how to prove that this determinant is a polynomial? Given two square matrices $A$ and $B$ of size $n\times n$. I am wondering how to prove that $\det(A+xB)$ is a polynomial function of $x$ ? does anyone has a (simple if possible) proof of this fact? What would be the degree of this polynomial?  Thanks.
 A: Depending on how you define determinant, it is more or less obvious that for an $n \times n$ matrix $M$, $\det(M)$ is a polynomial of total degree $n$ in the entries of $M$.  If the entries of $M$ are linear functions of a variable $x$, that implies the determinant is a polynomial of degree at most $n$ in $x$.
If $B$ has rank $r$, then Gaussian elimination  provides an invertible matrix $G$ such that $GB$ has only $r$ nonzero rows.  Now $\det(A+xB) = \det(G)^{-1} \det(GA + xGB)$, and in the expansion of 
$\det(GA + xGB)$ it is clear that there are no nonzero terms with more than $r$ $x$'s.  Therefore $\det(A+xB)$ has degree at most $r$.
A: Suppose the matrices are $\;n\times n\;$:
$$A=(a_{ij})\;,\;\;B=(b_{ij})\implies A+xB=(a_{ij}+xb_{ij})\implies$$
$$\det (A+xB):=\sum_{\sigma\in S_n}Syg(\sigma)\cdot \left(a_{1\sigma(1)}+xb_{1\sigma(1)}\right)\cdot\ldots\cdot \left(a_{n\sigma(n)}+xb_{n\sigma(n)}\right)\;---$$
indeed, a polynomial in $\;x\;$ of degree $\;n\;$ (unless the main diagonal of $\;B\;$ has zeros. In this last  case, we can only say the polynomial is of degree at most $\;n\;$)
A: Proceed by induction:
Base case, $n=2$:
$$
\begin{align*}A+xB$ &= \begin{pmatrix}a_{11}+xb_{11} & a_{12}+xb_{12} \\ a_{21}+xb_{21} & a_{22}+xb_{22} \end{pmatrix} \\
\det A+xB &= \left(a_{11}+xb_{11}\right)\left(a_{22}+xb_{22}\right) - \left(a_{12}+xb_{12}\right)\left(a_{21}+xb_{21}\right) \\
 &= a_{11}a_{22}-a_{12}a_{21}+\left(a_{22}b_{11}+a_{11}b_{22}-a_{21}b_{12}-a_{12}b_{21}\right)x+b_{11}b_{22}x^2.
\end{align*}$$
Induction step:
Assume the property holds true for all matrices of size $n\times n$. Then, for $A,B$ of order $n+1$:
$$\det A+xB = \sum_{k=1}^{n+1} (-1)^{k+1}\left(a_{1k}+xb_{1k}\right)\det M_{1k},$$
where $M_{1k}$ is the minor formed by excluding the first row and the $k$th column of $A+xB$.
Since $M_{1k}$ is $n\times n$, its determinant is a polynomial in $x$, and the result is clearly a polynomial in $x$ with a degree at most one greater than that of $M_{1k}$.
A: Suppose $A = (a_{i,j}) $ $B = (b_{i,j})$.
Then by definition of determinant $$\text{det}(A +xB) = \sum_{\sigma \in S_n}\text{sgn}(\sigma)\biggl(a_{1 \sigma(1)} + x b_{1 \sigma(1)}\biggr)\cdot \biggl(a_{2 \sigma(2)} + x b_{2 \sigma(2)}\biggr)\ldots \biggl(a_{n \sigma(n)} + x b_{n \sigma(n)}\biggr)$$ and this is a polynomial in $x$
