all &&!all &&!

# Why std::optional::value() &&; return &&?

const auto& value = getValue();
value.The reason of crash is that the method returns T&&.

# .then of Promise.all result never executes

stringify(item) + " *((*&*&*&*&^&*&*&*(&*(&*&*(&(&(&*( :::" + x);
if(item.address == "") {
/*if(!

# Finding eigenvalues of an 'almost-tridiagonal' circulant matrix

Consider the $2N\times 2N$ matrix

$$A=\begin{pmatrix} a &1 &0&0&0&\ldots&0&1 \\1 &-a&1 & 0 &0 & \ldots & 0&0 \\0 &1&a&1&0 &\cdots &0&0 \\ 0&0&1&-a &1 & \ldots &0&0 \\& & & \cdots \\ 1&0 &0&0&0&\ldots &1&-a\end{pmatrix}$$

Hopefully the structure is clear, but if not I can clarify further.There is a lot of literature exclusively on eigenvalues of tridiagonal matrices and circulant matrices, however $A$ is neither exactly circulant nor is it exactly tridaigonal.

# A generalization of Chebyshev polynomials

09216 \\
- & - & - & - & - & 0.09216 \\
- & - & - & - & - & - & 0.

# A question about $O(3,1)$

Recall that $O(3,1)$ is the collection of matrices $A\in M_4(\mathbb R)$ such that
$$A\begin{pmatrix}1 &&&\\&1&&\\&&1&\\&&&-1\end{pmatrix}A^T=\begin{pmatrix}1 &&&\\&1&&\\&&1&\\&&&-1\end{pmatrix}.Can we find a Q\in O(3,1) such that$$Q\psi Q^{-1}=\begin{pmatrix} 0&a&&\\-a&0&&\\&&0&b\\&&b&0\end{pmatrix}$$for some a,b\in \mathbb R? # Mean of a vector Then$$\mu_n\to(2,2,2,2,2,102,102,102,102,102).The exact rate of convergence will be determined by the second-largest singular value of the matrix
$$M = \begin{bmatrix} \tfrac13 & \tfrac13 & \tfrac13 & & & \\ \tfrac13 & \tfrac13 & \tfrac13 & & & \\ \tfrac13 & \tfrac13 & \tfrac13 & & & \\ & & & 1 & & \\ & & & & 1 & \\ & & & & & \ddots \end{bmatrix} \begin{bmatrix} 1 & & & & & \\ & \tfrac13 & \tfrac13 & \tfrac13 & & \\ & \tfrac13 & \tfrac13 & \tfrac13 & & \\ & \tfrac13 & \tfrac13 & \tfrac13 & & \\ & & & & 1 & \\ & & & & & \ddots \end{bmatrix} \begin{bmatrix} 1 & & & & & \\ & 1 & & & & \\ & & \tfrac13 & \tfrac13 & \tfrac13 & \\ & & \tfrac13 & \tfrac13 & \tfrac13 & \\ & & \tfrac13 & \tfrac13 & \tfrac13 & \\ & & & & & \ddots \end{bmatrix} \cdots$$
but I don't know how to compute that.

# Placing numbers $1,2,\ldots,n^3$ in a cube so that numbers of any two adjacent unit subcube are coprime

Basically, one can start with something like this (if $n$ was $11$ and $i_0$ was 3):

$\begin{array}{ccccccccccc} 30 & 1 & 2 & 15 & 2 & 1 & 10 & 3 & 2 & 5 & 6\\ 1 & 30 & 1 & 2 & 15 & 2 & 3 & 10 & 1 & 6 & 5\\ 30 & 1 & 2 & 15 & 2 & 1 & 10 & 3 & 2 & 5 & 6\\ \vdots & & & & & & & & & & \vdots\\ 1 & 30 & 1 & 2 & 15 & 2 & 3 & 10 & 1 & 6 & 5\\ 30 & 1 & 2 & 15 & 2 & 1 & 10 & 3 & 2 & 5 & 6\end{array}$

and then fix it so there aren't quite so many $30$s, etc.Peter Mueller's answer gives solutions for $(4,3,3)$, $(5,3,3)$ and $(4,4,4)$.

5ex\rlap{\scriptstyle#1}}$$$\begin{array}{c} & & 1 & & 1 & & \\ & & \da{} & & \da{} & & \\ & & C & \ra{=} & C\\ & & \da{} & & \da{} & & \\ 1 & \ra{} & A & \ra{} & G & \ra{} & B & \ra{} & 1 \\ & & \da{} & & \da{\pi} & & \da{=} \\ 1 & \ra{} & D & \ra{} & E & \ra{} & B & \ra{} & 1 \\ & & \da{} & & \da{} & & \\ & & 1 & & 1 & & \end{array}$$ in which all rows and columns are exact.Now$A/[C,D]$is the direct product$C/[C,D] \times D[C,D]/[C,D]$. # Eigenvalues of permutations of a real matrix: can they all be real? In rank 4 and 5, a small amount of trial and error yields$M = \pmatrix{1 & T^2 & T^3 & T^3 \\ 0 & 1 & T^2 & T^3 \\ 0 & 0 & 1 & T^2 \\ 0 & 0 & 0 & 1}$and$M = \pmatrix{1 & T^3 & T^5 & T^6 & T^3 \\ 0 & 1 & T^3 & T^5 & T^6 \\ 0 & 0 & 1 & T^3 & T^5 \\ 0 & 0 & 0 & 1 & T^3 \\ 0 & 0 & 0 & 0 & 1 }$.In rank 6, we have$M = \pmatrix{1 & T^{13} & T^{19} & T^{25} & T^{31} & T^{25} \\ 0 & 1 & T^7 & T^{25} & T^{27} & T^{31} \\ 0 & 0 & 1 & T^{21} & T^{25} & T^{25} \\ 0 & 0 & 0 & 1 & T^7 & T^{19} \\ 0 & 0 & 0 & 0 & 1 & T^{13} \\ 0 & 0 & 0 & 0 & 0 & 1 }$. # Verifying the correctness of a Sudoku solution We are thus looking for a solution of the following grid: $$\begin{array}{|ccc|ccc|ccc|ccc|ccc|} \hline 1&&&2&&&&&&&&&&&&\\ \strut&&&&&&&&&&&&&&&\\ \strut&&&&&&&&&&&&&&&\\ \hline &&&1&&&2&&&&&&&&&\\ &&&&&&&&&&&&&&\cdots&\\ &&&&&&&&\ddots&&&&&&&\\ \hline 2&&&&&&&&&1&&&&&&\\ \strut&&&&&&&&&&&&&&&\\ \strut&&&&&&&&&&&&&&&\\ \hline \strut&&&&&&&&&&&&&&&\\ \strut&&&&\vdots&&&&&&&&&&&\\ \strut&&&&&&&&&&&&&&&\\ \hline \end{array}$$ where the upper part is a$q'\times q'$subgrid,$q'=q/2$.$S$is not included in the complement of a Sudoku permutation of one of the following sets: a)$\{r_1,r_2\}$b)$\{r_1,c_1,b_1\}$c)$\{b_1,b_2,b_4,b_5\}$d)$\{r_1,r_4,b_1,b_4\}$e)$\{r_1,c_1,b_2,b_4,b_5\}$f)$\{b_2,b_3,b_4,b_6,b_7,b_8\}$g)$\{r_1,r_4,b_1,b_5,b_7,b_8\}$h)$\{r_1,c_1,b_3,b_5,b_6,b_7,b_8\}$Maximal incomplete sets http://math. # Matrix version of number theoretic integral lattice claim Given quadratic forms with symmetric matrices $$F \; = \; \left( \begin{array}{cccc} 2 & 1 & 0 & 1 \\\ 1 & 2 & 0 & 0 \\\ 0 & 0 & 2 & 1 \\\ 1 & 0 & 1 & 6 \end{array} \right)$$ and $$G \; = \; \left( \begin{array}{cccc} 2 & 1 & 1 & 0 \\\ 1 & 2 & 0 & 1 \\\ 1 & 0 & 2 & 0 \\\ 0 & 1 & 0 & 8 \end{array} \right)$$ Next, I take$r=3,$because the discriminant is$29,\$ and find
$$P \; = \; \left( \begin{array}{cccc} 1 & 1 & 1 & 7 \\\ 1 & -2 & 1 & -5 \\\ 1 & 1 & -2 & 1 \\\ 1 & 1 & 1 & -2 \end{array} \right)$$
that satisfies
$$P^T \; F \; P \; = \; 9 \; G = r^2 \; G.Indeed,$$ Q \; = \; \; \mbox{adj} \; P \; = \;
\left( \begin{array}{cccc}
-18 & -27 & -27 & -9 \\\
9 & 27 & 0 & -36 \\\
-9 & 0 & 27 & -18 \\\
-9 & 0 & 0 & 9
\end{array}
\right)
$$However, a miracle! # Eigenvectors and eigenvalues of Tridiagonal matrix Hi, is it possible to analytically evaluate the eigenvectors and the eigenvalues of a tridiagonal matrix of the form :$$
\mathcal{T}^{a}_n(p,q) = \begin{pmatrix}
0 & q & 0 & 0 &\cdots & 0 & 0 & 0 \\\
p & 0 & q & 0 &\cdots & 0 & 0 & 0 \\\
0 & p & 0 & q &\cdots & 0 & 0 & 0 \\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots& \vdots \\\
0 & 0 & 0 & 0 &\cdots & p & 0 & q \\\
0 & 0 & 0 & 0 & \cdots & 0 & p & 0
\end{pmatrix}
$$where p>0\ \ \& \ \ q > 0  and where there are n rows and n columns in the matrix above?$$
\mathcal{T}^{b}_n(p,q) = \begin{pmatrix}
0 & q & 0 & 0 &\cdots & 0 & 0 & p \\\
p & 0 & q & 0 &\cdots & 0 & 0 & 0 \\\
0 & p & 0 & q &\cdots & 0 & 0 & 0 \\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots& \vdots \\\
0 & 0 & 0 & 0 &\cdots & p & 0 & q \\\
q & 0 & 0 & 0 & \cdots & 0 & p & 0
\end{pmatrix}
$$Thanks Here is the calculation of the spectrum of the first matrix, which I write pJ+qK with K=J^T. # Bounding largest eigenvalue Hi all, do you know how to compute (as a function of n) the largest eigenvalue of this matrix (or at least to bound it)?$$
\left(\begin{array}{cccccc}
0 & 1 & & & & \cr
1 & 0 & \sqrt 2 & & & \cr
& \sqrt 2 & 0 & & & \cr
& & & \ddots & & & \cr
& & & & 0 & \sqrt n & \cr
& & & & \sqrt n & 0 &
\end{array}\right)
$$Thanks! # Is the following DNN matrix CP? Is the following Doubly Non-negative matrix Completely Positive: \frac{1}{6}\begin{bmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 1 \\0 & 2 & 0 & 1 & 0 & 1 & 1 & 0 & 1 \\0 & 0 & 2 & 1 & 1 & 0 & 1 & 1 & 0 \\0 & 1 & 1 & 2 & 0 & 0 & 0 & 1 & 1 \\1 & 0 & 1 & 0 & 2 & 0 & 1 & 0 & 1 \\1 & 1 & 0 & 0 & 0 & 2 & 1 & 1 & 0 \\0 & 1 & 1 & 0 & 1 & 1 & 2 & 0 & 0\\1 & 0 & 1 & 1 & 0 & 1 & 0 & 2 & 0\\1 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 2\end{bmatrix} Edit: I wanted the following matrix to be W.Robert Israel suggested I call it W^T instead. # Monotonicity of the hard EM algorithm. 9718]$$A^t = \left(\begin{array}{cccccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\\\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\\\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\\\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\\\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\\\
\end{array}\right)
$$B^t = (-2.546023113539513\\\\ 0\\\\ 0\\\\ -0. # Modify Odoo Discussions (mail) model &amp;&amp; (message.channel') &amp;&amp; options. # Regex - Getting Value from HTML with Changing Middle sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=32&amp;cad=rja&amp;uact=8&amp;ved=0ahUKEwjknIy87oHWAhXHi1QKHXQdAJsQ9zAIyQEwHw&amp;url=http%3A%2F%2Fwww.aft&amp;&amp;google. # why do these characters belong to the first group in this JS regex match? :Dokumentation&&&KKS-Nummer&&&Beschreibung&&&Seite)(&&&([^(&&&)]+)&&&([^(&&&)]+)&&&(\d+))+ The test string: %5B"Deckblatt: Anlagendokumentation&&&Produktdaten&&&KKS-Nummer&&&Hersteller&&&Typ&&&Artikelnummer&&&MA-KF1&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF11&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF12&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF13&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF14&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF15&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF16&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF17&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF18&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF19&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF20&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF21&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF22&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF23&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF24&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF25&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF26&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF27&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF28&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF29&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF30&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF31&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF32&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF33&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF34&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF35&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF36&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF37&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF38&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF39&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF40&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF41&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&Dokumentation&&&KKS-Nummer&&&Beschreibung&&&Seite&&&all&&&Vorwort&&&6&&&all&&&Produktübersicht&&&7&&&all&&&Grundlagen&&&8&&&all&&&Montage und Verdrahtung&&&9&&&all&&&Inbetriebnahme%2FAnwendungshinweise&&&10&&&all&&&Fehlerbehandlung und Diagnose&&&11&&&all&&&Anhang 1&&&12&&&all&&&Anhang 2&&&13&&&all&&&Anhang 3&&&14&&&all&&&Anhang 4&&&15&&&all&&&Anhang 5&&&16&&&all&&&Anhang 6&&&17&&&all&&&Anhang 7&&&18&&&all&&&Anhang 8&&&19&&&all&&&Anhang 9&&&20&&&all&&&Anhang 10&&&21&&&all&&&Anhang 11&&&22&&&all&&&Anhang 12&&&23&&&all&&&Anhang 13&&&24&&&all&&&Anhang 14&&&25&&&all&&&Anhang 15&&&26&&&all&&&Anhang 16&&&27&&&all&&&Anhang 17&&&28&&&all&&&Anhang 18&&&29&&&all&&&Anhang 19&&&30&&&all&&&Anhang 20&&&31&&&all&&&Anhang 21&&&32&&&all&&&Anhang 22&&&33&&&all&&&Anhang 23&&&34&&&all&&&Anhang 24&&&35&&&all&&&Anhang 25&&&36&&&all&&&Anhang 26&&&37&&&all&&&Anhang 27&&&38&&&all&&&Anhang 28&&&39&&&all&&&Anhang 29&&&40&&&all&&&Anhang 30&&&41&&&all&&&Anhang 31&&&42&&&all&&&Anhang 32&&&43&&&all&&&Anhang 33&&&44&&&all&&&Anhang 34&&&45&&&all&&&Anhang 35&&&46&&&all&&&Anhang 36&&&47&&&all&&&Anhang 37&&&48&&&all&&&Anhang 38&&&49&&&all&&&Anhang 39&&&50&&&all&&&Anhang 40&&&51&&&all&&&Anhang 41&&&52&&&all&&&Anhang 42&&&53"%5D The regex I wrote should get a first group, which appears after /Artikelnummer/ and before /Dokumentation&&&/ (etc), as well as a second group, which is what I'm having trouble with: It should consist of repetitions of this pattern: (&&&([^(&&&)]+)&&&([^(&&&)]+)&&&(\d+)+ By my reckoning, that should capture the entire substring: &&&all&&&Vorwort&&&6&&&all&&&Produktübersicht&&&7&&&all&&&Grundlagen&&&8&&&all&&&Montage und Verdrahtung&&&9&&&all&&&Inbetriebnahme%2FAnwendungshinweise&&&10&&&all&&&Fehlerbehandlung und Diagnose&&&11&&&all&&&Anhang 1&&&12&&&all&&&Anhang 2&&&13&&&all&&&Anhang 3&&&14&&&all&&&Anhang 4&&&15&&&all&&&Anhang 5&&&16&&&all&&&Anhang 6&&&17&&&all&&&Anhang 7&&&18&&&all&&&Anhang 8&&&19&&&all&&&Anhang 9&&&20&&&all&&&Anhang 10&&&21&&&all&&&Anhang 11&&&22&&&all&&&Anhang 12&&&23&&&all&&&Anhang 13&&&24&&&all&&&Anhang 14&&&25&&&all&&&Anhang 15&&&26&&&all&&&Anhang 16&&&27&&&all&&&Anhang 17&&&28&&&all&&&Anhang 18&&&29&&&all&&&Anhang 19&&&30&&&all&&&Anhang 20&&&31&&&all&&&Anhang 21&&&32&&&all&&&Anhang 22&&&33&&&all&&&Anhang 23&&&34&&&all&&&Anhang 24&&&35&&&all&&&Anhang 25&&&36&&&all&&&Anhang 26&&&37&&&all&&&Anhang 27&&&38&&&all&&&Anhang 28&&&39&&&all&&&Anhang 29&&&40&&&all&&&Anhang 30&&&41&&&all&&&Anhang 31&&&42&&&all&&&Anhang 32&&&43&&&all&&&Anhang 33&&&44&&&all&&&Anhang 34&&&45&&&all&&&Anhang 35&&&46&&&all&&&Anhang 36&&&47&&&all&&&Anhang 37&&&48&&&all&&&Anhang 38&&&49&&&all&&&Anhang 39&&&50&&&all&&&Anhang 40&&&51&&&all&&&Anhang 41&&&52&&&all&&&Anhang 42&&&53 But, for some reason, the only string in group 2 is: &&&Anhang 42&&&53 Why is this happening?You get &&&all&&&Anhang 42&&&53 in Group 2 because the (pattern)+ is a repeated capturing group that stores only the value captured at the last iteration. # Adjacency matrices of graphs This implies that their adjacency matrices are conjugate over \mathbb C.Consider the adjacency matrices$$ A = \left[\begin{array}{rrrrrrrrrrr}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
\end{array}\right] $$and$$ B = \left[ \begin{array}{rrrrrrrrrrr}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\\\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
\end{array}\right].