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Let $$A = \begin{bmatrix} 1 & 3 & 4\\ 3 & 6 & 9\\ 1 & 6 & 4 \end{bmatrix},$$ $B$ be a $3\times 3$ matrix and $$A \cdot A^{T} \cdot A +3B^{-1} =0$$

What would be the value of $ \det( \operatorname{adj} (A^{-1}(B^{-1}){2B^{T}}))$ ?

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  • $\begingroup$ Do you have some idea how you can start? $\endgroup$ – mathreadler Jun 14 at 11:49
  • $\begingroup$ no, I tried to find $$B^{-1}$$ and then the $$B$$ from that. then just solved it... but it was so long and probably isn't the way... + I did mistakes in the middle for sure. I will be glad to see full solution if possible $\endgroup$ – HelpMePlease Jun 14 at 12:04
  • $\begingroup$ I am not sure if this is the only way to solve this, but you may find $B^{-1}$ from the given equation and then just compute the matrix whose determinant you need to find. This is really tedious, but it solves your problem nevertheless. $\endgroup$ – Alexdanut Jun 14 at 12:05
  • $\begingroup$ yea that's what I did but I got the wrong answer... you have so many places to mistake once and then all this question can't be solved $\endgroup$ – HelpMePlease Jun 14 at 12:07
  • $\begingroup$ The question is not clear. We cannot take an arbitrary $B$. We should say that $B$ is given by $B=-3A^{-1}A^{-T}A^{-1}$ $\endgroup$ – Dietrich Burde Jun 14 at 12:14
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Let $$X = |( \operatorname{adj} (A^{-1}(B^{-1}){2B^{T}})| = |(\operatorname{adj}(2B^T) \operatorname{adj}(B^{-1})\operatorname{adj}(A^{-1}))| $$ $$ = |2B^T|^2 |B^{-1}|^2 |A^{-1}|^2 = 2^6 |B|^2 \cfrac{1}{(|A||B|)^2} = \cfrac{2^6}{|A|^2}$$

I hope now you can figure out $|A|$.

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  • $\begingroup$ $$|A| = 9$$ which mean it's $$2^6/81$$ but that's not the answer :S $\endgroup$ – HelpMePlease Jun 14 at 12:51
  • $\begingroup$ I really can't figure out where I did wrong? $\endgroup$ – Ajay Mishra Jun 14 at 12:56
  • $\begingroup$ Tell us what's the suggested answer, please. $\endgroup$ – Michael Hoppe Jun 14 at 17:07
  • $\begingroup$ @HelpMePlease We are right, See this imgur.com/a/SNcaFs6 A program to compute the same in python. $\endgroup$ – Ajay Mishra Jun 14 at 17:49
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$B$ is a red herring here and might be replaced by any invertible $3\times3$-matrix.

Since $\operatorname{adj}(C)=\det(C)C^{-1}$ for invertible $C$ we have $$\det(\operatorname{adj}(C)=\det(C^{-1}\cdot\det(C))=(\det(C))^3\det(C^{-1})=(\det(C))^2$$ if $C$ is of type $3\times3$.

Now happily compute \begin{align}\det( \operatorname{adj} (A^{-1}B^{-1}{2B^{T}}) &=\bigl(\det(A^{-1}B^{-1}2B^{T})\bigr)^2\\ &=\bigl(2^3\cdot\det(A^{-1}) \underbrace{\det(B^{-1})\det(B^{T})}_{=1}\bigr)^2\\ &=\left(2^3\cdot\frac{1}{9}\right)^2 =\frac{64}{81} \end{align} as $\det(A)=9$.

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We can calculate $B^{-1}$, since $$A \cdot A^{T} \cdot A +3B^{-1} =0 \leftrightarrow -3B^{-1} = A \cdot A^{T} \cdot A$$ So $$\begin{align} -3B^{-1} &= \begin{bmatrix} 1 & 3 & 4\\ 3 & 6 & 9\\ 1 & 6 & 4 \end{bmatrix} \cdot \begin{bmatrix} 1 & 3 & 1\\ 3 & 6 & 6\\ 4 & 9 & 4 \end{bmatrix} \cdot \begin{bmatrix} 1 & 3 & 4\\ 3 & 6 & 9\\ 1 & 6 & 4 \end{bmatrix} = \\ &= \begin{bmatrix} 26 & 57 & 35\\ 57 & 126 & 75\\ 35 & 75 & 53 \end{bmatrix} \cdot \begin{bmatrix} 1 & 3 & 4\\ 3 & 6 & 9\\ 1 & 6 & 4 \end{bmatrix} = \begin{bmatrix} 232 & 630 & 757\\ 510 & 1377 & 1662\\ 313 & 873 & 1027 \end{bmatrix} \end{align}$$ Dividing both sides by $-3$ we get $$B^{-1} = \frac{1}{3}\begin{bmatrix} -232 & -630 & -757\\ -510 & -1377 & -1662\\ -313 & -873 & -1027 \end{bmatrix}$$ From this, we can calculate $B$, since $B = (B^{-1})^{-1}$. We just need to invert the above matrix. $$B = \left(\frac{1}{3}\begin{bmatrix} -232 & -630 & -757\\ -510 & -1377 & -1662\\ -313 & -873 & -1027 \end{bmatrix}\right)^{-1} = \frac{1}{9}\begin{bmatrix} 1361 & -513 & -173\\ 132 & -49 & -18\\ -527 & 198 & 68 \end{bmatrix}$$ And from this we can easily calculate everything we need: $$A^{-1}(B^{-1}){2B^{T}} = \begin{bmatrix} 1 & 3 & 4\\ 3 & 6 & 9\\ 1 & 6 & 4 \end{bmatrix}^{-1} \cdot \frac{1}{3}\begin{bmatrix} -232 & -630 & -757\\ -510 & -1377 & -1662\\ -313 & -873 & -1027 \end{bmatrix} \cdot 2 \cdot \frac{1}{9}\begin{bmatrix} 1361 & -513 & -173\\ 132 & -49 & -18\\ -527 & 198 & 68 \end{bmatrix}^{T} = \\ = \frac{2}{27} \cdot \frac{1}{3}\begin{bmatrix} -10 & 4 & 1\\ -1 & 0 & 1\\ 4 & -1 & -1 \end{bmatrix} \cdot \begin{bmatrix} -232 & -630 & -757\\ -510 & -1377 & -1662\\ -313 & -873 & -1027 \end{bmatrix} \cdot \begin{bmatrix} 1361 & 132 & -527\\ -513 & -49 & 198\\ -173 & -18 & 68 \end{bmatrix} = \\ = -\frac{2}{81} \begin{bmatrix} 33& 81 & 105\\ 81 & 243 & 270\\ 105 & 270 & 339 \end{bmatrix} \cdot \begin{bmatrix} 1361 & 132 & -527\\ -513 & -49 & 198\\ -173 & -18 & 68 \end{bmatrix} = \frac{2}{9} \begin{bmatrix} 1645 & 167 & -643\\ 6792 & 675 & -2643\\ 6028 & 608 & -2353 \end{bmatrix}.$$ And now for the $\operatorname{adj}$: $$\operatorname{adj}(A^{-1}(B^{-1}){2B^{T}}) = \operatorname{adj}\frac{2}{9} \begin{bmatrix} 1645 & 167 & -643\\ 6792 & 675 & -2643\\ 6028 & 608 & -2353 \end{bmatrix} \stackrel{\text{adj is linear}}{=} \\ = \frac{2}{9} \cdot \operatorname{adj} \begin{bmatrix} 1645 & 167 & -643\\ 6792 & 675 & -2643\\ 6028 & 608 & -2353\end{bmatrix} = \frac{2}{9} \begin{bmatrix} 18669 & 2007 & -7356\\ 49572 & 5319 & -19521\\ 60636 & 6516 & -23889\end{bmatrix}$$ And finally: $$\det( \operatorname{adj} (A^{-1}(B^{-1}){2B^{T}})) \stackrel{\text{det is linear}}{=} \frac{2}{9} \cdot \det \begin{bmatrix} 18669 & 2007 & -7356\\ 49572 & 5319 & -19521\\ 60636 & 6516 & -23889\end{bmatrix} = \frac{2}{9} \cdot 6561 = 1458.$$

So the final answer is $1458$.

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Since $A$ is invertible, it follows that $B = -3A^{-1}A^{-T}A^{-1}$ and therefore that

\begin{align} \det( \operatorname{adj} (A^{-1}(B^{-1}){2B^{T}})) &=\det\left( \operatorname{adj} \left(A^{-1}\left(-\frac{1}{3}A A^{T} A \right){(-6A^{-T}A^{-1}A^{-T}})\right)\right). \end{align}

Since you already know $A$, this is just a computation.

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No need to find the value of matrix $B$ since \begin{equation} B^{-1}=-\frac{1}{3} A A^T A \end{equation} then \begin{equation} B=-\frac{1}{3}(A A^T A)^{-1} = 3 A^{-1} A^{-T} A^{-1} \end{equation} put into the desired system \begin{align} A^{-1}(B^{-1})2B^T &=A^{-1} (-\frac{1}{3} A A^T A) 2 (3 A^{-1} A^{-T} A^{-1}) \notag \\ &= 2(A^{-1}A)A^T (A A^{-1}) A^{-T} A^{-1} \notag \\ &= 2A^T A^{-T} A^{-1} \notag \\ &= 2A^{-1} \end{align} where \begin{equation} A^{-1}=-\frac{1}{3} \begin{pmatrix}-10 & 4 & 1\\ -1 & 0 & 1 \\ 4 & -1 & -1 \end{pmatrix} \end{equation} then find the result of $det(adj(A^{-1}))$.

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  • $\begingroup$ I got 1/81 using this which isn't the correct answer sadly :( $\endgroup$ – HelpMePlease Jun 14 at 12:46
  • $\begingroup$ what's the answer @HelpMePlease ? Are you sure you've typed the question correctly? $\endgroup$ – Ajay Mishra Jun 14 at 13:02
  • $\begingroup$ I don't know the answer the system is just telling me im put the wrong answer $\endgroup$ – HelpMePlease Jun 14 at 13:09
  • $\begingroup$ Check the equation after "put the desired..", you have to substitute $2B^T$ not $2B$ $\endgroup$ – Ajay Mishra Jun 14 at 17:16

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