Polynomial matrix is invertible if and only if its determinant is a nonzero constant. As you can see in the title, I want to prove that a polynomial matrix is invertible if and only if its determinant is a nonzero constant.
So far, I'm guessing that the proof would not be the same for the invertibility of a regular matrix. 
This is what I have constructed so far...
(if) Suppose that the determinant of an n by n matrix $A$ is a nonzero constant. Then by the invertibility matrix theorem, $A$ is an invertible matrix. 
(only if)
Suppose that an n by n polynomial matrix A is invertible. Then there exists an n by n polynomial matrix B such that 
$AB=I_n$. Then, 
$det(AB) = det(A) * det(B) = det(I_n) = 1$.
If $det(A)$ is zero, then we end up with $0 = 1$ , which is a contradiction. 
If $det(A)$ is a nonzero polynomial, then ...
...
This is where I'm stuck. 
Thank you for any help.
 A: Here's a proof.
If: Suppose that $A(x)$ has a non-zero, constant derivative.  We note that
$$
A(x) \operatorname{adj}(A(x)) = \det(A(x)) I
$$
where adj denotes the adjugate matrix.  Because $\det(A(x))$ is a non-zero constant, we can divide by it to get
$$
A(x) \cdot \frac{\operatorname{adj}(A(x))}{\det(A(x))} = I.
$$
We similarly find that $\frac{\operatorname{adj}(A(x))}{\det(A(x))} \cdot A(x) = I$, so that $\frac{\operatorname{adj}(A(x))}{\det(A(x))}$ is the inverse of $A(x)$.  So, $A(x)$ is indeed invertible.
Only if: Suppose that $A(x)$ is invertible.  That is, there exists a matrix polynomial $B(x)$ such that $A(x)B(x) = I$.  It follows that 
$$
\det(A(x)) \det(B(x)) = \det(I) = 1.
$$
We see that both $\det(A(x))$ and $\det(B(x))$ must be non-zero. For non-zero polynomials $p,q$, we have $\deg(pq) = \deg(p) + \deg(q)$.  Since $1$ is a degree-$0$ polynomial, both $\det(A(x)),\det(B(x))$ must be of degree $0$, which is to say that they are constant polynomials.  So, $\det(A(x))$ is indeed constant.
