# Thread: simplyfy the following as much as possible

1. ## simplyfy 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

2. Originally Posted by b0mb3rz
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
They are all correct.

3. Hello, b0mb3rz!

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

For matrices A and B, simplify: .$\displaystyle A\cdot\left(B + A^{-1}\right)\cdot B^{-1}A$

I got: .$\displaystyle A^2 + {\color{red}AB^{-1}}$ . . . . no

$\displaystyle A\cdot (B + A^{-1})\cdot B^{-1}A \;=\;\left[A\!\cdot\! B + A\!\cdot\! A^{-1}\right]\cdot B^{-1}A$

. . . . . . . . . . . . . $\displaystyle = \;\left[A\!\cdot\!B + I\right]\cdot B^{-1}A$

. . . . . . . . . . . . . $\displaystyle = \;(A\!\cdot\!B)\cdot\left(B^{-1}A\right) + I\!\cdot\!\left(B^{-1}A\right)$

. . . . . . . . . . . . . $\displaystyle = \;A\!\cdot\!\left(B\!\cdot\! B^{-1}\right)\cdot A + B^{-1}A$

. . . . . . . . . . . . . $\displaystyle = \;A\!\cdot\! I\!\cdot\!A + B^{-1}A$

. . . . . . . . . . . . . $\displaystyle = \;A^2 + {\color{blue}B^{-1}A}$

$\displaystyle (A + B)\left(A^{-1} + B^{-1}\right)$

I got: .$\displaystyle 2I + AB^{-1} + BA^{-1}$ . . . . Right!

We have: .$\displaystyle (A + B)\cdot(A^{-1} + B^{-1})$

Distribute: .$\displaystyle A\cdot(A^{-1} + B^{-1}) + B\cdot(A^{-1} + B^{-1})$

Distribute: .$\displaystyle A\!\cdot\! A^{-1} + A\!\cdot\! B^{-1} + B\!\cdot\! A^{-1} + B\!\cdot\! B^{-1}$

. . . . . . .$\displaystyle = \;I + A\!\cdot\!B^{-1} + B\!\cdot\!A^{-1} + I$

. . . . . . .$\displaystyle = \;2I + AB^{-1} + BA^{-1}$

$\displaystyle \bigg[A^3\left(A^2\right)^{-1}\bigg]^{-1}$

I got: .$\displaystyle A^{-1}$ . . . . Yes!
Do we dare to apply normal exponent rules to matrices?

Does $\displaystyle A^3\cdot(A^2)^{-1} \:=\:A^3\cdot A^{-2} \;=\;A$ ?
. . The answer is Yes.

$\displaystyle A^3\cdot(A^2)^{-1} \;=\;A\cdot A^2\cdot(A^2)^{-1} \;=\;A\cdot\underbrace{\bigg[A^2\cdot(A^2)^{-1}\bigg]}_{\text{Inverses}} \;=\;A\cdot I \;=\;A$