linear algebra - Generating a vector of dummy variables

So I'm the situation of needing analytical solutions to a family of equations of the form Ax=b, where A is an nxn matrix. I've written a function that does what I want, but I'm currently using a bit of a hack to generate arbitrary (dummy) variable names for the n-vector 'x' :

Map[Clear, Table["x" <> ToString[j], {j, n}]];
x = ToExpression[Table["x" <> ToString[j], {j, n}]];

Can anyone suggest a more elegant way of accomplishing this task?

Answer

You have several possibilities for this. The probably two easiest methods are first to use Unique

ClearAll[x];
x = Table[Unique["x"], {10}]
(* {x7, x8, x9, x10, x11, x12, x13, x14, x15, x16} *)

The good thing is, that when some of your variables xn are already defined, Unique will not return them. It always gives you fresh, unused symbols.

The other thing you should consider is, that x[1] can be used like a symbol too, although it isn't one. Therefore

ClearAll[x];
vars = Table[x[i], {i, 10}]
(* Out[15]= {x[1], x[2], x[3], x[4], x[5], x[6], x[7], x[8], x[9], x[10]} *)

Can be used as valid variables too. In any case you should watch out that your variables are not assigned to values accidently. This is a common source of errors if you use them in combination with Solve or its friends.

Why did I use x= in the first example and vars= in the second one?

Let's take a very simple example

a = b[1];
OwnValues[a]
(*
  {HoldPattern[a] :> b[1]}
*)

and now we assume that the b would be an a, than we would get an OwnValue-rule like

HoldPattern[a] :> a[1]

Therefore, the moment you use a an substitution process starts which is only stopped by the $RecursionLimit, because a is evaluated into a[1] which again contains an a in the front. This is repeatedly replaced.

Therefore, if you want to use the second approach, don't call it like x = Table[x[i], {i, 10}].