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Math Help - Vector and matrix multiplication

  1. #1
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    Vector and matrix multiplication

    Is this possible? If yes, how?

    Vector and matrix multiplication-rst8r.jpg
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  2. #2
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    Not as a usual matrix multiplication. In order to multiply "AB" where A and B are matrices, the number of columns of A must equal the number of rows of B. That is not true here and I do not know of any definition of "multiplcation" that will work.

    You could, of course, multiply the other way:
    \begin{bmatrix}1 & 0 & 0 \\ 0 & cos b & sen b \\ 0 & -sen b & cos b\end{bmatrix}\begin{bmatrix}0 \\ 0 \\ r\end{bmatrix}= \begin{bmatrix}0 \\ r sen b \\ cos b\end{bmatrix}.

    Added: you could do this as a "tensor multiplication". That would involve multiplying all three elements of A by each of the 9 elements of B giving a 27 element "third order tensor". It would have to be written as a "three dimensional matrix" consisting of the three layers:
    \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}

    \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}

    \begin{bmatrix} r & 0 & 0 \\ 0 & rcos b & r sen b \\ 0 & -rsen b & r cos b\end{bmatrix}
    but that is probably NOT what is intended!
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  3. #3
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    This is more about vector multiplication it seems, i thought it should be similar (sorry if i misled you). My teacher wrote this at the time, and i didn't get it. I know we can't multiply two matrices if the 1st doesn't have the same number of columns as the 2nd has of rows. So as he did it, then i suppose matrices rules don't apply here. Probably there is some other rule to multiply these two vectors...

    Forget the initial vector and matrix multiplication, this one reflects better my doubt.

    \begin{bmatrix}0 \\ 0 \\ z\end{bmatrix} \begin{bmatrix}0 \\ rsen b \\ rcos b\end{bmatrix}= \begin{bmatrix} zrsen b \\ 0 \\ 0\end{bmatrix}.
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  4. #4
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    Oh! That's the "cross product" of two three dimensional vectors. Given two vectors, \vec{X}= a\vec{i}+ b\vec{j}+ c\vec{k} and \vec{Y}= u\vec{i}+ v\vec{j}+ w\vec{k}, then the cross product is given by \vec{X}\times\vec{Y}= (bw- cv)\vec{i}- (cu- aw)\vec{j}+ (av- bu)\vec{k}. That can be remember by a mnemonic: write it as a "determinant":
    \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ a & b & c \\ u & v & w\end{array}\right|

    Now, cross-multiplication is "anti-symmetric": \vec{X}\times\vec{Y}= -\vec{Y}\times\vec{X} and, in fact, what you have is the other way around:
    \begin{bmatrix}0 \\ 0 \\ z\end{bmatrix}\times\begin{bmatrix}0 \\ rsen b\\ rcos b\end{bmatrix}= \begin{bmatrix}-zr senb \\ 0 \\ 0\end{bmatrix}
    while
    \begin{bmatrix}0 \\ r sen  \\ rcos b\end{bmatrix}\times\begin{bmatrix}0 \\ 0 \\ z \end{bmatrix}= \begin{bmatrix}zr senb \\ 0 \\ 0\end{bmatrix}

    The cross product cannot be generalized in any simple way to a product of a vector with a matrix or even to other dimensions. The cross product of two two-dimensional vectors is taken by treating them as three dimensional vectors with third component 0.
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