A, B are matrices nxn; If A is NON-invertible matrix, A times B IS non-invertible as well.
Please help me !
Stacey Warren - Expert brainly.com
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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|
WHICH alwas have to be zero given the condition of A
last step is to point out det(AB) = 0 -> AB is non-invertible
dude plz help me out with another question
show that A^2+2A = - I has inverse
A is nxn matrix;
I is the identity matrix
Hm, let me think.
I guess you can do this one with determinants too :)
i guess on this one i only can use propeties of matrix..
Do we need to prove that A has an inverse or that A^2+2A does?
Anyway, I need to go, so I'll leave with a short proof with determinants
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.
u^2 + 2u + 1 = 0, or detA = 1 or detA = -1,
u = -1, or detA = -1