$\displaystyle det(2A^8 B^{-1})$

I tried to break it up but I'm not sure where the 2 goes.

$\displaystyle det(2A^8) det(B^{-1})$ ?

Any help is appreciated!

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- Feb 19th 2010, 05:39 PMDarKFind determinant
$\displaystyle det(2A^8 B^{-1})$

I tried to break it up but I'm not sure where the 2 goes.

$\displaystyle det(2A^8) det(B^{-1})$ ?

Any help is appreciated! - Feb 19th 2010, 07:17 PMmath2009
$\displaystyle

\det(2A^{8}B^{-1})=

2^{3}\det(A^{8})\det(B^{-1})=

8\frac{\det(A)^{8}}{\det(B)}=

8\frac{ (-1)^{8}}{-2} = -4

$