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I need a hint. My task is to proof that

  • ((a11, a12), (b11, b12))

  • ((a21, a22), (b21, b22))

    ∈ R ⇔ a11 + a12 + a21 + a22 = b11 + b12 + b21 + b22

R is equivalnce relation. My problem is that I have no idea how to deal with matrix.

I tried to work like in this post and just make something like A={{a11,a12},{a21,a22}} instead of A={0,1}, but I got stuck.

I would be grateful for any help.

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    $\begingroup$ just check reflex, symmetric and transitivity. It is clear $\endgroup$ – Bhaskar Vashishth Nov 3 '14 at 11:13
  • $\begingroup$ Why don't you write what is meant by an equivalence relation? And try your self. $\endgroup$ – Babai Nov 3 '14 at 11:38
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    $\begingroup$ I'm pretty sure I saw this exact same question asked here several hours ago? Did you delete it an ask again with no changes? $\endgroup$ – hmakholm left over Monica Nov 3 '14 at 11:43
  • $\begingroup$ @Henning: math.stackexchange.com/questions/1003400/… also since this is a separate user account, you can't close as a duplicate (even though it is a word-for-word duplicate) unless there is an upvoted answer. $\endgroup$ – Asaf Karagila Nov 3 '14 at 12:16
  • $\begingroup$ @Asaf: Of course I tried to google for a phrase that included the one word there's in difference between the two questions. Thanks -- we'll just close it the other way around then. $\endgroup$ – hmakholm left over Monica Nov 3 '14 at 12:29
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Hint:

If $f:X\rightarrow Y$ is a function then the relation $R$ on $X$ defined by $xRy\iff f(x)=f(y)$ is an equivalence relation. So find a proper function here on $2\times 2$-matrices.

(Why? Simply because $f(x)=f(x)$ i.e. reflexitivity, $f(x)=f(y)\Rightarrow f(y)=f(x)$ i.e. symmetry, and $f(x)=f(y)\wedge f(y)=f(z)\Rightarrow f(x)=f(z)$ i.e. transitivity)

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