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I have a symbolic matrix given as

$$ A = \begin{bmatrix} 0 & 12.5 & 12.5 k_1-5. \\ 12.5 & 12.5 k_1-5. & 2.\, -5. k_1 \\ k_1 & 0. & 0. \\ \end{bmatrix} $$ when I calculate the inverse of this matrix with Inverse, I obtain the following matrix picture.

The problem with this matrix is that all $10^{-15}$ should be rounded to $0$. I do this by using Chop. Basically I just use

Chop[expr, 10^-1]

which then gives me picture2.

However, for example, (1,3) should just simplify to $\frac{1}{k_1}$, but I cannot achieve this with FullSimplify. Is there any other way?

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2 Answers 2

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Try define the matrix with exact numbers

a = {{0, 25/2, 25/2 k1 - 5}, {25/2, 25/2 k1 - 5, 2 - 5 k1}, {k1, 0, 0}}

and then

FullSimplify[Inverse[a]]
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  • $\begingroup$ Thanks. I have a long list of expressions that lead to this result. How should I approach it so I do not have to manually define my matrix? I mean, is there something like format short for Mathematica that I can default to, if that makes sense. $\endgroup$
    – ofey
    Commented May 1, 2021 at 16:54
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    $\begingroup$ you can do FullSimplify[ Inverse[Rationalize[{{0., 12.5, -5. + 12.5*k1}, {12.5, -5. + 12.5*k1, 2. - 5.*k1}, {k1, 0., 0.}}]]] $\endgroup$
    – Andreas
    Commented May 1, 2021 at 16:58
  • $\begingroup$ Aha. Much appreciated! $\endgroup$
    – ofey
    Commented May 1, 2021 at 16:59
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Mathematica can work with approximate numbers (as you do) or exact numbers. Exact numbers work much better in this case

M = {{0, 25/2, 25/2 k - 5}, {25/2, 25/2 k - 5, 2 - 5 k}, {k, 0, 0}};

Inverse[M] // FullSimplify
(* {{0, 0, 1/k}, {4/(125 k), 2/(25 k), -(1/k^2)}, {2/(25 k), 2/(
  5 (2 - 5 k) k), 5/(k^2 (-2 + 5 k))}} *)
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  • $\begingroup$ Thanks. I assume it is the fact that I have a decimal point makes the numbers approximate in my case? $\endgroup$
    – ofey
    Commented May 1, 2021 at 16:56

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