This is a degenerate case!
A = {{1/72, 1/63, 1/56}, {1/56, 1/63, 1/72}};
MatrixForm@A
A.{x, y, z}
MatrixRank[A]
Consider first the kernel of A: a vector V
V = First@NullSpace[A]
A.(k V) // Simplify
Complete V by other sandard vectors, and change of basis:
B = Transpose@{{1, 0, 0}, {0, 1, 0}, V};
MatrixForm@B
Det[B]
AI = A.B;
MatrixForm@AI
MatrixForm[AI.Inverse[B]]
The tranformed matrix AI is better-suited! Now, we can restrict ouselves to a 2-dimensional non-degenerated problem:
TA = Transpose@Drop[Transpose@AI, -1];
MatrixForm@TA
{X, Y} = (Inverse[TA]).{12/3, 14/3}
TA.{X, Y}
AI.{X, Y, z}
Come back in the original basis, and verify:
B.{X, Y, z}
A.B.{X, Y, z}
