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

**Orbent** · March 1st 2010, 03:02 AM

I have been asked to prove that the for a generic matrix A that, $(A^3)^{-1}$ is the same as $(A^{-1})^3$ . 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.

**HallsofIvy** · March 1st 2010, 03:30 AM

$\left(A^3\right)^{-1}$ is, by definition, the matrix, B, such that $B(A^3)= (A^3)B= I$ , the identity matrix. Show that this is true for $(A^{-1})^3A^3$ and $A^3(A^{-1})^3$ .

**Orbent** · March 1st 2010, 04:20 AM

Thank you very much, i know what to do. I'll do it just in case anyone else looks at this thread.
First $(A^{-1})^3A^3$
$(A^{-1})^3$ becomes $A^{-1}A^{-1}A^{-1}$
$(A^3)$ becomes $AAA$
and $AA^{-1} = I$ and $AI=A$ and $A^{-1}I=A^{-1}$
Combining these facts makes:
$AAAA^{-1}A^{-1}A^{-1}$ this becomes,
$AAIA^{-1}A^{-1}$ which becomes,
$AAA^{-1}A^{-1}$ , after we do that a few times we get,
$I$

This also works for,
$A^{-1}A^{-1}A^{-1}AAA$
Please let me know if i did that right

**HallsofIvy** · March 1st 2010, 06:22 AM

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 $(A^{-1})^3= (A^3)^{-1}$ .

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

Done.

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