Question

Let \(A\) be an \(n\times n\) matrix with \(a_{i,j}=\min\{i,j\}\) and \(1\le i\le n,~1\le j\le n\).

Find the determinant of \(A\).

Answer 1

Note: This question was adapted from this year's Putnam exam, which asked for the determinant of \(A\) but with \(a_{i,j}=\dfrac{1}{\min\{i,j\}}\). Also a fun problem.

Answer 2

If the notation isn't familiar, the matrix is
\[A=\begin{pmatrix}
1&1&1&\cdots&1&1&1\\
1&2&2&\cdots&2&2&2\\
1&2&3&\cdots&3&3&3\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\
1&2&3&\cdots&n-2&n-2&n-2\\
1&2&3&\cdots&n-2&n-1&n-1\\
1&2&3&\cdots&n-2&n-1&n
\end{pmatrix}\]

Answer 3

lol was trying to write that

Answer 4

the only thing i note that it sound symmetric to me :)

Answer 5

I don't have the precise solution for this version of the question (I'd actually started solving the original one while mistaking \(\min\{\}\) for \(\dfrac{1}{\min\{\}}\)... -_-)

Answer 6

... but I do have one method.

Answer 7

My approach involved writing \(A\) as a triangular matrix \(\bar{A}\) using a set of somewhat involved row operations, then finding the product of the diagonals.

Answer 8

isn't the determinant always 1 irrespective of order ?

Answer 9

with your approach you'd get all 1s in the diagonal, so yeah :)

Answer 10

i wish its a triangle matrices so det would be n! :P (just saying lol )

Answer 11

wondering how the problem with
\(a_{i,j}=\dfrac{1}{\min\{i,j\}}\)
would work :O

Answer 12

\[\prod_{i=2}^n\dfrac{-1}{i(i-1)} \]

Answer 13

-1/2, 1/12, -1/144,1/2880,...
is that what we get by that product .. .

Answer 14

yep, thats correct....how did u get that /

Answer 15

just used ur hint..

Answer 16

we do \(n-1\) row operations : \(R_i - R_{i-1}\)
\(2\le i \le n\)

Answer 17

cute :3

Answer 18

the inverse of this matrix looks interesting
its both symmetric and tridiogonal
http://www.wolframalpha.com/input/?i=inverse+%7B%7B1%2C1%2C1%2C1%7D%2C%7B1%2C2%2C2%2C2%7D%2C%7B1%2C2%2C3%2C3%7D%2C%7B1%2C2%2C3%2C4%7D%7D

Answer 19

i have tested upto n=5 i feel the pattern continues

Answer 20

yeah it looking nice

Answer 21

eh , i forgot all interesting things in matrices :P

Answer 22

n-1 column operations are also needed right ?
Rn - R(n-1)
and then
Cn - C(n-1)

Answer 23

*to make it a diagonal matrix

Answer 24

Oh yes, we can try finding the inverse step by step to see how it is tridiogonal
but for the determinant, triangular form is sufficient right ?

Answer 25

yes, correct

Answer 26

@ganeshie8 I'd ended up with the same,
\[\frac{(-1)^{n+1}}{n!(n-1)!}\]
Row operations and general procedure: Replace the \(j\)th column entry of the \(i\)th row, \(R_i\), with \(R_i-(i-1)!R_1\). This makes the last column all zero except for the first entry which is 1. A cofactor expansion along this column gives
\[\det(A)=(-1)^{n+1}\left(\prod_{k=1}^nk!\right)^{-1}\det(\bar{A})\]
where \(\bar{A}\) was used to denote the submatrix containing \(a_{i,j}\) with \(2\le i\le n\) and \(1\le j\le n-1\). All the while, \(\bar{A}\) is triangular and so \(\det(\bar{A})\) is the product of the diagonal entries, and that product looks like
\[\det(\bar{A})=\prod_{k=1}^{n-1}\left(\frac{(k+1)!}{k}-k!\right)\]
and so
\[\begin{align*}\det(A)&=(-1)^{n+1}\prod_{k=1}^{n-1}\frac{\dfrac{(k+1)!}{k}-k!}{(k+1)!}\\\\
&=(-1)^{n+1}\prod_{k=1}^{n-1}\frac{\dfrac{k+1}{k}-1}{k+1}\\\\
&=(-1)^{n+1}\prod_{k=1}^{n-1}\frac{1}{k(k+1)}\\\\
&=\frac{(-1)^{n+1}}{(n-1)!n!}
\end{align*}\]
Good to know I got that one right!

Answer 27

Wow! looks nice :) I think reducing the matrix from the bottom right corner would make it easy though

@SithsAndGiggles
I am not able to find putnam 2014 problems online. how did you get this problem ?

Answer 28

I had the luxury of taking it yesterday morning at my school. I can post the other problems if anyone is interested (unless there's some legal issue surrounding that?).

Answer 29

XD I wish you all the good luck for top place! let us know when they announce the results :) There should not be any issues as they will be publishing problems and solutions shortly
http://math.scu.edu/putnam/announcecJan.html

Answer 30

hhehe , never had something like this

Answer 31

I remember reading somewhere that the annual mean score is 0 out of 120, so I'm not too hopeful... and I was the only one taking the test this year.

Answer 32

here
http://www.artofproblemsolving.com/Forum/resources.php?c=2&cid=23&year=2014&sid=a13b46456e6c7f30ec620b3b0e10f03d

Answer 33

0 out of 120 ?! negative marks for wrong attempts is it! I read something like they rank based on the overall performance of a 3 member team ? @Marki looks artofproblemsolving is super fast in releasing keys xD

Answer 34

Patniss and Keeta in 7th Annual Putnal Games!
ROFL

Answer 35

*75th

Answer 36

since its art works ;) so no doubt its fast