0
$\begingroup$

How I can calculate product of series matrices: A= Im...In. In-1 ...I1

which In = {{n,7n-2,3,2n-1},{3,4,n,6n-1},{9,3,n,n-3},{n-2,1,8,7n/3}}

thanks!

$\endgroup$
3
$\begingroup$

Here is a neat way:

Array[Function[n, {{n, 7 n - 2, 3, 2 n - 1}, {3, 4, n, 6 n - 1}, {9, 3, n, n - 3}, {n - 2, 1, 8, 7 n/3}}], 10, 1, Dot]
$\endgroup$
  • $\begingroup$ Thanks alot! It takes me alot of time to do. That's great! $\endgroup$ – Haimuoi Jul 12 '16 at 13:09
1
$\begingroup$

You can do

j=10;
mat = Table[{{n, 7 n - 2, 3, 2 n - 1}, {3, 4, n, 6 n - 1}, {9, 3, n, 
     n - 3}, {n - 2, 1, 8, 7 n/3}}, {n,j}];
Dot @@ Table[mat[[i]], {i,j}]//N

Where j is the order to which you want to calculate the product, this gives as answer:

{{7.28971*10^13, 1.09566*10^14, 7.92596*10^13, 2.71886*10^14}, {8.76822*10^13, 1.31681*10^14, 9.53649*10^13, 3.27209*10^14}, {9.74138*10^13, 1.46673*10^14, 1.05908*10^14, 3.63366*10^14}, {6.90965*10^13, 1.03702*10^14, 7.50509*10^13, 2.57084*10^14}}

$\endgroup$
  • $\begingroup$ thanks, it done. You are very kind, thanks alot! :) $\endgroup$ – Haimuoi Jul 12 '16 at 13:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.