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Let $A=\left(\begin{matrix}1&1&-1\\-1&1&1\\1&-1&1\end{matrix} \right)$,and ${A}^{T}B{\left( \cfrac{1}{2}{A}^{T}\right)}^{T}-8{A}^{-1}B=I$, How to compute $\left|B \right|$?。

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    What exactly is $E$?2012-11-20
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    @CameronBuie I'm assuming identity. $I$, $E$, whatever.2012-11-20
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    @Herp: I've seen $E$ as identity, and I've seen $E$ as a square matrix of all $1$s. I didn't want to assume.2012-11-20
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    A rather strange problem! Why write $A^T B (\frac{1}{2} A^T)^T$ instead of $\frac{1}{2} A^T B A$? Why ask for the determinant rather than the matrix itself?2012-11-20
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    @RobertIsrael I think to compute determinant may be easier than to find out the matrix itself?2012-11-20

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