Matrix Rotation Query: Inertia Tensor D (+90y, -90z)

[bugatti79]

I have an inertia tensor D in the old Cartesian system which i need to rotate through +90 in y and -90 in z to translate to the new system. I am using standard right hand rule notation for this Cartesian rotation.


$$D= \mathbf{\left(\begin{array}{lll}I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I_{zy}&I_{zz}\\\end{array}\right)}$$, $$N_y(+90)=\mathbf{\left(\begin{array}{lll}0&0&1\\0&1&0\\-1&0&0\\\end{array}\right)}$$, $$N_z(-90)=\mathbf{\left(\begin{array}{lll}0&1&0\\-1&0&0\\0&0&1\\\end{array}\right)}$$


If we let


$$N_R=N_z N_y$$ (I am pre-multiplying $$N_y$$ by $$N_z$$ because that is the order) and the transpose $$N'_R=N_R^T$$.


Is the the new system tensor $$N_RDN'_R$$ or $$N'_RDN_R$$...?

 

[Ackbach]

Is the the new system tensor $$N_RDN'_R$$ or $$N'_RDN_R$$...?

This page would seem to indicate the former.