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Best answer by ValeryOchkov

Solution:

But I think not all is correct in Prime! Jakov Kucan, where are yuo!?

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17 replies

ValeryOchkov24-Ruby IVAuthor
24-Ruby IV
December 20, 2013

I think it is an error. Or?

rankUnitMatrixSimpl.png

AlanStevens19-Tanzanite
19-Tanzanite
December 21, 2013

I don't think it's an error. To find the rank of a matrix one has to perform some operations that involve subtraction of terms from different rows. This only makes sense if all the terms in the matrix are of the same type. The determinant works for your examp;les because items of different type are only multiplied together (which can make sense).

Alan

Werner_E
Werner_E25-Diamond I
25-Diamond I
December 21, 2013

Fully agreed. I, too, won't consider this an error.

This only makes sense if all the terms in the matrix are of the same type.

And thats the reason the numeric processor fails - its aware of units. While I don't consider it a bug or error I can see that it could be convenient if Prime would see the unit not as a real unit but rather as a simple factor in that case.

The symbolic processor is not aware of units and treats it as normal (unknown) variables. Consequently a symbolic evaluation works.

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ValeryOchkov24-Ruby IVAuthorAnswer
24-Ruby IV
December 21, 2013

Solution:

But I think not all is correct in Prime! Jakov Kucan, where are yuo!?

17-17-EMF-symbol.png

Werner_E
Werner_E25-Diamond I
25-Diamond I
December 21, 2013

As the symbolics is not aware of units its right.

21.12.png

But using a modifier like "fully" I would have expected muPad to give me a more complete result by telling me, that the rank = 2 only if V is not Omega*A and its 1 otherwise.

ValeryOchkov24-Ruby IVAuthor
24-Ruby IV
December 21, 2013

See too http://communities.ptc.com/message/220306

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