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SORRY, I made a typo. it should be $D \not= 0$,not $D>0$.

It is a bit like Vandermonde determinant $$D=$$ $$\begin{vmatrix} 1 & 2 & 3&\cdots &2008&2009 & 2010 & 2011\\ 2^2 & 3^2 &4^2& \cdots&2009^2&2010^2 & 2011^2 &2012^2 \\ 3^3 & 4^3 &5^3&\cdots &2010^3&2011^3&2012^3 &2012^3 \\ \cdots &\cdots &\cdots &\cdots&\cdots&\cdots &\cdots &\cdots\\ k^k&(k+1)^k&\cdots&2011^k&2012^k&\cdots&2012^k&2012^k\\ \cdots &\cdots &\cdots &\cdots &\cdots &\cdots\\ 2010^{2010}&2011^{2010}&2012^{2010}&\cdots&2012^{2010}&2012^{2010}&2012^{2010}&2012^{2010}\\ 2011^{2011} &2012^{2011} &2012^{2011}&\cdots &2012^{2011} &2012^{2011} &2012^{2011}&2012^{2011} \end{vmatrix}$$

Is the above determinant $D\not= 0$?

the exam is only need to show $D \not= 0$,maybe figure out $D$ is impossible. and I edit it, maybe more clearly.

thanks for comments and answer.

FIRST PAGE OF EXAM:

enter image description here

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2  
The pattern isn't clear. It looks like the diagonals are increasing powers of a fixed base, but also it looks like the lower right half is all powers of $2012$. By the way, this isn't a contest problem, is it? –  Gerry Myerson Nov 23 '12 at 3:48
    
yes,the diagonals are increasing powers of a fixed base $2011$, thanks for your comment,I will modify it. and it is a contest problem. –  Laura Nov 23 '12 at 4:04
3  
A problem from an ongoing contest? Is there a link to a contest website? –  Gerry Myerson Nov 23 '12 at 4:31
    
OK,it is a graduate entrance exam of Peking university in 2012. but there is no official version. and there is a link to its original which is recalled by candidate:math.org.cn/…. you can find it from the PDF file. :) –  Laura Nov 23 '12 at 5:31

2 Answers 2

up vote 4 down vote accepted

I am posting this as an answer since it is too long for a comment. In general, note that $$A_{n+1} = \begin{bmatrix} A_n & u_n\\ v_n & (2n+3)^{n+2}\end{bmatrix}$$ where $$u_n = \begin{bmatrix}(n+2)\\(n+3)^2\\ (n+4)^3\\ \vdots \\ (2n+2)^{n+1} \end{bmatrix}$$ $$v_n = \begin{bmatrix} (n+2)^{n+2} & (n+3)^{n+2} & (n+3)^{n+2} & \cdots & (2n+1)^{n+2} & (2n+2)^{n+2}\end{bmatrix}$$ and $A_0 = \begin{bmatrix} 1 \end{bmatrix}$. I computed the value of determinant for $A_k$ for $k=0$ to $k=5$ using MATLAB and below are the values.

$$\det(A_0) = 1; \det(A_1) = 1; \det(A_2) = 4; \det(A_3) = 72; \det(A_4) = 6912; \det(A_5) = 4147200$$ $$\det(A_6) = 17915904000;$$

Feeding this sequence in OEIS gave that a possible sequence is $$\det(A_{n-1}) = (n-1)! \times \prod_{k=1}^{n-2} (n-k)!$$ where $n \in \mathbb{Z}^+$. Once we have this, we can hope for a proof by induction.

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Would you mind clarifying the pattern for me? I interpreted $A_2$ as $$A_2=\begin{pmatrix}1 & 2 & 3 \\ 2^2 & 3^2 & 4^2 \\ 3^3 & 4^3 & 4^3\end{pmatrix}$$ but that gives me a determinant of $-57$. –  EuYu Nov 23 '12 at 5:17
    
Agreeing with EuYu. OP has not made things completely clear, but the bottom row in the body of the question doesn't look like the $v_n$ in the answer. However, the strategy of calculating the first few and then consulting the oracle at oeis may well pay off. –  Gerry Myerson Nov 23 '12 at 5:59
    
The sequence as I interpreted it seems to give $$1,\ 1,\ -57,\ -49099,\ 820631141,\ \cdots$$ Doesn't seem familiar to oeis. –  EuYu Nov 23 '12 at 6:30
    
@EuYu, your $++--$ pattern seems simple enough. I pasted the first page of the examination into the question. –  Will Jagy Nov 23 '12 at 6:41
    
I am sorry. I made a typo. it need to prove $D \not=0$,not $D >0$. –  Laura Nov 23 '12 at 7:04

For the record, the first 20 items in the obvious sequence, computed using

f[n_] := Det[Table[If[i + j - 1 > n, n + 1, i + j - 1]^i, {i, 1, n}, {j, 1, n}]]

in Mathematica, are

\begin{gather}\small 1 \\\small 1 \\\small -57 \\\small -49099 \\\small 820631141 \\\small 327729323411785 \\\small -3722810907570484463905 \\\small -1395732893854465247614690459535 \\\small 19654898955250800722122617078850379547649 \\\small 11654863933592492659392873622726534581699247698016601 \\\scriptsize -322373512908450688046702815032229983463944019597828165058661862873 \\\scriptsize -456322749650163764951140063242869266114020813743510155958815030863103530198278707 \\\scriptsize 359763590088436766228887856876796313253226027493473716114855323607493415839311215 63741918002361805 \\\scriptsize 170784854835456576916145215980211059998029046033398578352565914083160320685101869 469195969039220374230020741584750601 \\\scriptsize -52473397006405670917113840563365195251372913721092487498198489291680306518569632 237546150574105804810447098763276030563961357045407349377 \\\scriptsize -11161338095346579407742610541224509857161891767489700245561198039424133307461760 19001249905947908531883035800858849726377707025535086543207728368627630155179671 \\\tiny 175043317755348307253812637289721345473677623476953037216431178436787653927346563 419897174832441679689898012101336598764499805173923767791098038693711420817887509 1111661312135016606441 \\ \end{gather}

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thank you. your calculation have reminded me that i made a typo. it should to prove $D\not=0$,not $D>0$. I am sorry for my mistake... –  Laura Nov 23 '12 at 7:09
1  
@R_bot, I posted the first page of the examination at the end of your question. Please make your very best, most careful translation of the question 2 text and edit it into your question just before the JPEG. If you need help with a good quality translation get it. –  Will Jagy Nov 23 '12 at 19:38
    
thank you. Will Jagy. I had modified a typo. now I can ensure there is no translate error. –  Laura Nov 24 '12 at 4:13

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