Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\left|B \right|$?。

share|improve this question
4  
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

2 Answers 2

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 $\det(B)=-1/90$.

share|improve this answer

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 $\det(B)=\frac1{-90}$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.