My text says$ \left\{\begin{pmatrix}a&a\\a&a\end{pmatrix}:a\ne0,a\in\mathbb R\right\}$ forms a group under matrix multiplication. My text says$$\left\{\begin{pmatrix}a&a\\a&a\end{pmatrix}:a\ne0,a\in\mathbb R\right\}$$ forms a group under matrix multiplication.
But I can see $I\notin$ the set and so not a group.
Am I right?
 A: It's important to note that this set of matrices forms a group but it does NOT form a subgroup of the matrix group $GL_2(\mathbb{R})$ (the group we are most familiar with as being a matrix group - the group of invertible $2\times 2$ matrices) as no elements in this set have non-zero determinant. In particular, we are looking at a subset of $Mat(\mathbb{R},2)$ which is disjoint from $GL_2(\mathbb{R})$.
The identity of the group will then be the matrix $\pmatrix{\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}}$ and the inverse of the element $\pmatrix{a&a\\ a&a}$ will be $\dfrac{1}{4}\pmatrix{a^{-1}&a^{-1}\\ a^{-1}&a^{-1}}$ (you should check this).
A: If the structure be a group so it has an identity element, so for any matrix called $A$ we have: $$A:=\begin{pmatrix}
  a & a \\
 a & a \\
  \end{pmatrix}$$ for some $a\neq 1$ and if $$\text{id}_G=\begin{pmatrix}
  b & b \\
 b & b \\
  \end{pmatrix}, ~~b\neq 0$$ then $A\times\text{id}_G=\text{id}_G\times A=A$. Now $$A\times\text{id}_G=\begin{pmatrix}
  2ab & 2ab \\
 2ab & 2ab \\
  \end{pmatrix}$$ which should be equal to $A$ itself. So $a=2ab$ and since $a\neq 0$ so $b=0.5$.
A: The identity element in this group is $\pmatrix{\tfrac12&\tfrac12\\ \tfrac12&\tfrac12}$, not $I_2$.
A: Call that matrix $(a)$ and you get $(a)(b)=(2ab)$ thus it's closed under multiplication and $(1/2)$ is its unit.
