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Stuck on problem with determinants

  1. Aug 9, 2006 #1
    hi all,

    have a rather sticky problem on determinants to deal with...hope someone can offer some help.

    [In what follows, the numbers in brackets denote suffixes, so that, for example, A(s)(j) refers to the element in the sth row of A and jth column of A.]

    Let A be an n x n matrix. Let s be a fixed integer with 1=<s=<n. Suppose that A(s)(j) = b(s)(j) + c(s)(j) for j = 1, 2, ... n. Let B and C be the n x n matrixes defined by B(i)(j) = C(i)(j) = A(i)(j) for j = 1, 2, ..., n, if i is not equal to s; B(s)(j) = b(s)(j), j = 1, 2, ..., n; and C(s)(j) = c(s)(j), j = 1, 2, ..., n. Then prove that det (A) = det (B) + det (C).

    First, we let c(s)(1), c(s)(2), ..., c(s)(n) be fixed and let d(B) = det (A) - det (C), where the rows of B, A, and C are the same except for row s; C(s)(j) = c(s)(j), j = 1, 2, ... n; A(s)(j) = B(s)(j) + C(s)(j), j = 1,2 , ...n.
    The main difficulty I face is in proving that d(B) satisfies the properties of a determinant function without assuming the expansion formula for a determinant. The properties in question are:

    1) If A' is obtained from A by an interchange of two rows of A, then d(A') = -d(A).

    2) If A' is obtained from A by multiplying one row of A by a real number p, then d(A') = p d(A).

    3) If A' is obtained from A by multiplying one row of A by a number and adding to another row, then d(A') = d(A).

    4) d(I) = 1.

    That d(B) satisfies these properties is fairly clear if row s is not involved. [for example, to prove d(b) satisfies 1): if we exchange two rows of B, neither of which is row s, this involves an interchanging of two rows of A and two rows of C, giving d(B') = -det(A) + det(C) = -d(B).]

    but what if we consider the case when row s is interchanged with another row...?? I can't think of any way other than to expand the determinants of A' and C'! hope someone can help me out here..thanks.
    Last edited: Aug 10, 2006
  2. jcsd
  3. Aug 10, 2006 #2
    Suppose you row-reduce all the other rows (not s) in the three matrices (A, B, and C) to echelon form using the same sequence of row operations (just sort of ignore the s row as this happens). Notice that the these rows are still equal among the matrices (the i-th row of A is the same as the i-th row of B, etc.). Then do necessary row exchanges to slide the s-th row into k-th row, where k is the column (or one of the columns) that lacks a pivot position. So each matrix now has been left multiplied by some matrix E (made from elementary ones). Then do some "multiplying one row of A by a number and adding to another row" till that special s row has 0's up till the k-th entry. Then consider the diagonal determinants.

    And now you can fill in the blanks. Maybe there's a better way to do this problem, but I don't know enough algebra. :smile: Or maybe I do, and I'm just not seeing it....
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