
for all $y$, $z$, clearly not possible. 
