Show that the set of all matrices of the form $\bigl(\begin{smallmatrix} a & 0 \\ 0 & 0 \end{smallmatrix} \bigr) $, $a \in R$ − {0} forms an abelian group under matrix multiplication.

Answer 1

Toolbox:

• A non-empty set G, together with an operation $*$ i.e., $(G, *)$ is said to be a group if it satisfies the following axioms
• (1) Closure axiom : $a,b\in G\Rightarrow a*b\in G$
• (2) Associative axiom : $\forall a, b, c \in G, (a * b) * c = a * (b * c)$
• (3) Identity axiom : There exists an element $e \in G$ such that $a * e = e * a = a, \forall a^ {-1} G.$
• (4) Inverse axiom : $\forall a \in G$ there exists an element $a^{-1}\in G$ such that $a^{-1}*a=a*a^{-1}=e.$
• e is called the identity element of $G$ and $a^{?1}$ is called the inverse of a in $G.$

Step 1:

Let $G=\{\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr),a\in R-\{0\}\}$

To show $(G,o)$ is an abelian group where $'o'$ stands for matrix multiplication.

Step 2:

Closure :

Consider $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr),\bigl(\begin{smallmatrix}b &0\\0&0\end{smallmatrix} \bigr)\in G$

$\therefore a,b\neq 0$

$\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}b &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}ab &0\\0&0\end{smallmatrix} \bigr)\in G$

Since $a\neq 0,b\neq 0\Rightarrow ab\neq 0$

Closure is satisfied.

Step 3:

Associativity :

Matrix multiplication is associative.

Step 4:

Existence of identity :

Consider $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr),\bigl(\begin{smallmatrix}e &0\\0&0\end{smallmatrix} \bigr)(a\neq 0)$

Such that $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}e &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$

$\Rightarrow \bigl(\begin{smallmatrix}ae &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$

$\Rightarrow \bigl(\begin{smallmatrix}e &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$

$\Rightarrow \bigl(\begin{smallmatrix}e &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)$

It can be seen that $\bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$

$\therefore \bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)\in G$ is the identity element.

Step 5:

Existence of inverse :

Let $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)\in G(a\neq 0)$

Consider $\bigl(\begin{smallmatrix}a' &0\\0&0\end{smallmatrix} \bigr)$ such that

$\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}a' &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)$

$\Rightarrow \bigl(\begin{smallmatrix}aa' &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)$

$\Rightarrow aa'=1\Rightarrow a'=\large\frac{1}{a}$(which exists since $a\neq 0)$

Now $\large\frac{1}{a}\neq 0$

$\bigl(\begin{smallmatrix}\large\frac{1}{a} &0\\0&0\end{smallmatrix} \bigr)\in G$

$\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}\large\frac{1}{a} &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}\large\frac{1}{a} &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}1 &0\\0&0\end{smallmatrix} \bigr)$

$\therefore$ the inverse of $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$ is $\bigl(\begin{smallmatrix}\large\frac{1}{a} &0\\0&0\end{smallmatrix} \bigr)$ in $G$

Thus every element in G has an inverse.

Step 6:

Let $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr),\bigl(\begin{smallmatrix}b &0\\0&0\end{smallmatrix} \bigr)\in G$

Consider $\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}b &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}ab &0\\0&0\end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix}ba &0\\0&0\end{smallmatrix} \bigr)$

$\Rightarrow \bigl(\begin{smallmatrix}b &0\\0&0\end{smallmatrix} \bigr)o\bigl(\begin{smallmatrix}a &0\\0&0\end{smallmatrix} \bigr)$

Since multiplication of real numbers is commutative.

$\therefore$ commutativity is satisfied.

The four group axioms being satisfied and the commutative property being satisfied $(G,0)$ is an abelian group.