I need to prove that $2\times 2$ matrices in a specific format form a group under matrix multiplication. One of the formats is this one:
$$
\left\{ \left.
\begin{pmatrix} a & b \\ b & c \end{pmatrix}
\right| \, ac \neq b^2 \land a,b,c\in\mathbb{R}
\right\}
$$
To prove they form a group under matrix multiplication we prove the product is in the original set, the elements of the set multiply associatively, there is and identity element, and an inverse element for each matrix in the set.
Matrix multiplication is associative. The identity matrix (ones on the diagonal, $0$ for the other two entries) serves as the identity element for this set.
So I'm left with proving the closure and proving and showing how to find an inverse of any element in the set.
I tried multiplying two matrices:
$\begin{pmatrix} a_1 & b_1 \\ b_1 & c_1 \end{pmatrix}\times\begin{pmatrix} a_2 & b_2 \\ b_2 & c_2 \end{pmatrix}=\begin{pmatrix} a_1a_2 + b_1b_2 & a_1b_2 + b_1c_2 \\ b_1a_2 + c_1b_2 & b_1b_2 + c_1c_2 \end{pmatrix}$
What I think I notice is that the result is not of the given format anymore, the "b"s are not the same in the product ($a_1b_2 + b_1c_2 \neq b_1a_2 + c_1b_2$) (I think this is where I go wrong), does this mean that this format does not create a group regardless of the extra rule "$ac\neq b^2$"
Best Answer
Let $A=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1 & 1 \\ 1 & 0\end{pmatrix}$
Then $AB = \begin{pmatrix} 2 & 2 \\ 1 & 0\end{pmatrix}$.
Clearly, $ab_{12}\neq ab_{21}$. So the set is not closed under multiplication