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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 (n-1) x (n-1) 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

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# Definition of the determinant i = 1

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