Factor of determinant with identical row

Question

How the following fact applies to determinants (I came across it while solving problems):

Consider A is a n× n 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?

Comments

@Suraj

Answer 1

Elements of A are polynomial in X, so the determinant(A)(let's say f(x)) would also be polynomial in x.

Now consider the case when r=2 and x=a, so it'll produce the two identical rows in A and determinant of A would be 0.

(Let's assume that ith and jth rows are identical with x=a, so if we replace the ith row with ith row -jth row, determinant would be same. But now new ith row would be zero with x=a, it states that (x-a) is one of the factor of ith row)

   f(a) = 0  => (x -a) would be factor of f(x).

Similarly, when r=3, (x-a)² would be factor of f(x).

and to generalize this you can say that, (x-a)^r-1 would be factor of f(x).

PS:  It's a bit tricky when r>2, but if you are not convinced I can prove that also.