Let $n\geq 2$, $A=[a_{ij}]\in Mat_n(\mathbb{R})$ the matrix defined by $a_{ij}=0$ if $i=j$ and $a_{ij}=1$ if $i\not =j$   Calculate $\det(A)$  Proof:  Sylvester's theorem:   Using Sylvester's determinant theorem, which says that for any rectangular matrices of mutually transposed shapes $A\in\mathrm M_{n,m}(K)$ and $B\in \mathrm M_{m,n}(K)$ one has $$\det(I_n+AB)=\det(I_m+BA).$$  If $N$ is your matrix then $-N=I_n-AB$ where $A\in\mathrm M_{n,1}(K)$ is a one column all-one matrix and $B$ is its transpose. Then  $$ \det(N)=(-1)^n\det(-N)=(-1)^n\det(I_1-BA)=(-1)^n(1-n). $$ I need an explanation about one part of the proof  Question:     I don't see this step:      $(-1)^n\det(I_1-BA)=(-1)^n(1-n).$
How do I calculate the determinant of the following $n\times n$ matrices  $$\begin {bmatrix} 0 &amp; 1 &amp; \ldots &amp; 1 \\ 1 &amp; 0 &amp; \ldots &amp; 1 \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 1 &amp; 1 &amp; ... &amp; 0 \end {bmatrix}$$  and the same matrix but one of columns replaced only with $1$s?  In the  above matrix all off-diagonal elements are $1$ and diagonal elements are $0$.
Install button in the software center is grayed out when trying to install .debs without an internet connection I just installed ubuntu 11.10 from a usb stick.   After installation, none of the software which I had collected and run on 11.10 or 11.04 runs on it. Even my newly downloaded opera or chrome installations does not work.  I do not have a internet connection.  Whenever i try to install any .deb file by double clicking it,software center opens up but the install button is grayed out. Continuous Images of $GL_n(\mathbb{R})$ Which of the following is a continuous image of $GL_n(\mathbb{R})$?  The real line $\mathbb{R}$  The subspace $\{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$ of $\mathbb{R}^2$  $\mathbb{R}^2 \setminus \{(x,\frac1x):x\in \mathbb{R},x \ne 0\}$  $\{(x_1,x_2,...,x_n):x_1^2+x_2^2+...+x_n^2 \le 1\}$ of $\mathbb{R}^n$   How do I proceed? $GL_n(\mathbb{R})$ is not compact, not connected, so I cannot rule out some options using topological properties. How to use unnumbered chapters with KOMA-script? I am having some issues with the hyperref package. I would like to create a Introduction chapter without a number so I use \chapter*{Introduction} and \addcontentsline{toc}{chapter}{Introduction} because I want it in the table of contents (see example).  But then the link in my PDF viewer gets it wrong and also the automatically created header is no longer correct (see picture). The table of contents shows the right page number (p.1) but the link is also wrong.   What is my mistake here?  \documentclass{scrbook}  \usepackage{hyperref} \usepackage{blindtext} \usepackage{glossaries}   \makeglossaries  \newacronym{eg}{e.g.}{% for example }   \begin{document}   \frontmatter  % abstract before the table of contents \chapter*{Abstract} \addcontentsline{toc}{chapter}{Abstract} \blindtext  \tableofcontents  % chapter with acronyms \cleardoublepage \phantomsection  \printglossary[type=\acronymtype] \addcontentsline{toc}{chapter}{Glossary}  \cleardoublepage \mainmatter   \phantomsection % \chapter{Introduction} % with this everything is correct but I don't want the number \chapter*{Introduction} \addcontentsline{toc}{chapter}{Introduction}  \blindtext  \pagebreak  \blindtext  % first chapter \chapter{Chapter1}  \gls{eg} \blindtext  \pagebreak  \blindtext   \end{document} There is some conflict or error I don't understand. I can skip it and nothing visible happens. The output is okay and everything seems to be fine. Still I would like to get rid of the error message. If I run the following minimal example:   \documentclass[]{scrbook}  \usepackage[applemac, grafiken]{tahbiblatex} \usepackage[ngerman]{babel}   \begin{document}  Testing äüöaüä.  \end{document}   I get the error message:      ! Missing = inserted for \ifnum. &lt;to be read again&gt;                     , l.4 \usepackage                [ngerman]{babel} ?  ! Missing number, treated as zero. &lt;to be read again&gt;                     , l.4 \usepackage                [ngerman]{babel}   Here is my .sty file:  \NeedsTeXFormat{LaTeX2e}\relax \ProvidesPackage{tahbiblatex}[2012, by tah modified by sm]  \RequirePackage{ifthen}  \DeclareOption{latin1}{\PassOptionsToPackage{\CurrentOption}{inputenc}} \DeclareOption{applemac}{\PassOptionsToPackage{\CurrentOption}{inputenc}} \DeclareOption{cp850}{\PassOptionsToPackage{\CurrentOption}{inputenc}} \DeclareOption{utf8}{\PassOptionsToPackage{\CurrentOption}{inputenc}}  \DeclareOption{dvi}{% \newif\ifpdf \ifx\pdfoutput\undefined     \pdffalse \else     \pdfoutput=0     \pdftrue \fi}  \newcounter{Grafiken}\setcounter{Grafiken}{0} \DeclareOption{grafiken}{\setcounter{Grafiken}{1}}  \newcounter{Randausgleich}\setcounter{Randausgleich}{0} \DeclareOption{randausgleich}{\setcounter{Randausgleich}{1}}  \newcounter{Tabellen}\setcounter{Tabellen}{0} \DeclareOption{tabellen}{\setcounter{Tabellen}{1}}  \newcounter{Sync}\setcounter{Sync}{0} \DeclareOption{sync}{\setcounter{Sync}{1}}  \ProcessOptions  \RequirePackage{fixltx2e} \RequirePackage{inputenc} \RequirePackage[T1]{fontenc} \RequirePackage[T1]{url} \RequirePackage{setspace} %\RequirePackage[german]{varioref} \RequirePackage{textcomp}  \newcommand{\urlangabe}[2]{% $\langle$\url{#1}$\rangle$ -- Zugriff am #2}  \def\hierlockern {\begingroup \def\par{\endgraf\endgroup}%   \tolerance 1414%   \hbadness 1414%   \emergencystretch 1.5em%   \hfuzz 0.3pt%   \vfuzz \hfuzz%   \raggedbottom% }  \newcommand{\alleslockern}{%   \tolerance 1414%   \hbadness 1414%   \emergencystretch 1.5em%   \hfuzz 0.3pt%   \vfuzz \hfuzz%   \raggedbottom% } \ifthenelse{\value{Grafiken}=1}   {   \ifx\pdfoutput\undefined \newcount\pdfoutput \fi   \ifcase\pdfoutput       \RequirePackage[dvips]{graphicx}       \DeclareGraphicsExtensions{.eps}       \typeout{Paket tah Meldung: Die Option `grafiken' fuer DVI wurde gewaehlt.}% \else     \RequirePackage[pdftex]{graphicx}     \DeclareGraphicsExtensions{.jpg,.tif,.png,.pdf}     \typeout{Paket tah Meldung: Die Option `grafiken' fuer PDF wurde gewaehlt.}%   \fi}   {}  \ifthenelse{\value{Sync}=1}   {     \ifx\pdfoutput\undefined \newcount\pdfoutput \fi     \ifcase\pdfoutput     \RequirePackage{srcltx}     \typeout{Paket tah Meldung: Die Option `sync' fuer DVI wurde gewaehlt.}%     \else     \RequirePackage{pdfsync}     \typeout{Paket tah Meldung: Die Option `sync' fuer PDF wurde gewaehlt.}%     \fi }   {}   \ifthenelse{\value{Randausgleich}=1}   {\typeout{Paket tah Meldung: Die Option `randausgleich' wurde gewaehlt.}%   \RequirePackage[activate]{pdfcprot}}   {}  \ifthenelse{\value{Tabellen}=1}   {\typeout{Paket tah Meldung: Die Option `tabellen' wurde gewaehlt.}%   \RequirePackage{tabularx}   \RequirePackage{ragged2e}   \RequirePackage{dcolumn}   \RequirePackage{booktabs}   \newcolumntype{L}{&gt;{\RaggedRight\hspace{0pt}\arraybackslash}l}   \newcolumntype{Y}{&gt;{\RaggedRight\hspace{0pt}\arraybackslash}X}   \newcolumntype{d}[1]{&gt;{\DC@{.}{,}{#1}}l&lt;{\DC@end}}   }   {}    \endinput   I also don't really need all the features. So it would be nice to reduce it to minimal features, which are needed.. Find $\space\ \begin{align*} \lim_ {x \to+\infty} \left [ \frac{\log_{2}(x-1)}{x}\right]   \end{align*}$.   After some minutes around this limit I did it this way:  $\log_{2}(x-1)=y \Leftrightarrow 2^y=x-1$  So,$\space x=2^y+1$.  When $x \to +\infty$,$\space y \to +\infty$ also. By substitution:  $\begin{align*} \lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y+1-1)}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y)}{2^y+1}\right]=\end{align*}$  $\begin{align*}\lim_ {y \to+\infty} \left [ \frac{y}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y+1}{y}}  \right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y}{y}+\frac{1}{y}}\right]= \frac{1}{+\infty+0}=0  \end{align*}$  Is this correct?Are there any other easy way to find this limit?Thanks Can a smaller sensor's "crop factor" be used to calculate the exact increase in depth of field? If APS-C and similar crop-sensor digital cameras have a focal length multiplying effect such that a 50mm lens has an apparent focal length closer to the field of view of an 80mm on a full frame camera, and yet at the same time the depth of field for the smaller sensor camera is more like the depth of field a 50mm lens would produce on a full frame camera (using the same aperture), then this would seem to suggest the concept of an "aperture dividing effect."  In other words, a 50mm f/1.8 lens on an APS-C camera would act more like a 80mm f/2.8 (approx. 1.8 * 1.6x) lens in 35mm equivalent — for depth of field, not considering exposure.  Can someone with a better understanding of the physics involved clarify this for me.  I've never seen this concept explicitly mentioned anywhere, so I am a bit suspect of it. How to convert image file data in a byte array to a Bitmap? A fair coin is tossed until a head comes up for the first time. The probability of this happening on an odd number toss is?  How do i approach this problem ? Root of continuous function using Rolle's Theorem $$f(x)= 1 + e^x\cos(x)$$   $$g(x)= 1 + e^x\sin(x)$$      Show that there is at least one root of $g$ between two roots of $f$   How do I start to answer this question and do I need to use Intermediate Value Theorem? Ideal charging / discharging percentage for maximum battery life? One of my friends has a relatively new Samsung smartphone. He has this app that notifies him whenever the charging is up to 80%, and tells him to unplug the charger.  He tells me that it will increase the number of charging cycles for his battery. He says that if he lets it charge all the way up to 100%, it will mean he has to replace the battery of the phone more often.  Is this true?  Somehow, I feel this is a thing of the past. I thought the internal architecture of batteries these days was made so that it wouldn't drop too low or too high. (In other words, when the phone is at 100%, it isn't really 100%. It's what the battery can safely be charged to in order not to degenerate too much.)  Here's a  which taught me something about this. Long-distance Mind Trick Why is Severus Snape's Patronus the same as Lily Potter's? Are Snape and Lily related somehow? Are Snape and Harry related? I am technically knowledgeable enough to know that this most likely isn't possible, BUT, I am curious just the same (it must be put to rest). Is there anything preventing us from creating a means to use two different internet connections, like two different ethernet cards or an ethernet card and a wireless card, and using the two IP addresses at once for more internet speed?  My specific situation is that I would like to use both a wireless card and my wired connection on my machine running Kubuntu 13.10. Using two internet connections would allow me to do things like:   Torrent and browse simultaneously without lag Open Firefox with multiple tabs faster (?) Download a couple or a few large files at once double time Increase overall internet connection speed (?) KeyListener not working Can I still use logistic regression if some variables are not applied to all the observational units? Some of the variables are questions that are not being asked to every single person in the questionaire depending on how they answered previous question(s). So would I still be able to use a logistic regression if some of the questions (the independent variables) are only measuring the responses of SOME of the respondents? Modelling the "Moving Sofa" I believe that many of you know about the moving sofa problem; if not you can find the description of the problem .     In this question I am going to rotate the L shaped hall instead of moving a sofa around the corner. By rotating the hall $180^{\circ}$ what remains between the walls will give the shape of the sofa. Like this:    The points on the hall have the following properties:  \begin{eqnarray} A &amp; = &amp; \left( r\cos { \alpha  } ,t\sin { \alpha  }  \right)  \\ { A }' &amp; = &amp; \left( r\cos { \alpha  } +\sqrt { 2 } \cos { \left( \frac { \pi  }{ 4 } +\frac { \alpha  }{ 2 }  \right)  } ,t\sin { \alpha  } +\sqrt { 2 } \sin { \left( \frac { \pi  }{ 4 } +\frac { \alpha  }{ 2 }  \right)  }  \right)  \\ { B } &amp; = &amp; \left( r\cos { \alpha  } -\frac { t\sin { \alpha  }  }{ \tan { \left( \frac { \alpha  }{ 2 }  \right)  }  } ,0 \right)  \\ { B }' &amp; = &amp; \left( r\cos { \alpha  } -\frac { t\sin { \alpha  }  }{ \tan { \left( \frac { \alpha  }{ 2 }  \right)  }  } -\frac { 1 }{ \sin { \left( \frac { \alpha  }{ 2 }  \right)  }  } ,0 \right)  \\ C &amp; = &amp; \left( r\cos { \alpha  } +t\sin { \alpha  } \tan { \left( \frac { \alpha  }{ 2 }  \right)  } ,0 \right)  \\ { C }' &amp; = &amp; \left( r\cos { \alpha  } +t\sin { \alpha  } \tan { \left( \frac { \alpha  }{ 2 }  \right)  } +\frac { 1 }{ \cos { \left( \frac { \alpha  }{ 2 }  \right)  }  } ,0 \right)  \end{eqnarray}  Attention: $\alpha$ is not the angle of $AOC$, it is some angle $ADC$ where $D$ changes location on $x$ axis for $r\neq t$. I am saying this because images can create confusion. Anyways I will change them as soon as possible.  I could consider $r=f(\alpha)$ and $t=g(\alpha)$ but for this question I am going to take $r$ and $t$ as constants. If they were functions of $\alpha$ there would appear some interesting shapes. I experimented for different functions however the areas are more difficult to calculate, that's why I am not going to share. Maybe in the future.  We rotate the hall for $r=t$ in the example above: In this case:   point A moves on a semicircle  The envelope of lines between A' and C' is a circular arc. One has to prove this but I assume that it is true for $r=t$.   If my second assumption is correct the area of sofa is $A= 2r-\frac { \pi r^{ 2 } }{ 2 } +\frac { \pi  }{ 2 } $. The maximum area is reached when $r = 2/\pi$ and it's value is: $$A = 2/\pi+\pi/2 = 2,207416099$$  which matches with Hammersley's sofa. The shape is also similar or same:    Now I am going to increase $t$ with respect to $r$. For $r=2/\pi$ and $t=0.77$:    Well, this looks like .   I believe thearea can be maximized by finding the equations of envelopes above and below the sofa. Look at  where @Aretino has computed the area below $ABC$.   I don't know enough to find equations for envelopes. I am afraid that I will make mistakes. I considered to calculate area by counting number of pixels in it, but this is not a good idea because for optimizing the area I have to create many images.  I will give a bounty of 200 for whom calculates the maximum area. As I said the most difficult part of the problem is to find equations of envelopes. @Aretino did it.  PLUS: Could following be the longest sofa where $(r,t)=((\sqrt 5+1)/2,1)$ ?   If you want to investigate further or use animation for educational purposes here is the Geogebra file:      Ok, I had some free time and I count number of pixels in the sofa and I am sure that I have something bigger than Hammersley's constant.  First, I made a simulation for Hammersley's sofa where $r=t=2/\pi$ and exported the image to png in 300 dpi (6484x3342 pixels) and using Gimp counted number of pixels which have exactly same value. For Hammersley I got $3039086$ pixels.   For the second case $r=0.59$ and $t=0.66$ and I got $3052780$ pixels. To calculate area for this case:  $$\frac{3052780}{3039086}(2/\pi + \pi/2)=2.217362628$$  which is slightly less than Gerver's constant which is $2.2195$. Here is the sofa: I have an Ubuntu machine that I'm using as my home Internet router.  It uses a USB mobile broadband modem as its Internet connection, and shares this with another router (via the router's "Internet" port) which I use as my LAN.  How do I redirect requests from within my LAN for a given URL to a custom IP address (ex have google.com redirect to x.x.x.x)? Ownership of content in Community Wiki posts In Lubuntu, there's no 'Add New Program...' button in Desktop Session Settings. Is there an easy way to add new autostart programs in Lubuntu? For example, I can't install package qcad because of error: "can't find qdict.h". How can I find the package which provides the file qdict.h so I can install the package qcad?