Question

A, B are matrices nxn; If A is NON-invertible matrix, A times B IS non-invertible as well.
Please help me !

Answer 1

Can you use determinants?

Answer 2

oh i see...

Answer 3

if A INVERSE does not exit then |A| =0

Answer 4

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|

Answer 5

WHICH alwas have to be zero given the condition of A

Answer 6

right ?

Answer 7

right

Answer 8

last step is to point out det(AB) = 0 -> AB is non-invertible

Answer 9

dude plz help me out with another question

Answer 10

Sure

Answer 11

show that A^2+2A = - I has inverse
WHERE:
A is nxn matrix;
I is the identity matrix

Answer 12

Hm, let me think.

Answer 13

of course

Answer 14

I guess you can do this one with determinants too :)

Answer 15

i guess on this one i only can use propeties of matrix..

Answer 16

Do we need to prove that A has an inverse or that A^2+2A does?

Answer 17

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.

Answer 18

u^2 + 2u + 1 = 0, or detA = 1 or detA = -1,

CORRECTION

u = -1, or detA = -1

Answer 19

thanks dude