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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
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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