# 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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Edit. Revisiting this question, I wonder why I couldn't solve for $B$ directly by paper and pencil in my previous answer. The equation in the OP can be rewritten as $$\frac12A^TBA - 8A^{-1}B = I.\tag{1}$$ We first show that it has a unique solution. Consider the homogeneous linear equation $\frac12A^TBA - 8A^{-1}B = 0$. Let $J$ be the all-one matrix. Then $AA^T=4I-J$. Therefore \begin{align} \frac12A^TBA - 8A^{-1}B = 0 &\Rightarrow AA^TBA - 16B = 0\\ &\Rightarrow (4I-J)BA - 16B=0\\ &\Rightarrow 4B(A-4I) = JBA\tag{2}\\ &\Rightarrow 4B = JBA(A-4I)^{-1},\tag{3} \end{align} where $A-4I$ has an inverse because it is strictly diagonally dominant. Therefore, $B=JX$ for some matrix $X$. It follows that $JB=J^2X=3B$. Hence $(2)$ gives $4B(A-4I) = 3BA$, or $B(A-16I)=0$. Again, $A-16I$ is strictly diagonally dominant and hence nonsingular. Thus the equation $\frac12A^TBA - 8A^{-1}B = 0$ possesses only the trivial solution and $(1)$ is uniquely solvable.

Note that $AA^T$ is also equal to $2A-A^2$. So, we may rewrite $(1)$ as $$(2A-A^2)BA - 16B = 2A.\tag{4}$$ Therefore, it makes sense to try some $B$ that is a polynomial in $A$, but in this case, as indicated in the other answer, $A$ and $B$ commute. Therefore $(1)$ can be rewritten as $B=\left(\frac{A^T A}{2} - 8 A^{-1}\right)^{-1}$. As $\det\left(\frac{A^T A}{2} - 8 A^{-1}\right)=-90\ne0$, this $B$ exists and it is indeed the unique solution to $(1)$. Hence $\det(B)=\frac1{-90}$.

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Not something I'd want to do by hand, but a computer algebra system will do it nicely. – Robert Israel Nov 20 '12 at 6:25
Not something I'd do by hand even if $A$ is 2-by-2 ... – user1551 Nov 20 '12 at 6:44