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| Hopefully the modified code below more closely matches the initial description: s = 3; diagonal = A + Table[i (s - i + 1), {i, 2 s + 1}] offDiagonal = (2 s)^(1/2) \[Epsilon]^(-1) Table[-i, {i, 1, 2 s}]; m = Table[ If[i |
| Ah okay. Thank you very much! Why can't the solution be substituted into B[u] normally though? |