# Tag Info

9

Here's a nice one-liner: TransformationFunction[{{1, 0, 0, 0}, {0, 0, -1, 0}, {0, 1, 0, 3}, {0, 0, 0, 1}}] // InverseFunction TransformationFunction[{{1, 0, 0, 0}, {0, 0, 1, -3}, {0, -1, 0, 0}, {0, 0, 0, 1}}] Note that TransformationFunction[] is the head of the results returned by geometric *Transform functions, which take a homogeneous transformation ...

6

If you have a homogenous transformation matrix of the form $$\begin{bmatrix} \mathrm{R_{3 \times 3}} & \mathrm{d}_{3 \times 1} \\ 0_{1\times 3} & 1_{1\times 1} \end{bmatrix}$$ Then the inverse is given by $$\begin{bmatrix} \mathrm{R}^{-1} & -\mathrm{R}^{-1}\mathrm{d} \\ 0 & 1 \end{bmatrix}$$ Therefore, if your ...

3

Reduce[expr, x, Reals] will be your friend here, but it can take a bit of work to parse its result. Here's a solution that should work for any expression, not just polynomials (at least for the small set of examples I tried). RealInverse[a_. x_^q_Integer?Positive + b_., x_] /; FreeQ[{a, b}, x] := Surd[(x-b)/a, q] RealInverse[expr_, x_] := Module[{y, ...

0

What about this? inverseFunc[x_] = x /. Solve[x^3 == y, x, Reals][[1]] /. y -> x; N@inverseFunc[3] N@inverseFunc[-1] (*1.44225*) (*-1.*)

0

One way to find and plot the real root of x^3 == y is x /. Solve[x^3 == y, x, Reals] (* {Root[-y + #1^3 &, 1]} *) Plot[%, {y, -10, 10}, AxesLabel -> {y, x}] This works for any polynomial f[x] == y that has a real root, as requested in the question. Is this what you had in mind? Addendum In response to the OP's comment below, a do-it-yourself ...

2

One possibility is to use a rectangular matrix with one block being the identity matrix. Let us consider the example in block matrix form. \$\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{pmatrix} \cdot \begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} ...

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