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

4,4301926




4,4301926












  • $begingroup$
    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












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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

3,7421721




3,7421721












  • $begingroup$
    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


















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