SOLUTION: Let n ∈ N and B is a n × n matrice with real entries and has determinant 1. Show that there exist n × n matrices K, A and N such that B = KAN.

Algebra ->  Finance -> SOLUTION: Let n ∈ N and B is a n × n matrice with real entries and has determinant 1. Show that there exist n × n matrices K, A and N such that B = KAN.       Log On


   

Question 1183381: Let n ∈ N and B is a n × n matrice with real entries and has determinant 1. Show that there exist n × n matrices
K, A and N such that B = KAN.
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

Surely,  for any  nxn-matrix  B  there are  nxn-matrices  K,  A  and  N  such that   B = KAN.

Simply take  A = B,  K = I  (the identity matrix),  N = I.


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In your post,  you missed some important properties of matrices  A,  B,  K  and  N,  that make the problem  SPECIAL.

As it is worded in your post,  the statement is  TRIVIAL.

It is valid for any square  nxn-matrix  B,  independently of its determinant.

So, twice and thrice check with your source.


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The term  "matrice"  in English is  "matrix".

Use  "matrix"  for single.

Use  "matrices"  for plural.


Happy learning (!)