# Prove a cubed inverse matrix is the same as an inverse cubed matrix

• Mar 1st 2010, 03:02 AM
Orbent
Prove a cubed inverse matrix is the same as an inverse cubed matrix
I have been asked to prove that the for a generic matrix A that, $\displaystyle (A^3)^{-1}$ is the same as $\displaystyle (A^{-1})^3$. I have no idea how to start this, any ideas would be much appreciated.

Pretty much the wording is prove that A cubed, then inversed is the same as A inversed then cubed.
• Mar 1st 2010, 03:30 AM
HallsofIvy
$\displaystyle \left(A^3\right)^{-1}$ is, by definition, the matrix, B, such that $\displaystyle B(A^3)= (A^3)B= I$, the identity matrix. Show that this is true for $\displaystyle (A^{-1})^3A^3$ and $\displaystyle A^3(A^{-1})^3$.
• Mar 1st 2010, 04:20 AM
Orbent
Done.
Thank you very much, i know what to do. I'll do it just in case anyone else looks at this thread.
First $\displaystyle (A^{-1})^3A^3$
$\displaystyle (A^{-1})^3$ becomes $\displaystyle A^{-1}A^{-1}A^{-1}$
$\displaystyle (A^3)$ becomes $\displaystyle AAA$
and $\displaystyle AA^{-1} = I$ and $\displaystyle AI=A$ and $\displaystyle A^{-1}I=A^{-1}$
Combining these facts makes:
$\displaystyle AAAA^{-1}A^{-1}A^{-1}$ this becomes,
$\displaystyle AAIA^{-1}A^{-1}$ which becomes,
$\displaystyle AAA^{-1}A^{-1}$, after we do that a few times we get,
$\displaystyle I$

This also works for,
$\displaystyle A^{-1}A^{-1}A^{-1}AAA$
Please let me know if i did that right :)
• Mar 1st 2010, 06:22 AM
HallsofIvy
Yes, that's exactly what you needed to do.

Now you use the fact that a matrix has at most one inverse to argue that $\displaystyle (A^{-1})^3= (A^3)^{-1}$.