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I have a given Matrix equation

$R(s)^{'}_{3\times 3} = \psi(s)_{3\times 3}R(s)\tag 1$

Conditions

  • R(s) is orthogonal and determinent 1. Can say in the format of rotation matrix
  • $R^{'}(s)$ derivative w.r.t s,has determinent $0$ . So it cant have inverse

Question

  • What I need to do is to switch the terms of $R^{'}(s),R(s)$ between L.H.S and R.H.S using their inverse. More precisely I need to make $R(s)^{-1}$ alone on L.H.S. But the issue is $\left({\frac{\mathrm{d}R(s) }{\mathrm{d} s}} \right )^{-1}$ doest not have an inverse so I am struggling to take that to R.H.S to make a form I desire. How do we do that? Is there anyway to make this $\left({\frac{\mathrm{d}R(s) }{\mathrm{d} s}} \right )$ to a determinent non zero matrix by addition?
Nirvana
  • 1,687
  • I changed that.. – Nirvana Oct 04 '14 at 06:30
  • Can you explain why you want to bring your equation into a different form? – Urgje Oct 04 '14 at 09:43
  • @Urgje I was doing one work. So ended up a closed form for $R(s)$. But unfortunately it contains terms that cant be integrated analytically(A form look like this clickhere). Then I have this relation equation (1) for my problem. Since I am able to find derivative of $R(s)$, I thought it would be helpful by writing terms in equation 1 using integration by parts may end up a formula for $\int R(s) \ ds$ – Nirvana Oct 04 '14 at 10:18

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