Rotational Matrix

[Laganum]

Hello all. Not the best mathematician here, thus asking help about a small problem I have.
We were given an assignment in our robotics class, and I got stuck in a certain rotational matrix multiplication.
Could someone please guide me on how the person got this answer in the screenshot below ? I did not get it in symbolic or numeric form, I am puzzled.
C and S meaning sine and cosine ofcourse.
Thank you for reading.

 

[stapel]

Hello all. Not the best mathematician here, thus asking help about a small problem I have.
We were given an assignment in our robotics class, and I got stuck in a certain rotational matrix multiplication.
Could someone please guide me on how the person got this answer in the screenshot below ? I did not get it in symbolic or numeric form, I am puzzled.
C and S meaning sine and cosine ofcourse.
Thank you for reading.
View attachment 4075

You started, of course, by evaluating the sines and cosines, reducing things to simple numerics. And then you multiplied. What did you get for your evaluated forms? What did you get for your result?

Please be complete. Thank you!

 

[Laganum]

You started, of course, by evaluating the sines and cosines, reducing things to simple numerics. And then you multiplied. What did you get for your evaluated forms? What did you get for your result?

Please be complete. Thank you!

The screenshot above is not mine, it is an example of how we should start doing our assignment.

I did the same multiplication in MathCad, however I did not recieve the same answer, not in numeric or nowhere near in symbolics.
At the moment of replying I do not have access to my work.
I doubt doing this by hand yields different results aswell.

 

[HallsofIvy]

Of course, cos(-90)= 0 and sin(-90)= -1 so
\(\displaystyle \begin{bmatrix}cos(-90) & 0 & sin(-90) \\ 0 & 1 & 0 \\ -sin(-90) & 0 & cos(-90)\end{bmatrix}= \begin{bmatrix}0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}\)

and \(\displaystyle \begin{bmatrix}1 & 0 & 0 \\ 0 & cos(-90) & -sin(-90) \\ 0 & sin(-90) & cos(-90)\end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{bmatrix}\)
so the matrix multiplication is
\(\displaystyle \begin{bmatrix}0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{bmatrix}= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}\)
as given.

Now, is that what you got for the two matrices being multiplied or is it the multiplication you are asking about?

You say you used Mathcad. Did you enter the matrices in the correct order?

 

[Laganum]

Of course, cos(-90)= 0 and sin(-90)= -1 so
\(\displaystyle \begin{bmatrix}cos(-90) & 0 & sin(-90) \\ 0 & 1 & 0 \\ -sin(-90) & 0 & cos(-90)\end{bmatrix}= \begin{bmatrix}0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}\)

and \(\displaystyle \begin{bmatrix}1 & 0 & 0 \\ 0 & cos(-90) & -sin(-90) \\ 0 & sin(-90) & cos(-90)\end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{bmatrix}\)
so the matrix multiplication is
\(\displaystyle \begin{bmatrix}0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{bmatrix}= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}\)
as given.

Now, is that what you got for the two matrices being multiplied or is it the multiplication you are asking about?

You say you used Mathcad. Did you enter the matrices in the correct order?

*hangs head in shame*
I have found the error in my ways
Thank you