Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: ion, n det(A ) = a 1p C 1p p=1 where the 1p cofactor of A is C1p = (−1)(1+p) det A1p and A1p is the k × k matrix obtained by deleting the 1st row and pth column of A .5 Since the first row of A is c times the first row of A, we have a1p = c a1p . In addition, since the remaining rows of A are identical to those of A, A1p = A1p . (To obtain these matrices, the first row of A is removed.) Hence det A1p = det (A1p ), so that C1p = C1p . As a result, we get n n a1p C1p = det(A ) = p=1 n c a1p C1p = c p=1 a1p C1p = c det(A), p=1 as required. Hence, P (k + 1) is true in this case, which means the result is true in this case for all natural numbers n ≥ 1. (You’ll note that we did not use the induction hypothesis at all in this case. It is possible to restructure the proof so that induction is only used where 4 5 See Exercise 6 in Section 3.4. See Section 8.5 for a review of this notation. 9.3 Mathematical Induction 577 it is needed. While mathematically more elegant, it is less intuitive, and we stand by our approach because of its pedagogical value.) Case 2: The row R bei...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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