# Matrix Operations Query

1. Nov 20, 2014

### bugatti79

Hi Folks,

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 $N'_RDN_R$...?

Thanks

2. Nov 21, 2014

### Staff: Mentor

Imagine how a vector in the new system would come in: the matrix on the right side would transform this vector to your old system, then the old matrix is applied, then the matrix on the left side transforms it back to your new coordinate system.

3. Nov 21, 2014

### bugatti79

Hi mbf,

Not sure I follow. Can you clarify a bit?
Thanks

4. Nov 21, 2014

### Staff: Mentor

If v is a vector in your new coordinate system, does $N_R v$ or $N'_R v$ represent the vector in the original coordinate system?
This will be used in the product $N_R D N_R v$ (with the right ' added).

5. Nov 22, 2014

### bugatti79

I still havent grasp your idea of a vector to cross-check. However, I know from a clue that the value of $I_{zz}$ in the new system has to be the same as $I_{xx}$ in the old system because "z axis new" lines up with "x axis old" and so $N_R D N'_R$ does this for me.

However, in the event of no clue, im still not clear how to use a vector....

Thanks