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I have the following matrix:

$$ \left( \begin{array}{ccc} 13 & 9.1 & 8.19 & 8.281 & 8.9271\\ 9.1 & 8.19 & 8.281 & 8.9271 & 10.02001\\ 8.19 & 8.281 & 8.9271 & 10.02001 & 11.562759\\ 8.281 & 8.9271 & 10.02001 & 11.562759 & 13.6147921\\ 8.9271 & 10.02001 & 11.562759 & 13.6147921 & 16.27802631\end{array} \right) $$

When I find the inverse of this matrix on Excel, I get:

$$ \left( \begin{array}{ccc} 5.657342657 & -48.60139861 & 124.4172494 & -122.3776224 & 40.79254079\\ -48.6013986 & 467.9916897 &-1266.958328 & 1288.221582 & -438.9197404\\ 124.4172494 & -1266.958328 & 3565.198078 & -3723.319165 & 1293.334933\\ -122.3776224 & 1288.221582 & -3723.319165 & 3967.84588 & -1399.744047\\ 40.79254079 & -438.9197404 & 1293.334933 & -1399.744047 & 499.9085881\end{array} \right) $$

These values are not exact values. I can indefinitely keep adding more and more decimal places on Excel - However, I need exact values (in the form of a fraction).

I tried converting the values in the initial matrix into fractions before inverting it (the following matrix), but had no luck:

$$ \left( \begin{array}{ccc} 130/10 & 91/10 & 819/10^2 & 8281/10^3 & 89271/10^4\\ 91/10 & 819/10^2 & 8281/10^3 & 89271/10^4 & 1002001/10^5\\ 819/10^2 & 8281/10^3 & 89271/10^4 & 1002001/10^5 & 11562759/10^6\\ 8281/10^3 & 89271/10^4 & 1002001/10^5 & 11562759/10^6 & 136147921/10^7\\ 89271/10^4 & 1002001/10^5 & 11562759/10^6 & 136147921/10^7 & 1627802631/10^8\end{array}\right) $$

How can I get an exact inverse? I'd like to do some matrix multiplication with the exact inverse, and so a method to do that would also be much appreciated.

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  • $\begingroup$ why do you need it? usually there are faster and more stable computational methods than working with the inverse directly $\endgroup$ – gt6989b Jul 31 '16 at 2:31
  • $\begingroup$ Multiplying the inverse with another matrix gives me a 1x5 matrix with the coefficients of a fifth order polynomial (regression). Seven decimal places (as in the second matrix above) were not enough for my purposes. $\endgroup$ – StopReadingThisUsername Jul 31 '16 at 2:33
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The inverse is:

$$\begin{pmatrix} \dfrac{809}{143} & -\dfrac{6950}{143} & \dfrac{53375}{429} & -\dfrac{17500}{143} & \dfrac{17500}{429} \\ -\dfrac{6950}{143} & \dfrac{95565775}{204204} & -\dfrac{110879125}{87516} & \dfrac{9395000}{7293} & -\dfrac{9603125}{21879} \\ \dfrac{53375}{429} & -\dfrac{110879125}{87516} & \dfrac{312011875}{87516} & -\dfrac{81462500}{21879} & \dfrac{28296875}{21879} \\ -\dfrac{17500}{143} & \dfrac{9395000}{7293} & -\dfrac{81462500}{21879} & \dfrac{28937500}{7293} & -\dfrac{30625000}{21879} \\ \dfrac{17500}{429} & -\dfrac{9603125}{21879} & \dfrac{28296875}{21879} & -\dfrac{30625000}{21879} & \dfrac{10937500}{21879} \end{pmatrix} $$

In Mathematica code, this is simply:

 mat = {{130/10,91/10,819/10^2,8281/10^3,89271/10^4}, {91/10,819/10^2,8281/10^3,89271/10^4,1002001/10^5},{819/10^2,8281/10^3,89271/10^4,1002001/10^5,11562759/10^6},{8281/10^3,89271/10^4,1002001/10^5,11562759/10^6,136147921/10^7},{89271/10^4,1002001/10^5,11562759/10^6,136147921/10^7,1627802631/10^8}}

 Inverse[mat]
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  • $\begingroup$ How have you calculated this that quickly? What software? =). $\endgroup$ – Physicist137 Jul 31 '16 at 2:42
  • $\begingroup$ Mathematica ${}{}$ You should be able to use one of the free CAS programs to do the same thing in an instant for such a small matrix. $\endgroup$ – Moo Jul 31 '16 at 2:42
  • $\begingroup$ I see.. Well... +1. (assuming this is correct, of course..). $\endgroup$ – Physicist137 Jul 31 '16 at 2:43
  • $\begingroup$ Very easy to verify, multiply it with the matrix above and you should get the identity $I$. $\endgroup$ – Moo Jul 31 '16 at 2:45
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    $\begingroup$ You can try any of a number of them from Computer Algebra Systems - like SAGE. You might also want to try Octave (basically, a Matlab clone) and maybe Symbolic Python. $\endgroup$ – Moo Jul 31 '16 at 3:06

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