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4

Iterating @J.M.'s comment: This problem has no exact solution. With[{ matrix = {{0.8111, 0.4867, -0.3244}, {a, b, 0}, {c, d, e}} }, Print[matrix.Transpose[matrix]]; Solve[ matrix.Transpose[matrix] == IdentityMatrix[3], {a, b, c, d, e} ] ] (* {{0.999995,... *) (* {} *) That is, the first column's first entry is not $1$, so there is ...


2

Using Smith Normal Form you can get two integer matrices with determinant 1 that would satisfy the equation $$X1.K.X2 = K_2$$ May be this is a good start you can use... resK = SmithDecomposition[K]; MatrixForm /@ resK resK2 = SmithDecomposition[K2]; MatrixForm /@ resK2 resK[[2]] == resK2[[2]] (* True *) K2 == ...


2

I learned a lot from WReach's and Leonid's answers and I'd like to make a small contribution: It seems worth emphasizing that the primary intention of the list-valued second argument of Flatten is merely to flatten certain levels of lists (as WReach mentions in his List Flattening section). Using Flatten as a ragged Transpose seems like a side-effect of ...


1

SeedRandom[5] pts = RandomReal[1, {6, 2}]; pts = pts[[FindShortestTour[pts][[2]]]]; am = RandomChoice[{.7, .3} -> {0, 1}, {6, 6}]; AdjacencyGraph using the polygon vertices as vertex coordinates: Labeled[AdjacencyGraph[am, VertexCoordinates -> pts, DirectedEdges -> False, Vertexlabels->"Name", Prolog -> {Yellow, ...



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