I have a mapping $f:R^3\rightarrow R^3$ such that: $$x_1=(1+\eta)X_1$$ $$x_2=(1+\gamma)X_2$$ $$x_3=(1+\gamma)X_3$$ and I have to obtain the transformation matrix $\varepsilon$ corresponding to this mapping. I know the matrix has to be$$\varepsilon=\begin{pmatrix}1+\eta& 0& 0\\ 0& 1+\gamma& 0\\ 0& 0& 1+\gamma\end{pmatrix}.$$ I want to get this matrix using Mathematica but I don't know how. I tried the following:

x1 = (1 + \[Eta]) X1;
x2 = (1 + \[Gamma]) X2;
x3 = (1 + \[Gamma]) X3;
\[Epsilon] = {{\[Epsilon]11, \[Epsilon]12, \[Epsilon]13}, \
{\[Epsilon]21, \[Epsilon]22, \[Epsilon]23}, {\[Epsilon]31, \
\[Epsilon]32, \[Epsilon]33}};
Solve[{x1, x2, x3} == \[Epsilon] .{X1, X2, X3}, {\[Epsilon]11, \[Epsilon]12, 
\[Epsilon]13, \[Epsilon]21, \[Epsilon]22, \[Epsilon]23, \[Epsilon]31, 
\[Epsilon]32, \[Epsilon]33}] // Simplify

but I got the message Solve::svars: Equations may not give solutions for all "solve" variables. and some strange output.

How to let Mathematica correctly solve for $\varepsilon$ ?

  • $\begingroup$ It's giving you the correct solution. It's just that you have three equations and 9 variables, so there's infinitely many solutions. Let sol be your solution and try (\[Epsilon] /. sol).{X1, X2, X3} // Simplify. Basically, it means that you're free to set 6 of your variables to whatever you want (like zero). $\endgroup$ Commented Oct 7, 2017 at 10:37
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    – Michael E2
    Commented Oct 7, 2017 at 12:56

2 Answers 2


In general, for a linear form, you can use the linear component (part [[2]]) of CoefficientArrays:

Normal@CoefficientArrays[{x1, x2, x3}, {X1, X2, X3}][[2]]

Mathematica graphics


Solved the problem in the following way:

Subscript[x, 1] = (1 + \[Eta]) Subscript[X, 1];
Subscript[x, 2] = (1 + \[Gamma]) Subscript[X, 2];
Subscript[x, 3] = (1 + \[Gamma]) Subscript[X, 3];
\[Epsilon] = Table[Coefficient[Subscript[x, i], Subscript[X, j]], {i, 3}, {j, 3}]
\[Epsilon] // MatrixForm

It looks like I shouldn't have dealt with it as an equation to be solved.

  • $\begingroup$ If solving isn't necessary, you could also use DiagonalMatrix[1 + \[Eta], 1 + \[Gamma], 1 + \[Gamma]]. $\endgroup$ Commented Oct 7, 2017 at 11:06
  • $\begingroup$ Ok, thank you for this shortcut! $\endgroup$
    – Tofi
    Commented Oct 7, 2017 at 12:18
  • 1
    $\begingroup$ Right, no need to solve for anything, it's really a matter of converting from explicit equations to matrix algebra. $\endgroup$ Commented Oct 7, 2017 at 15:16

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