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Is the amount of arrangements of the elements of a matrix equal to its area factorial?


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Posted

I am assuming that since the amount of arrangements possible for a set of elements is equal to the amount of elements within the set to the factorial, that for a matrix it would be the area of the matrix factorial. Is this true?

Posted

Yes the Area.

 

A matrix can be a k by d, where k is the width and d is the height. Multiply them to get the area.

 

[math]\begin{bmatrix} 3&2 &5 \\ 1& 3 &5 \\ 6&3 & 4 \end{bmatrix}=A_{3\times 3}[/math]

Posted

Unity, this is why you have to be very careful with your terminology. I don't think I had ever heard anyone call the total number of elements in a matrix the 'area' before. It is especially confusing since there are matrix operations that can calculate areas, e.g. usually via a determinant.

 

And, yes, if there are no restrictions on the ordering of the elements (i.e. no symmetry requirements or similar), then it would be the factorial of the product of rows x columns

Posted

Unity, this is why you have to be very careful with your terminology. I don't think I had ever heard anyone call the total number of elements in a matrix the 'area' before. It is especially confusing since there are matrix operations that can calculate areas, e.g. usually via a determinant.

 

And, yes, if there are no restrictions on the ordering of the elements (i.e. no symmetry requirements or similar), then it would be the factorial of the product of rows x columns

I apologize for this. I guess terminology for me is a bit confusing sometimes.

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