# Group problem with matrices

• Aug 24th 2007, 09:27 PM
Group problem with matrices
Let $G = \left[\begin{array}{cc}a&a\\a&a\end{array}\right] | a \in R, a \ne 0$. Show that G is a group under matrix multiplication. Explain why each element of G has an inverse even though the matrices have 0 determinant.

Why isn't my latex code working? I want to produce a matrix with two rows, each rows with [ a a ]...
• Aug 25th 2007, 12:07 AM
red_dog
$\displaystyle G=\left\{X_a=\left.\left(\begin{array}{cc}a & a\\a & a\end{array}\right)\right|a\in\mathbf{R},a\neq 0\right\}$

Let $X_a,X_b\in G\Rightarrow X_a\cdot X_b=\left(\begin{array}{cc}2ab & 2ab\\2ab & 2ab\end{array}\right)=X_{2ab}\in G$
We have $X_a=X_b\Leftrightarrow a=b$
Comutativity:
$X_a\cdot X_b=X_{2ab}=X_{2ba}=X_b\cdot X_a$
Asociativity:
$(X_a\cdot X_b)\cdot X_c=X_{2ab}\cdot X_c=X_{4abc}$
$X_a\cdot(X_b\cdot X_c)=X_a\cdot X_{2bc}=x_{4abc}$
Then $(X_a\cdot X_b)\cdot X_x=X_a\cdot(X_b\cdot X_c)$
Identity element:
Let $X_e$ be the identity element:
Then $X_a\cdot X_e=X_e\cdot X_a=X_a$
We have $X_a\cdot X_e=X_a\Leftrightarrow X_{2ae}=X_a\Leftrightarrow 2ae=a\Leftrightarrow e=\frac{1}{2}$
So the identity element is $\displaystyle X_{\frac{1}{2}}=\left(\begin{array}{cc}\frac{1}{2} & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2}\end{array}\right)$
Reciprocal element:
Let $X_a\in G$ and $X_{a'}\in G$ the reciprocal element.
Then $X_a\cdot X_{a'}=X_{a'}\cdot X_a=X_{\frac{1}{2}}$
$X_{2aa'}=X_{\frac{1}{2}}\Rightarrow 2aa'=\frac{1}{2}\Rightarrow a'=\frac{1}{4a}$
So, the reciprocal element of $X_a$ is $X_{\frac{1}{4a}}$

Each element has an inverse because the identity element is not $I_2$ and we can't apply the same rule for an inverse of a matrix.
• Aug 25th 2007, 12:52 AM
CaptainBlack
Quote:

Let $G = \left[\begin{array}{cc}a&a\\a&a\end{array}\right] | a \in R, a \ne 0$. Show that G is a group under matrix multiplication. Explain why each element of G has an inverse even though the matrices have 0 determinant.