2
$\begingroup$

This question already has an answer here:

I have the folllowing array

$$h = \left(\begin{array}{cccc}1 & 0 & 0 & 0\\ 0 & f^2 & 0 & f^2 \cos\theta\\ 0 & 0 & g^2 & 0 \\ 0 & f^2\cos\theta & 0 & f^2\cos^2\theta + g^2\sin^2\theta\end{array}\right)$$

defined in Mathematica using

h = {{1,0,0,0},{0,f^2,0,f^2 Cos[theta],{0,0,g^2,0},{0,f^ Cos[theta],0,f^2 Cos[theta]^2 + g^2 Sin[theta]^2}};

If I now have a matrix $M$ of the same dimension as $h$, and if I do the following operation:

Mprime = Simplify[M.Inverse@g];

what will this achieve?

More precisely what does the code

Matrix1.Inverse@Matrix2

actually end up doing?

$\endgroup$

marked as duplicate by Kuba Jun 23 '17 at 8:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See linked topic,especially: What the @#%^&*?! do all those funny signs mean?. $\endgroup$ – Kuba Jun 23 '17 at 8:08
  • $\begingroup$ See also: when is f@g not the same as f[g]? $\endgroup$ – Kuba Jun 23 '17 at 8:09
  • $\begingroup$ p.s. lookup Cos and documentation in general because f^2\cos\theta is not close to a valid syntax. $\endgroup$ – Kuba Jun 23 '17 at 8:11
  • 1
    $\begingroup$ @Kuba, thanks for the links. As for the commands, that was a typo which I've fixed now. I entered the LaTeX code instead of the Mathematica code. :) $\endgroup$ – leastaction Jun 23 '17 at 8:34