- #1
xalvyn
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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.
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.
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