linear algebra - How to Solve or LinearSolve $A = I$ matrix equation?

I'd like to solve this equation for $A = B$ where $B = I$, which represents 3 systems of 3 linear equations, for $a, b, c, d, e, f$, without writing LinearSolve 3 times. What is a simple way to accomplish this?

Input:

   A = {
   {a + b, -a + b, a + 2 b}, 

   {c + d, -c + d, c + 2 d}, 
   {e + f, -e + f, e + 2 b}
   };
MatrixForm[A]
B = IdentityMatrix[3]
(*want to solve A == B, but probably wrong...*)
M = LinearSolve[A, IdentityMatrix[3]]
MatrixForm[M]

Answer

As the documentation says : LinearSolve[m,b] finds an x which solves the matrix equation m.x == b, i.e. in your case it finds x such that A.x == B (Dot[A,x] == B). However your task is to find A solving an adequate system of 9 linear equations for 6 variables a,b,c,d,e,f knowing that B is an IdentityMatrix. You are trying to solve an overdetermined system of linear equations and there could exist any solutions only if certain compatibility conditions were satisfied.

For your task use simply Solve :

Solve[A == B, {a, b, c, d, e, f}]

{}

or

Reduce[A == B, {a, b, c, d, e, f}]

False

This means that there are no solutions, i.e. the above equation is contradictory. You could use Variables[A] instead of specifying variables {a, b, c, d, e, f}.

Consider a different matrix equation where we have 4 unknowns and 4 independent equations e.g. :

A1 = {{a + b, a - 2 b}, {a - c, c + d}};
Solve[ A1 == IdentityMatrix[2], {a, b, c, d}] 

{{a -> 2/3, b -> 1/3, c -> 2/3, d -> 1/3}}

i.e. there is only one solution.

Edit

Inverse[A] could be a solution assuming that you wanted x in the matrix equation A.x == IdentityMatrix[3] when A was given. There exists an inverse matrix to A under this condition :

Det[ A] != 0

i.e.

-4 b^2 c + 4 a b d + 4 b c f - 4 a d f != 0

Neither A exists nor this assumption can be satisfied when A is defined as in your question and B is an identity matrix.