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Prove that the matrices $\bigl(\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \bigr) $, $\bigl(\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \bigr) $ form a group under matrix multiplication

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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 $a * e = e * a = a, \forall a ∈ G.$
  • (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 $S=\{I,A\}$
Where $I=\bigl(\begin{smallmatrix}1 &0\\0 &1\end{smallmatrix} \bigr)$ and $I=\bigl(\begin{smallmatrix}0 &1\\1 &0\end{smallmatrix} \bigr)$
Under matrix multiplication $IoA=AoI=A$ and $IoI=I$
$I=\bigl(\begin{smallmatrix}1 &0\\0 &1\end{smallmatrix} \bigr)$ is the identity matrix.
Now consider $AoA=\bigl(\begin{smallmatrix}0 &1\\1 &0\end{smallmatrix} \bigr)\bigl(\begin{smallmatrix}0 &1\\1 &0\end{smallmatrix} \bigr)$
$\qquad\qquad\qquad\quad=\bigl(\begin{smallmatrix}0+1 &0+0\\0+0 &1+0\end{smallmatrix} \bigr)$
$\qquad\qquad\qquad\quad=\bigl(\begin{smallmatrix}1 &0\\0 &1\end{smallmatrix} \bigr)$
$\qquad\qquad\qquad\quad=I$
Step 2:
Drawing up cayley's Table,we have
Step 3:
From the table ,it is evident that $S$ is closed under matrix multiplication.
Step 4:
Associativity :
Matrix multiplication is associative.
Step 5:
Existence of identity :
$I$ is the identity element and $I\in S$
Step 6:
Existense of inverse :
It is evident that $I^{-1}=I$ and $A^{-1}=A$.
$\therefore$ every element has an inverse in $S$
The four group axioms being satisfied,$(S,o)$ is a group where $o$ is the operation of matrix multiplication.
answered Sep 13, 2013 by sreemathi.v
 

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