This preview shows page 1. Sign up to view the full content.
Unformatted text preview: det A = det(( E 1 E m ) E m +1 )det A and the induction goes through. Hence det BA = det B det A if B is nonsingular. If B is singular, problem 26, Chapter 2.7 tells us that BA is also singlular. However singular matrices have zero determinant, so det B = 0 det BA = 0 , so the equation det BA = det B det A holds trivially in this case. 16. f f f f f f f f a + b + c a + b a a a + b a + b + c a a a a a + b + c a + b a a a + b a + b + c f f f f f f f f 51...
View Full Document
This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.
- Fall '10