Re: System Ax=b!

*To*: mathgroup at smc.vnet.net*Subject*: [mg32231] Re: [mg32224] System Ax=b!*From*: BobHanlon at aol.com*Date*: Wed, 9 Jan 2002 03:17:19 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

In a message dated 1/7/02 3:44:05 AM, irc_rebecca at yahoo.com writes: >I would really be happy, if someone could help me. >I was trying to write a program in mathematica on how >to solve the system Ax=b with partial pivoting, where >A is a three-diagonal matrix(a matrix that only has >numbers on the main diagonal and on two diagonals that >are lying on both sides from that main diagonal, every >other element is 0), but came nowhere. >That is why it would be nice if someone has any >suggestions how I could do that. >I am looking forward on your answer and am thanking >you in advance! > n = 3; A = Table[If[Abs[i-j]<=1, ToExpression[ "a" <> ToString[i]<>ToString[j]], 0],{i,n},{j,n}]; x = Table[ToExpression[ "x" <> ToString[i]],{i,n}]; b = Table[ToExpression[ "b" <> ToString[i]],{i,n}]; Solve[A.x==b, x] Bob Hanlon Chantilly, VA USA