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rock_mit182
 one year ago
A, B are matrices nxn; If A is NONinvertible matrix, A times B IS noninvertible as well.
Please help me !
rock_mit182
 one year ago
A, B are matrices nxn; If A is NONinvertible matrix, A times B IS noninvertible as well. Please help me !

This Question is Closed

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1Can you use determinants?

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0if A INVERSE does not exit then A =0

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0So there are two cases when: B inverse exist and B inverse does't exist, in other words: B =! 0 OR B = 0 in any case : AB = A*B

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0WHICH alwas have to be zero given the condition of A

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1last step is to point out det(AB) = 0 > AB is noninvertible

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0dude plz help me out with another question

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0show that A^2+2A =  I has inverse WHERE: A is nxn matrix; I is the identity matrix

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1Hm, let me think.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1I guess you can do this one with determinants too :)

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0i guess on this one i only can use propeties of matrix..

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1Do we need to prove that A has an inverse or that A^2+2A does?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1Anyway, I need to go, so I'll leave with a short proof with determinants From A^2 + 2A = I, we have det(A)*det(A) + 2det(A) + det(I) = 0 substituting u = det(A) u^2 + 2u + 1 = 0, or detA = 1 or detA = 1, so in either case A is invertible. I don't have a nondeterminant proof, but here's a hint: If you take the matrix A^2+2A = I its rank is n. A^2 + 2A = (A +2I)(A), and now you can use some linear algebra that both A and A+2I are rank n.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1u^2 + 2u + 1 = 0, or detA = 1 or detA = 1, CORRECTION u = 1, or detA = 1
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