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Definition of the determinant i = 1by jeff1evesque
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#1
May109, 05:16 PM

P: 312

Lemma:Let B be an element in [tex]M_n_x_n(F)[/tex], where n >= 2. If row i of B equals [tex]e_k[/tex] for some k (1<= k <= n ), then det(B) = (1)[tex]^i^+^kdet(B_i_k).[/tex]
Proof: The proof is by mathematical induction on n. The lemma is easily proved for n = 2. Assume that for some integer n >= 3, the lemma is true for (n1) x (n1) matrices, and let B be an nxn matrix in which row i of B equals [tex]e_k[/tex]for some k (1<=k<=n). The result follows immediately from the definition of the determinant i = 1. Questions: 1. Can someone help me understand/prove the italized text above.. "...follows from immediately from the definition of the determinant i = 1" 2. And how does det(B) = (1)[tex]^i^+^kdet(B_i_k).[/tex]? Why isn't there a coefficient in that equation [tex]B_i_,_k[/tex]? This questions is not necessary I guess since If i follow and understand the proof in the book the logic will work out but I always thought that in calculating a determinant there would be a scalar coefficient thus: det(B) = [tex]B_i_,_k[/tex](1)[tex]^i^+^kdet(B_i_k).[/tex] thanks, JL 


#2
May109, 06:40 PM

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Thanks
P: 26,148

Hi jeff1evesque!
(try using the X_{2} tag just above the Reply box ) but in this case, B_{ik} is 1. 


#3
May109, 10:19 PM

P: 312

Thanks so much, JL 


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