Quiz 5 Solutions April 24, 2005 Thursday (1) This was a homework problem (HW 9). Let d n be the determinant of the n × n matrix shown. Then by Laplace expanding along the ﬁrst row, we see that d n = (-1) n +1 d n-1 for n > 1 . Using this recursion, we ﬁnd that d n = (-1) ∑ n j =2 j +1 The summation in the exponent is equal to n ( n + 3) / 2. This is an even number when n ( n + 3) / 2 is odd and an odd number otherwise. Elementary arithmetic shows that n ( n + 3) is divisible by four whenever n or n + 3 is, and not divisible by four when n and n + 3 are not. Therefore d n is equal to one whenever n is congruent to 1 or 4 (mod 4) and-1 otherwise. (2) (a) Recalling that det AB = (det A )(det B ) for n × n matrices A and B , we see that A 2 = A implies that (det A ) 2 = (det A ) Thus (det
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