the A and B's represent matrices

A(B + A^-1)((B^-1) A) i got A^2 +AB^-1

(A + B)(A^-1 + B^-1) i got 2I + AB^-1 + BA^-1

and [A^3(A^2)^-1]^-1 i got A^-1

last one is A to power of 3 times A to power of 2 to the power of -1, all to the -1.

thanks

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- Apr 2nd 2009, 05:52 AMb0mb3rzsimplyfy the following as much as possible
the A and B's represent matrices

A(B + A^-1)((B^-1) A) i got A^2 +AB^-1

(A + B)(A^-1 + B^-1) i got 2I + AB^-1 + BA^-1

and [A^3(A^2)^-1]^-1 i got A^-1

last one is A to power of 3 times A to power of 2 to the power of -1, all to the -1.

thanks - Apr 2nd 2009, 09:41 AMHallsofIvy
- Apr 2nd 2009, 10:03 AMSoroban
Hello, b0mb3rz!

I don't agree with your first answer . . .

Quote:

For matrices A and B, simplify: .

I got: . . . . . no

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

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Quote:

I got: . . . . . Right!

We have: .

Distribute: .

Distribute: .

. . . . . . .

. . . . . . .

Quote:

I got: . . . . . Yes!

*dare*to apply normal exponent rules to matrices?

Does ?

. . The answer is Yes.