rock_mit182
  • rock_mit182
A, B are matrices nxn; If A is NON-invertible matrix, A times B IS non-invertible as well. Please help me !
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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beginnersmind
  • beginnersmind
Can you use determinants?
rock_mit182
  • rock_mit182
oh i see...
rock_mit182
  • rock_mit182
if A INVERSE does not exit then |A| =0

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rock_mit182
  • rock_mit182
So 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
  • rock_mit182
WHICH alwas have to be zero given the condition of A
rock_mit182
  • rock_mit182
right ?
beginnersmind
  • beginnersmind
right
beginnersmind
  • beginnersmind
last step is to point out det(AB) = 0 -> AB is non-invertible
rock_mit182
  • rock_mit182
dude plz help me out with another question
beginnersmind
  • beginnersmind
Sure
rock_mit182
  • rock_mit182
show that A^2+2A = - I has inverse WHERE: A is nxn matrix; I is the identity matrix
beginnersmind
  • beginnersmind
Hm, let me think.
rock_mit182
  • rock_mit182
of course
beginnersmind
  • beginnersmind
I guess you can do this one with determinants too :)
rock_mit182
  • rock_mit182
i guess on this one i only can use propeties of matrix..
beginnersmind
  • beginnersmind
Do we need to prove that A has an inverse or that A^2+2A does?
beginnersmind
  • beginnersmind
Anyway, 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 non-determinant 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
  • beginnersmind
u^2 + 2u + 1 = 0, or detA = 1 or detA = -1, CORRECTION u = -1, or detA = -1
rock_mit182
  • rock_mit182
thanks dude

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