# simplyfy the following as much as possible

• Apr 2nd 2009, 05:52 AM
b0mb3rz
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
• Apr 2nd 2009, 09:41 AM
HallsofIvy
Quote:

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.
• Apr 2nd 2009, 10:03 AM
Soroban
Hello, b0mb3rz!

Quote:

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

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

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

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

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

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

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

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

Quote:

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

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

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

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

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

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

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

Quote:

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

I got: . $A^{-1}$ . . . . Yes!

Do we dare to apply normal exponent rules to matrices?

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

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