The number of $7×7$ matrices with specific conditions












2












$begingroup$


I am asked to find the number of $7×7$ matrices with entries $0,1,2,...,9$ whose determinant is not a multiple of $10$.



Expanding the determinant along rows (or columns) gives us a messy expression and, henceforth difficult to deal with.



Is it attackable using recurrences or we have to choose another approach? Please help me and give a hint to start with this problem.



Thank you.






linear-algebra combinatorics matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try to get an idea here.
    $endgroup$
    – metamorphy
    Dec 22 '18 at 10:47










  • $begingroup$
    The results in the linked page of wikipedia are for fileds $mathbb{Z}_p$ ( p is a prime). Can we generalize this results to fileds of order $pq$ ?
    $endgroup$
    – Fermat
    Dec 22 '18 at 11:24












  • $begingroup$
    I didn't mean it to be a ready-to-use answer. But OK, see mine below.
    $endgroup$
    – metamorphy
    Dec 22 '18 at 12:07
















2












$begingroup$


I am asked to find the number of $7×7$ matrices with entries $0,1,2,...,9$ whose determinant is not a multiple of $10$.



Expanding the determinant along rows (or columns) gives us a messy expression and, henceforth difficult to deal with.



Is it attackable using recurrences or we have to choose another approach? Please help me and give a hint to start with this problem.



Thank you.






linear-algebra combinatorics matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try to get an idea here.
    $endgroup$
    – metamorphy
    Dec 22 '18 at 10:47










  • $begingroup$
    The results in the linked page of wikipedia are for fileds $mathbb{Z}_p$ ( p is a prime). Can we generalize this results to fileds of order $pq$ ?
    $endgroup$
    – Fermat
    Dec 22 '18 at 11:24












  • $begingroup$
    I didn't mean it to be a ready-to-use answer. But OK, see mine below.
    $endgroup$
    – metamorphy
    Dec 22 '18 at 12:07














2












2








2


2



$begingroup$


I am asked to find the number of $7×7$ matrices with entries $0,1,2,...,9$ whose determinant is not a multiple of $10$.



Expanding the determinant along rows (or columns) gives us a messy expression and, henceforth difficult to deal with.



Is it attackable using recurrences or we have to choose another approach? Please help me and give a hint to start with this problem.



Thank you.






linear-algebra combinatorics matrices







share|cite|improve this question











$endgroup$




I am asked to find the number of $7×7$ matrices with entries $0,1,2,...,9$ whose determinant is not a multiple of $10$.



Expanding the determinant along rows (or columns) gives us a messy expression and, henceforth difficult to deal with.



Is it attackable using recurrences or we have to choose another approach? Please help me and give a hint to start with this problem.



Thank you.






linear-algebra combinatorics matrices




linear-algebra combinatorics matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 16:31







Fermat

















asked Dec 22 '18 at 10:33









FermatFermat

4,4351926




4,4351926












  • $begingroup$
    Try to get an idea here.
    $endgroup$
    – metamorphy
    Dec 22 '18 at 10:47










  • $begingroup$
    The results in the linked page of wikipedia are for fileds $mathbb{Z}_p$ ( p is a prime). Can we generalize this results to fileds of order $pq$ ?
    $endgroup$
    – Fermat
    Dec 22 '18 at 11:24












  • $begingroup$
    I didn't mean it to be a ready-to-use answer. But OK, see mine below.
    $endgroup$
    – metamorphy
    Dec 22 '18 at 12:07


















  • $begingroup$
    Try to get an idea here.
    $endgroup$
    – metamorphy
    Dec 22 '18 at 10:47










  • $begingroup$
    The results in the linked page of wikipedia are for fileds $mathbb{Z}_p$ ( p is a prime). Can we generalize this results to fileds of order $pq$ ?
    $endgroup$
    – Fermat
    Dec 22 '18 at 11:24












  • $begingroup$
    I didn't mean it to be a ready-to-use answer. But OK, see mine below.
    $endgroup$
    – metamorphy
    Dec 22 '18 at 12:07
















$begingroup$
Try to get an idea here.
$endgroup$
– metamorphy
Dec 22 '18 at 10:47




$begingroup$
Try to get an idea here.
$endgroup$
– metamorphy
Dec 22 '18 at 10:47












$begingroup$
The results in the linked page of wikipedia are for fileds $mathbb{Z}_p$ ( p is a prime). Can we generalize this results to fileds of order $pq$ ?
$endgroup$
– Fermat
Dec 22 '18 at 11:24






$begingroup$
The results in the linked page of wikipedia are for fileds $mathbb{Z}_p$ ( p is a prime). Can we generalize this results to fileds of order $pq$ ?
$endgroup$
– Fermat
Dec 22 '18 at 11:24














$begingroup$
I didn't mean it to be a ready-to-use answer. But OK, see mine below.
$endgroup$
– metamorphy
Dec 22 '18 at 12:07




$begingroup$
I didn't mean it to be a ready-to-use answer. But OK, see mine below.
$endgroup$
– metamorphy
Dec 22 '18 at 12:07










1 Answer
1






active

oldest

votes


















3












$begingroup$

Let $M_n(m)$ be the set of all $ntimes n$ matrices over $mathbb{Z}/mmathbb{Z}$, and let
$$Z_n(m)={Ain M_n(m) : det A=0}.$$
Chinese remainder theorem says that, for $m_1,m_2$ coprime, the map
$$Amapsto(Abmod m_1,Abmod m_2)$$
is a bijection between $M_n(m_1 m_2)$ and $M_n(m_1)times M_n(m_2)$, and also between $Z_n(m_1 m_2)$ and $Z_n(m_1)times Z_n(m_2)$. From the link given in the comment above, for prime $p$ we have
$$F(n,p):=#(M_n(p)setminus Z_n(p))=prod_{k=0}^{n-1}(p^n-p^k).$$
Thus the answer is $(pq)^{n^2}-big(p^{n^2}-F(n,p)big)big(q^{n^2}-F(n,q)big)$ for $n=7, p=2, q=5$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your detailed answer. +1
    $endgroup$
    – Fermat
    Dec 22 '18 at 15:16












Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049292%2fthe-number-of-7%25c3%25977-matrices-with-specific-conditions%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Let $M_n(m)$ be the set of all $ntimes n$ matrices over $mathbb{Z}/mmathbb{Z}$, and let
$$Z_n(m)={Ain M_n(m) : det A=0}.$$
Chinese remainder theorem says that, for $m_1,m_2$ coprime, the map
$$Amapsto(Abmod m_1,Abmod m_2)$$
is a bijection between $M_n(m_1 m_2)$ and $M_n(m_1)times M_n(m_2)$, and also between $Z_n(m_1 m_2)$ and $Z_n(m_1)times Z_n(m_2)$. From the link given in the comment above, for prime $p$ we have
$$F(n,p):=#(M_n(p)setminus Z_n(p))=prod_{k=0}^{n-1}(p^n-p^k).$$
Thus the answer is $(pq)^{n^2}-big(p^{n^2}-F(n,p)big)big(q^{n^2}-F(n,q)big)$ for $n=7, p=2, q=5$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your detailed answer. +1
    $endgroup$
    – Fermat
    Dec 22 '18 at 15:16
















3












$begingroup$

Let $M_n(m)$ be the set of all $ntimes n$ matrices over $mathbb{Z}/mmathbb{Z}$, and let
$$Z_n(m)={Ain M_n(m) : det A=0}.$$
Chinese remainder theorem says that, for $m_1,m_2$ coprime, the map
$$Amapsto(Abmod m_1,Abmod m_2)$$
is a bijection between $M_n(m_1 m_2)$ and $M_n(m_1)times M_n(m_2)$, and also between $Z_n(m_1 m_2)$ and $Z_n(m_1)times Z_n(m_2)$. From the link given in the comment above, for prime $p$ we have
$$F(n,p):=#(M_n(p)setminus Z_n(p))=prod_{k=0}^{n-1}(p^n-p^k).$$
Thus the answer is $(pq)^{n^2}-big(p^{n^2}-F(n,p)big)big(q^{n^2}-F(n,q)big)$ for $n=7, p=2, q=5$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your detailed answer. +1
    $endgroup$
    – Fermat
    Dec 22 '18 at 15:16














3












3








3





$begingroup$

Let $M_n(m)$ be the set of all $ntimes n$ matrices over $mathbb{Z}/mmathbb{Z}$, and let
$$Z_n(m)={Ain M_n(m) : det A=0}.$$
Chinese remainder theorem says that, for $m_1,m_2$ coprime, the map
$$Amapsto(Abmod m_1,Abmod m_2)$$
is a bijection between $M_n(m_1 m_2)$ and $M_n(m_1)times M_n(m_2)$, and also between $Z_n(m_1 m_2)$ and $Z_n(m_1)times Z_n(m_2)$. From the link given in the comment above, for prime $p$ we have
$$F(n,p):=#(M_n(p)setminus Z_n(p))=prod_{k=0}^{n-1}(p^n-p^k).$$
Thus the answer is $(pq)^{n^2}-big(p^{n^2}-F(n,p)big)big(q^{n^2}-F(n,q)big)$ for $n=7, p=2, q=5$.






share|cite|improve this answer











$endgroup$



Let $M_n(m)$ be the set of all $ntimes n$ matrices over $mathbb{Z}/mmathbb{Z}$, and let
$$Z_n(m)={Ain M_n(m) : det A=0}.$$
Chinese remainder theorem says that, for $m_1,m_2$ coprime, the map
$$Amapsto(Abmod m_1,Abmod m_2)$$
is a bijection between $M_n(m_1 m_2)$ and $M_n(m_1)times M_n(m_2)$, and also between $Z_n(m_1 m_2)$ and $Z_n(m_1)times Z_n(m_2)$. From the link given in the comment above, for prime $p$ we have
$$F(n,p):=#(M_n(p)setminus Z_n(p))=prod_{k=0}^{n-1}(p^n-p^k).$$
Thus the answer is $(pq)^{n^2}-big(p^{n^2}-F(n,p)big)big(q^{n^2}-F(n,q)big)$ for $n=7, p=2, q=5$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 12:19

























answered Dec 22 '18 at 12:02









metamorphymetamorphy

3,8721721




3,8721721












  • $begingroup$
    Thank you for your detailed answer. +1
    $endgroup$
    – Fermat
    Dec 22 '18 at 15:16


















  • $begingroup$
    Thank you for your detailed answer. +1
    $endgroup$
    – Fermat
    Dec 22 '18 at 15:16
















$begingroup$
Thank you for your detailed answer. +1
$endgroup$
– Fermat
Dec 22 '18 at 15:16




$begingroup$
Thank you for your detailed answer. +1
$endgroup$
– Fermat
Dec 22 '18 at 15:16


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049292%2fthe-number-of-7%25c3%25977-matrices-with-specific-conditions%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
Read more

In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
Read more

How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

Image
1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
Read more