Reduce equations formed of symbolic matrices and vectors

Question

I was defining the following assumptions

$Assumptions = {A ∈ Matrices[{3, 3}, Reals], 
  x ∈ Vectors[3, Reals], b ∈ Vectors[3, Reals]}

Reduce[A . x == b, x]

or

Solve[A . x == b, x]

None of them is working. I was expecting something in the lines of x = Inv(A).b

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I was expecting something in the lines of x = Inv(A).b presupposes that A is non-singular. — Syed, Jan 12 '23 at 18:58
Can I define A as non singular and get then the desired result. I just want to reduce equations... — Andrei Chalapco, Jan 13 '23 at 19:17
If you want others to see your comments you should tag them. For instance you should write something like @Syed — bmf, Jan 14 '23 at 10:57
I don't know of a way to restrict 3x3 matrices to the non-singular ones only. Thanks @bmf for the heads up. — Syed, Jan 14 '23 at 11:02

score 1 · Answer 1 · answered Jan 12 '23 at 16:50

1

You can proceed in the following way to obtain an analytic solution, albeit not as compact as one might have hoped for.

n = 3;
amat = Array[a, {n, n}];
xvec = Array[x, n];
bvec = Array[b, n];
sltn = Solve[amat . xvec - bvec == 0, xvec] // Flatten // 
   FullSimplify;

And now you can check your solution

xvec /. sltn // MatrixForm

You can, also, check explicitly with the expected compact answer

LinearAlgebra`Private`ZeroArrayQ[(Inverse[amat] . bvec // 
    FullSimplify) - (xvec /. sltn)]

answered Jan 12 '23 at 16:50

bmf

This seems overly complicated for such a simple linear equations ... where I just want a reduce of the equation – Andrei Chalapco Jan 12 '23 at 16:58
@AndreiChalapco I understand that it seems like that, but perhaps you want to have a look here and understand why I suggested the above solution :-) – bmf Jan 12 '23 at 17:01

1 Answers1