Euler transform matrix multiplication help

[weedannycool]

!Euler transform matrix multiplication help!

Homework Statement

This may be rather simple but i am really struggling to complete a 3 3x3 matrix multiplication. I NEED STEP BY STEP WORKING!. This would really help me
I understand the theory. Basically I have three matrices

T1= cosψ sin ψ 0
-sinψ cosψ 0
0 0 1

T2= cosθ 0 - sin θ
0 1 0
sin 0 cosθ

T3= 1 0 0
0 cosω sin
0 -sinω cosω

Homework Equations

The formulae used for the transform is X=T3*T2*T1 Xe

I just need to know how to evaluate T3*T2*T1

Also do they have to be done in order?

I just need the new transform matrix

The Attempt at a Solution

I have endless attempts at this. all wrong

I've looked for worked examples but they all skip the actual working part that would gain me the marks I need.

I know this will be a pain to type but it give me the help a need

 

[HallsofIvy]

My question is, why are you being given homework like this if you have never been taught how to multiply matrices?

I have always felt that the best way to think about matrix multiplication to think of it in terms of "dot products". In terms of 3-vectors, the dot product is given by \begin{pmatrix}a & b & c \end{pmatrix}\begin{pmatrix}x & y & z\end{pmatrix}= ax+ by+ cz

To multiply two 3 by 3 matrices, think of each row of the first matrix and each column of the second matrix as vectors. The number in the "ith row", "jth column" of the product matrix is the dot product of the ith row of the first matrix with the jth column of the second matrix.

For example, to multiply
A= \begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}
and
B= \begin{pmatrix}j & k & l\\ m & n & o \\ p & q & r\end{pmatrix}

the number in the "second row, third column" will be the dot product of the second row of A, \begin{pmatrix}d & e & f\end{pmatrix} and the third column of B, \begin{pmatrix}l & o & r\end{pmatrix}, dl+ eo+ fr.

AB= \begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}\begin{pmatrix}j & k & l\\ m & n & o \\ p & q & r\end{pmatrix}= \begin{pmatrix}aj+bm+ cp & ak+ bn+ cq & al+ bo+ cr \\ dj+ ek+ fl & dm+ en+ fq & dl + eo+ fr \\ gj+ hm+ ip & gk+ hn+ iq & gl+ ho+ ir\end{pmatrix}

 

[weedannycool]

Cheers. Just gaps in my learning really. Will the answer be the same if it was B*A? I think i must have been doing the order of calculation wrong. In my other tries i have done T1T2T3, which is not the same as the formula . So i guess it must be the other way around

 

[HallsofIvy]

No, matrix multiplication is NOT "commutative". In general, AB\ne BA. In terms of matrices representing transformations, the order of multiplication depends upon the order in which you are applying the transformations. If you first apply the transformation B to vector or point x, then apply transformation A, you multiply the matrices as ABx.

(Note this is based on the convention of writing x as a column. If you are writing x as a row rather than a column then the multiplication would be xBA. But in either case, B, the transformation first applied to x, is the matrix next to x.)