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A in M2(prime), A=B^2 and det(B)=p^2
jasperE3   1
N 3 hours ago by KAME06
Source: VJIMC 2012 1.2
Determine all $2\times2$ integer matrices $A$ having the following properties:

$1.$ the entries of $A$ are (positive) prime numbers,
$2.$ there exists a $2\times2$ integer matrix $B$ such that $A=B^2$ and the determinant of $B$ is the square of a prime number.
1 reply
jasperE3
May 31, 2021
KAME06
3 hours ago
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Σ to ∞
phiReKaLk6781   3
N Yesterday at 6:12 PM by Maxklark
Evaluate: $ \sum\limits_{k=1}^\infty \frac{1}{k\sqrt{k+2}+(k+2)\sqrt{k}}$
3 replies
phiReKaLk6781
Mar 20, 2010
Maxklark
Yesterday at 6:12 PM
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Geometric inequality
ReticulatedPython   0
Yesterday at 5:12 PM
Let $A$ and $B$ be points on a plane such that $AB=n$, where $n$ is a positive integer. Let $S$ be the set of all points $P$ such that $\frac{AP^2+BP^2}{(AP)(BP)}=c$, where $c$ is a real number. The path that $S$ traces is continuous, and the value of $c$ is minimized. Prove that $c$ is rational for all positive integers $n.$
0 replies
ReticulatedPython
Yesterday at 5:12 PM
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Inequalities
sqing   27
N Yesterday at 3:51 PM by Jackson0423
Let $ a,b $ be reals such that $ a^2+ab+b^2 =3$ . Prove that
$$ \frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$$$$ \frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$$$$ \frac{19}{ 6}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$$Let $ a,b $ be reals such that $ a^2-ab+b^2 =1 $ . Prove that
$$ \frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$$$$ \frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$$$$ \frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$$
27 replies
sqing
Apr 16, 2025
Jackson0423
Yesterday at 3:51 PM
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Problem of the Week--The Sleeping Beauty Problem
FiestyTiger82   1
N Yesterday at 3:24 PM by martianrunner
Put your answers here and discuss!
The Problem
1 reply
FiestyTiger82
Yesterday at 2:30 PM
martianrunner
Yesterday at 3:24 PM
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Inequalities
sqing   4
N Yesterday at 1:09 PM by sqing
Let $ a,b,c $ be real numbers such that $ a^2+b^2+c^2=1. $ Prove that$$ |a-b|+|b-2c|+|c-3a|\leq 5$$$$|a-2b|+|b-3c|+|c-4a|\leq \sqrt{42}$$$$ |a-b|+|b-\frac{11}{10}c|+|c-a|\leq \frac{29}{10}$$
4 replies
sqing
Yesterday at 5:05 AM
sqing
Yesterday at 1:09 PM
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Inequalities
nhathhuyyp5c   2
N Yesterday at 12:38 PM by pooh123
Let $a, b, c$ be non-negative real numbers such that $a^2 + b^2 + c^2 = 3$. Find the maximum and minimum values of the expression
\[ P = \frac{a}{a^2 + 2} + \frac{b}{b^2 + 2} + \frac{c}{c^2 + 2}. \]
2 replies
nhathhuyyp5c
Apr 20, 2025
pooh123
Yesterday at 12:38 PM
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Challenging Optimization Problem
Shiyul   5
N Yesterday at 12:28 PM by exoticc
Let $xyz = 1$. Find the minimum and maximum values of $\frac{1}{1 + x + xy}$ + $\frac{1}{1 + y + yz}$ + $\frac{1}{1 + z + zx}$

Can anyone give me a hint? I got that either the minimum or maximum was 1, but I'm sure if I'm correct.
5 replies
Shiyul
Monday at 8:20 PM
exoticc
Yesterday at 12:28 PM
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Radical Axes and circles
mathprodigy2011   4
N Yesterday at 7:53 AM by spiderman0
Can someone explain how to do this purely geometrically?
4 replies
mathprodigy2011
Yesterday at 1:58 AM
spiderman0
Yesterday at 7:53 AM
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Combinatoric
spiderman0   0
Yesterday at 7:46 AM
Let $ S = \{1, 2, 3, \ldots, 2024\}.$ Find the maximum positive integer $n \geq 2$ such that for every subset $T \subset S$ with n elements, there always exist two elements a, b in T such that:

$|\sqrt{a} - \sqrt{b}| < \frac{1}{2} \sqrt{a - b}$
0 replies
spiderman0
Yesterday at 7:46 AM
0 replies
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BMT 2018 Algebra Round Problem 7
IsabeltheCat   5
N Yesterday at 6:56 AM by P162008
Let $$h_n := \sum_{k=0}^n \binom{n}{k} \frac{2^{k+1}}{(k+1)}.$$Find $$\sum_{n=0}^\infty \frac{h_n}{n!}.$$
5 replies
IsabeltheCat
Dec 3, 2018
P162008
Yesterday at 6:56 AM
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determinant of matrix
jokerjoestar   7
N Mar 31, 2025 by loup blanc
Calculate the determinant of the matrix below: \[ A = \begin{bmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 3 & 4 & \cdots & n & 1 \\ 3 & 4 & 5 & \cdots & 1 & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n-1 & n & 1 & \cdots & n-3 & n-2 \\ n & 1 & 2 & \cdots & n-2 & n-1 \end{bmatrix} \]
7 replies
jokerjoestar
Mar 30, 2025
loup blanc
Mar 31, 2025
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determinant of matrix
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jokerjoestar
150 posts
jokerjoestar
#1 h  Mar 30, 2025, 3:32 PM • 1 Y
Y by MihaiT
Calculate the determinant of the matrix below: \[ A = \begin{bmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 3 & 4 & \cdots & n & 1 \\ 3 & 4 & 5 & \cdots & 1 & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n-1 & n & 1 & \cdots & n-3 & n-2 \\ n & 1 & 2 & \cdots & n-2 & n-1 \end{bmatrix} \]
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paxtonw
28 posts
paxtonw
#2 h  Mar 30, 2025, 4:31 PM
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Let A be the given n x n matrix.

The last two rows are cyclic shifts of each other, meaning one is just a shifted version of the other.

So, the last two rows are linearly dependent.

A matrix with two linearly dependent rows has determinant 0.

By the way.. can anyone tell me why people ask these questions? Is it for school, work, etc? Or for fun.
This post has been edited 1 time. Last edited by paxtonw, Mar 30, 2025, 4:33 PM
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loup blanc
3583 posts
loup blanc
#3 h  Mar 30, 2025, 7:14 PM
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When $n=10$, $\det(A)=-55\times 10^8$.
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Soupboy0
339 posts
Soupboy0
#4 h  Mar 30, 2025, 7:20 PM
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When $n = 2, \det{A} = 3$
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vanstraelen
8983 posts
vanstraelen
#5 h  Mar 30, 2025, 7:46 PM
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$\det A_{n}=-\frac{n(n+1)}{2} \cdot n^{n-2}$.
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ysharifi
1672 posts
ysharifi
#8 h  Mar 30, 2025, 10:37 PM
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If you switch the rows (2, n), (3,n-1), ..., you'll get a circulant matrix and there's a formula for the determinant of circulant matrices. In your case, the answer is $(-1)^{\lfloor n/2 \rfloor}\frac{n^{n-1}(n+1)}{2}.$
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ddot1
24516 posts
ddot1
#9 h  Mar 31, 2025, 2:04 AM • 1 Y
Y by MihaiT
paxtonw wrote:
The last two rows are cyclic shifts of each other, meaning one is just a shifted version of the other.

So, the last two rows are linearly dependent.
The last two rows can't be linearly dependent, since that would mean that one is a multiple of the other.

As others mentioned, there's a nice formula for the determinant of a circulant matrix, where each row is a shift of the previous:
$$A=\begin{pmatrix}a_1& a_2&a_3&\cdots& a_{n-1}&a_n\\ a_n& a_1&a_2&\cdots& a_{n-2}&a_{n-1}\\ a_{n-1}& a_n&a_1&\cdots& a_{n-3}&a_{n-2}\\ \cdots\\ a_2& a_3&a_4&\cdots &a_n &a_1 \end{pmatrix}.$$It turns out that it has eigenvalues $$\begin{pmatrix}1\\ \zeta\\ \zeta^2\\ \vdots \\ \zeta^{n-1} \end{pmatrix},$$where $\zeta$ is any $n$th root of unity. The corresponding eigenvectors are $a_1+a_2\zeta+a_3\zeta^2+\cdots+a_n\zeta^{n-1},$ so that gives a nice factorization for the determinant: $$\det A=\prod_{\zeta^n=1}\left(a_1+a_2\zeta+a_3\zeta^2+\cdots+a_n\zeta^{n-1}\right).$$
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loup blanc
3583 posts
loup blanc
#10 h  Mar 31, 2025, 1:46 PM • 2 Y
Y by MS_asdfgzxcvb, MihaiT
Here $A$ is an anti-circulant matrix; to find a circulant matrix it is enough to consider $JA$, where $J$ is the permutation
$J_{1,1}=1$, $J_{i,n-i+2}=1$ for $i=2,\cdots, n$ and the others $J_{i,j}$ are $0$ (note that $J^2=I_n$).
Indeed, $JA=I+2P+3P^2+\cdots+nP^{n-1}$, where $P$ is the cycle $(n,n-1,\cdots,2,1)$. Let $f(x)=1+2x+\cdots+nx^{n-1}$.
Let $u=\exp(2i\pi/n)$; then $spectrum(P)=\{u^k;k=0,\cdots,n-1\}$ and $spectrum(JA)=(\lambda_k)_k=(f(u^k))_k$.
$\lambda_0=n(n+1)/2$. $f(x)=(\dfrac{x(1-x^n)}{1-x})'=\dfrac{(1-(n+1)x^n)(1-x)+x-x^{n+1}}{(1-x)^2}$ and if $x^n=1$ and $x\not= 1$, then
$f(x)=\dfrac{-n}{1-x}$. Up to the signum, $\det(JA)=\Pi_k \lambda_k=\pm \dfrac{n(n+1)n^{n-1}}{2\Pi_{k>0}(1-u^k)}$.
Exercise: $\Pi_{k>0}(1-u^k)= n$. Finally $\det(JA)=\pm \dfrac{n(n+1)n^{n-2}}{2}$.
This post has been edited 4 times. Last edited by loup blanc, Mar 31, 2025, 1:52 PM
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