# Trying to find $\det (B)$

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 $E$? –  Cameron Buie Nov 20 '12 at 5:42
@CameronBuie I'm assuming identity. $I$, $E$, whatever. –  Herp Derpington Nov 20 '12 at 5:48
@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. –  Cameron Buie Nov 20 '12 at 5:50
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? –  Robert Israel Nov 20 '12 at 6:25
@RobertIsrael I think to compute determinant may be easier than to find out the matrix itself? –  Leitingok Nov 20 '12 at 7:11

Consider the permutation matrix $$P = \pmatrix{0 & 1 & 0\cr 0 & 0 & 1\cr 1 & 0 & 0\cr}$$ Note that $P^3 = I$ and $A = I + P - P^2$. Since $A P^2 = P^2 + I - P = -A + 2 I$, we see that $A^{-1} = (P^2 + I)/2$, while $A^T = I + P^2 - P$. It seems reasonable, then, that a solution $B$ might also be a linear combination of $I$, $P$ and $P^2$. In fact, if $B = a I + b P + c P^2$ we get \eqalign{\frac{1}{2} A^T B A &- 8 A^{-1} B = \frac{1}{2} (I + P^2 - P)(a I + b P + c P^2)(I + P - P^2) - 4 (P^2 + I) (aI + b P + cP^2)\cr &= \frac{-5 a - 9 b-c}{2} I + \frac{-a-5b-9c}{2} P + \frac{-9 a - b - 5 c}{2} P^2} For this to be $I$, we just have to solve the $3 \times 3$ system \eqalign{ 5a + 9 b + c &= -2\cr a + 5 b + 9 c &= 0\cr 9 a + b + 5 c &= 0\cr} obtaining $a = -2/45, b = -19/90, c = 11/90$. Now note that the eigenvalues of $P$ are the three cube roots of $1$, say $1, \omega, \omega^2$; the eigenvalues of $B = a I + b P + c P^2$ are then $a+b+c$, $a+b\omega + c \omega^2$ and $a + b\omega^2 + c \omega$, so (using $1 + \omega + \omega^2 = 0$) \eqalign{\det(B) &= (a+b+c)(a+b\omega + c \omega^2)(a + b \omega^2 + c \omega)\cr &= a^3 + b^3 + c^3 - 3 a b c\cr} Substituting the values of $a,b,c$ gives us the answer $-1/90$.

EDIT: Note that this shows that there is a $B$ with determinant $-1/90$, but not that this is the only solution.

The calculation can be shortened: if $B$ is of the form $aI + bP + cP^2$, then it commutes with $A$. So we can simplify the equation to $$\left(\frac{A^T A}{2} - 8 A^{-1}\right) B = I$$ Thus $B = \left(\dfrac{A^T A}{2} - 8 A^{-1}\right)^{-1}$, and $\det B = 1/\det \left(\dfrac{A^T A}{2} - 8 A^{-1}\right)$. We just have to calculate $\det \left(\dfrac{A^T A}{2} - 8 A^{-1}\right) = -90$ to say $\det(B)=-1/90$.

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No brainer: \begin{align} A^TB{\left( \frac12 A^T\right)}^T - 8A^{-1}B=I,\\ \frac12 A^TBA-8{A}^{-1}B=I,\\ \left(\frac12 A^T\otimes A^T - 8I\otimes A^{-1}\right)\mathrm{vec}(B)=\mathrm{vec}(I). \end{align} Solving the last line (9 linear equations with 9 unknowns), you get $B$. Then compute the determinant. See the Wikipedia entries for vectorisation and Kronecker product for the meanings of $\mathrm{vec}(\cdot)$ and $\otimes$.
Not something I'd do by hand even if $A$ is 2-by-2 ... –  user1551 Nov 20 '12 at 6:44