Let A and B be two n×n matrices with real entries.
Define the function f : R → R by
f(x) = det(A + Bx)
(i) Show that f^{(3)}(x) = 3! det B.
(ii) Show that in general f^{(n)}(x) = n! det B.
f^{(n)}(x) is the nth derivative of f(x).
Let C = A + Bx. Then Det C is the sum of n! terms, each such term being the product of n components of the matrix, C.
Upon taking the nth derivative of Det C, each such term will vanish, except for the terms of degree n or higher. But each component of C is either constant or linear in x. Hence, Det C is an nth degree polynomial in x, where the term in the nth power of x is made from the product of all linear terms in x; in other words, the product along the diagonal: Product( ann + x bnn).
This product itself is an nth degree polynomial, and we only need consider the term in x to the n; namely Product (x bnn). Taking the nth derivative gives n! Det B.

Posted by FrankM
on 20190527 13:35:53 