Question
then 060 5 e det b 5 C 5 f a a det b P 2 d 12 = e = 0 f and
then 060 5 e det b 5 C 5 f a a det b P 2 d 12 = e = 0 f and 1 d a det b 1 e = 1 f - -4, and det a 1 d 62 e = 1, 3 f
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Linear Algebra And Its Applications
Authors: David Lay, Steven Lay, Judi McDonald
6th Global Edition
978-1292351216, 1292351217
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