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Hi this is my first post and this is a problem that has bothered me a lot.

Suppose we have the $2012\times2012$ square matrix:

$\begin{bmatrix}-3&1&1&1&1&1&\ldots&1&1&1&1\\1&2&1&1&1&1&\ldots&1&1&1&1\\1&1&-3&1&1&1&\ldots&1&1&1&1\\1&1&1&2&1&1&\ldots&1&1&1&1\\1&1&1&1&-3&1&\ldots&1&1&1&1\\1&1&1&1&1&2&\ldots&1&1&1&1\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ldots&\vdots&\vdots&\vdots&\vdots\\1&1&1&1&1&1&\ldots&-3&1&1&1\\1&1&1&1&1&1&\ldots&1&2&1&1\\1&1&1&1&1&1&\ldots&1&1&-3&1\\1&1&1&1&1&1&\ldots&1&1&1&2\end{bmatrix}$

How can I write it using Mathematica? Of course I need to define something like a function which will assign elements to the positions of the matrix.

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    $\begingroup$ You shouldn't be in such a hurry to Accept (green check-mark) an answer, as it may discourage other, potentially better answers. $\endgroup$
    – Mr.Wizard
    Sep 24, 2012 at 19:19

4 Answers 4

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Try something like

matrix = SparseArray[{{i_, i_} /; OddQ[i] -> -3, {i_, i_} /; EvenQ[i] -> 
 2}, {2012,2012}, 1] // Normal;

If you are interested in the formal solution for it determinant,

Clear[matrix];
matrix[n_] :=  SparseArray[{{i_, i_} /; OddQ[i] -> -3, {i_, i_} /; EvenQ[i] -> 
 2}, {n, n}, 1];

you can start guessing a recursion from 

Table[matrix[n] // Det, {n, 2, 32, 2}]

Thanks to Mr Wizard's advice you can in fact do

   f = Table[matrix[n] // Det, {n, 2, 32, 2}] // FindSequenceFunction;
   f[n]

(* (-1)^n 4^(n-1) (3 n+4) *)

And for the Odd matrices (following R.M's request)

   Table[matrix[n] // Det, {n, 1, 31, 2}] //FindSequenceFunction // #[n] &

(* 3 (-1)^n 4^(n-1) n *)

   f/@ Range[16]
   f[2012/2] // N

(* 3.552922584185648*10^608 *)

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  • $\begingroup$ Ok thanks, but I need to find the determinant, how can this be done? $\endgroup$
    – Papadopoulos Geοrgios
    Sep 24, 2012 at 19:10
  • $\begingroup$ you type Det[matrix] though i) it might take a while and ii) its probably a big number $\endgroup$
    – chris
    Sep 24, 2012 at 19:10
  • $\begingroup$ for instance for a 200x200 matrix of this type the determinant is 122127291363683260941189119017928357791687427527492255482904576 $\endgroup$
    – chris
    Sep 24, 2012 at 19:13
  • $\begingroup$ Thanks again, if I replace the {2012,2012} with {n,n} will I get the general formula of the determinant? $\endgroup$
    – Papadopoulos Geοrgios
    Sep 24, 2012 at 19:13
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    $\begingroup$ Less guessing if you use FindSequenceFunction ;-) $\endgroup$
    – Mr.Wizard
    Sep 24, 2012 at 19:21
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SparseArray[Band[{1, 1}, {12, 12}] -> {-3, 2}, Automatic, 1] // MatrixForm

Mathematica graphics

Replace {12, 12} with {2012, 2012} for the full array.

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    $\begingroup$ +1. And to implement your FindSequenceFunction hint, take a look at a[n_] := SparseArray[Band[{1, 1}, {n, n}] -> {-3, 2}, Automatic, 1]; Partition[Table[ 2^(2 - n - Mod[n, 2]) Det[a[n]] , {n, 2, 50, 1}] , 2] // MatrixForm. $\endgroup$
    – whuber
    Sep 24, 2012 at 19:58
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    $\begingroup$ @Mr.Wizard beyond readability, any advantage of using Band? $\endgroup$
    – chris
    Sep 24, 2012 at 20:27
  • $\begingroup$ @chris It is also about 1400 times faster for this specific problem on my system. $\endgroup$
    – Mr.Wizard
    Sep 24, 2012 at 20:32
  • $\begingroup$ @Mr.Wizard ah! I see your point about a better answer then ;-) Well I already had given you my vote so... $\endgroup$
    – chris
    Sep 24, 2012 at 20:33
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    $\begingroup$ @chris My comment was not to slight your answer. Your answer is more complete than mine and you were engaging the OP to discover and address his real needs, both of which merit the Accept IMHO. Rather, I frequently encourage users to wait a day before choosing an answer as there are people in different time zones who will not even have a chance to see the question before it is "concluded" (if you will). $\endgroup$
    – Mr.Wizard
    Sep 24, 2012 at 20:59
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Just for completeness sake, here is an even more efficient way to construct the SparseArray

n = 123013;
AbsoluteTiming[
 pos = Transpose[{#, #}] &[Range[n]];
 vals = ConstantArray[-3, {n}];
 vals[[2 ;; -Mod[n, 2] - 1 ;; 2]] = -2;
 SparseArray[pos -> vals, {n, n}, 1]
 ]
(*{0.009, Null}*)
AbsoluteTiming[
 SparseArray[Band[{1, 1}, {n, n}] -> {-3, 2}, Automatic, 1]
 ]
(*{0.482, Null}*)

Using one rule is generally a very efficient way to generate a SparseArray

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  • $\begingroup$ You can make this twice as fast by using pos = {#, #}\[Transpose] &@Range@n; vals = ConstantArray[-3, n]; :-) (+1 btw) $\endgroup$
    – Mr.Wizard
    Sep 25, 2012 at 5:59
  • $\begingroup$ @Mr.Wizard, you are right, why not go all the way. Fixed. $\endgroup$
    – user21
    Sep 25, 2012 at 6:05
  • $\begingroup$ OK, I also did, that, but that seemed small compared to the Transpose. $\endgroup$
    – user21
    Sep 25, 2012 at 6:11
  • $\begingroup$ Interestingly I get about a 50% speed improvement of the entire operation by using ConstantArray in place of Table, over the intermediate version. Maybe Table is faster in v8? $\endgroup$
    – Mr.Wizard
    Sep 25, 2012 at 6:17
  • $\begingroup$ Hm, yes this may be due to version or OS. It's sure good to have both. But an improvement of Table is also conceivable. Something like this can be very hard to track down to a (or several) courses. $\endgroup$
    – user21
    Sep 25, 2012 at 6:25
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Yet another way (with timings similar to ruebenko's):

 values = ArrayPad[{-3, 2}, {0, n - 2}, "Periodic"];
 diag = Transpose[Range[{n, n}]];
 SparseArray[diag -> values, {n, n}, 1]]
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  • $\begingroup$ a quick note: ArrayPad does not return a packed array, so, it is less efficient in space. $\endgroup$
    – user21
    Sep 25, 2012 at 10:13
  • $\begingroup$ if you see crashes, please report them to the support such that they can be fixed. $\endgroup$
    – user21
    Sep 25, 2012 at 11:43
  • $\begingroup$ that said, the fact that some function does/can not return a packed array does not mean that it is more susceptible to crashes. Out of memory, and it's subsequent kernel shut down yes. $\endgroup$
    – user21
    Sep 25, 2012 at 11:50

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