i have a proplem with some matrix

Respuestas (2)

Geoff Hayes
Geoff Hayes el 23 de Dic. de 2014
Ali - if you assume that all elements in your matrix are positive, then you could use the sqrt function as
A = [100 0 ; 0 100];
R1 = sqrt(A);
R2 = -sqrt(A);

1 comentario

ali
ali el 23 de Dic. de 2014
Editada: ali el 23 de Dic. de 2014
in this case i should have two result , but in case A.1/3 i should have 3 results ,and what if they are negitive ?

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mohammad
mohammad el 23 de Dic. de 2014
You can use this
A = [100 0 ; 0 100];
syms B
solve('B*B-A')
this gives you two answers and if arrays of A are negative, there will be no problem.

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Preguntada:

ali
el 23 de Dic. de 2014

Respondida:

el 23 de Dic. de 2014

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