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0

To clear up ambiguities, the following code finds the transformation from the points in xy space $(0,1),(1,0)$ and $(0,0)$ to the points in uv space $a,b$ and $c$ projected onto $\overline{ab}$ respectively. With[{a = {ax, ay}, b = {bx, by}, c = {cx, cy}, m = {{mx, mxy}, {myx, myy}}}, With[{v = Projection[c - b, a - b, Dot] + b}, m /. Simplify@ First@...

8

Your question seems to miss some points in What topics can I ask about here?. This answer is for helping you to write the code by yourself and if you encounter any problem, feel free to post another question. In Mathematica use: (* comment *) syntax for commenting your code LowerTriangularize function for accessing the lower triangular part of a matrix ...

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If you have got all your matrices in matrices then use ResourceFunction["LinearlyIndependent"] for example: LinearlyIndependent = ResourceFunction["LinearlyIndependent"]; matrices = RandomInteger[20, {16, 4, 4}]; LinearlyIndependent[Flatten /@ matrices] To extract your matrices you can do this: onebigmatrix = RandomInteger[20, {4, 64}]; ...

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As suggested by @bill s, you can write x = Array[a, 3]; deriv = D[x . x, {x}] (* {2 a[1], 2 a[2], 2 a[3]} *) Note that the you need to put {x} rather than x (otherwise it will attempt to interpret the 2nd item in the list as the order of the derivative - see the help for D - the syntax is rather over loaded). The easiest way to substitute values is perhaps ...

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A simple way to solve your equations is: e1 == W1 /. {x -> 0}; e2 == (D[W1, {x, 1}]) /. {x -> 1}; e3 = (W1 /. {x -> z}) - (W2 /. {x -> z}); e4 = ((D[W1, {x}]) /. {x -> z}) - ((D[W2, {x}]) /. {x -> z}); Solve[{e1, e2, e3, e4} == {0, 0, 0, 1}, {A1, A2, B1, B2}] However, look at your equations: TableForm[Transpose[Append[Transpose[R], {0, 0, ...

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