If $G=\{\begin{pmatrix}x & x\\0 & 0\end{pmatrix}:x\in \mathbb R\setminus \{0\}\}$ show that G is abelian (commuative) group. I have to show "." is a Binary opreation, Associative and commuative. but as for a Identity element and Inverse, Is this $E=\begin{pmatrix}1 & 1\\0 & 0\end{pmatrix}$ and $A^{-1}=\begin{pmatrix}1/x & 1/x\\0 & 0\end{pmatrix}$ correct answer ?

  • $\begingroup$ Yes, it is the correct answer. It only remains to show the operation is associative and commutative. Hint: If you forget about the matrices and think about the entries of the matrix, it looks a lot like the commutative group $(\mathbb{R}\setminus\{0\},\cdot)$ $\endgroup$ – Darío G Mar 23 '16 at 10:29
  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures cannot be searched and are inaccessible to those using screen readers. I have edited your question to reflect this principle $\endgroup$ – Shahab Mar 23 '16 at 10:34

Since matrix multiplication is associative, we only have to show that the multiplication is commutative. This follows from


This multiplication also shows that the product of two elements of the set is in the set again.


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