How the following fact applies to determinants (I came across it while solving problems):
Consider A is a nxn matrix, the elements of which are real (or complex) polynomials in x. If r rows of the determinant become identical when x = a, then the determinant has a factor of order < r.
The only additional information I found in the solutions key is that
(x-a)$^{r-1}$ is a factor of det. A.
Can I know how logically connected is collapsing of rows of matrix (into one row) with order of its factors.
Am I missing some stupid fact here?
I edited this post. Now the solution is much simpler. It only uses some properties of the determinant.
Let $L_1(x),\ldots,L_n(x)$ be the rows of $A(x)$ and let us write $A(x)=(L_1(x),\ldots,L_n(x))$. Assume $L_1(a)=\ldots=L_{k}(a)$
Let $B(x)$ be the matrix obtained from $A(x)$ subtracting the first row of $A(x)$ from the rows $2,3,\ldots,k$. Using our notation, we have $B(x)=(L_1(x),L_2(x)-L_1(x),\ldots,L_k(x)-L_1(x), L_{k+1}(x),\ldots,L_n(x))$.
Now, by property 13 of the determinant here, $det(B(x))=det(A(x))$.
Next, the rows $2,3,\ldots,k$ of $B(a)$ are zero. Thus the coefficients $B_{ij}(x)$ of $B(x)$, for $2\leq i\leq k$ and $1\leq j\leq n$, have one root equal to $a$. Thus, we can write $B_{ij}(x)=(x-a)q_{ij}(x)$, where $q_{ij}(x)$ is a polynomial.
Now, since the determinant is a linear function on the rows and the rows $2,3,\ldots,k$ of $B(x)$ are all multiplied by $(x-a)$, we obtain $det(B(x))=(x-a)^{k-1}det(C(x))$, where $C(x)=(L_1(x),\frac{L_2(x)-L_1(x)}{(x-a)},\ldots,\frac{L_k(x)-L_1(x)}{x-a}, L_{k+1}(x),\ldots,L_n(x))$.
Since $C(x)$ is a matrix with polynomial coordinates then $det(C(x))$ is a polynomial and we obtained the result.