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Unformatted text preview: i.e a ik = a jk for all k = 1 , 2 ,...,n + 1. Now we compute det( A ) by expand by the rth row of A with r 6 = i or j . ⇒ det( A ) = n +1 ∑ k =1 (1) r + k a rk det( M rk ) Note that M rk are n × n matrix and M rk has two identical rows. By induction hypothesis, det M rk =0 for k = 1 , 2 ,...,n + 1. ⇒ det(A) = 0 ....
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This note was uploaded on 04/18/2010 for the course FINANCE 1231854365 taught by Professor Wuyiling during the Spring '10 term at Nashville State Community College.
 Spring '10
 wuyiling

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