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 $\left|B \right|$?。
Trying to find $\det (B)$
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matrices
determinant
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4What exactly is $E$? – 2012-11-20
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0@CameronBuie I'm assuming identity. $I$, $E$, whatever. – 2012-11-20
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0@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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0A 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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0@RobertIsrael I think to compute determinant may be easier than to find out the matrix itself? – 2012-11-20