Is {S,+,.} with S = {matrix(a,b,a-b,a)|a,b ∊ R} a Field?

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SUMMARY

The system {S,+,.} with S = {matrix(a,b,a-b,a)|a,b ∊ R} is definitively not a field under matrix addition (+) and matrix multiplication (.). The discussion confirms that while {S,+} forms an Abelian group, {S,.} does not maintain the necessary closure property, as demonstrated by the multiplication of matrices resulting in elements outside the defined set. Specifically, the example of squaring the matrix \(\left( \begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array} \right)\) yields \(\left( \begin{array}{cc} 5 & 4 \\ 4 & 5 \end{array} \right)\), which does not conform to the original form.

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sam_0017
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true or false ..
The system {S,+,.} with S = { matrix (a,b,a-b,a)|a,b ∊ R)
is not a field under matrix addition (+) and matrix multiplication (.)





i find that the statement is false .
since : 1. {S,+} is Abelian group.
2. {S,.} : is Abelian group.

is my finding is true ?
 
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Hi, Sam,
if by matrix(a,b,a-b,a) you mean[tex]\left( \begin{array}{cc} a & b \\ a-b & a \end{array} \right)[/tex]and multiplication is the usual matrix multiplication, then try to multiply two such matrices, to see that the result may not have the same form. For example,[tex]\left( \begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array} \right)^2 = \left( \begin{array}{cc} 5 & 4 \\ 4 & 5 \end{array} \right)[/tex]but 5-4 is not 4.
 

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