Question

Let G be the set of all 3×3 non-singular matrices with entries from Q. Let X be a fixed matrix in G.prove that operation *defined by A*B= X^(-1)ABX be a binary operation. How do you prove that G is a group with respect to binary operation defined by A*B=X^(-1) ABX?

Answer 1

$G=\{A=\begin{pmatrix} a_{11} & a_{12}&a_{13} \\ a_{21} & a_{22}&a_{23} \\ a_{31} & a_{32}&a_{33} \end{pmatrix}, a_{ij}\in Q,\\ i=1,2,3,j=1,2,3, detA\neq0 \},\\ X=\begin{pmatrix} x_{11} & x_{12}&x_{13} \\ x_{21} & x_{22}&x_{23} \\ x_{31} & x_{32}&x_{33} \end{pmatrix}, detX\neq0$

X-fixed matrix in G

$detX\neq0\to\exists X^{-1}, detX^{-1}\neq0$

$X^{-1}=\frac{1}{detX}\begin{pmatrix} y_{11} & y_{12}&y_{13} \\ y_{21} & y_{22}&y_{23} \\ y_{31} & y_{32}&y_{33} \end{pmatrix}$

where $y_{ij}$ - algebraic addition to the element $x_{ij}$ .

$B\in G: B=\begin{pmatrix} b_{11} & b_{12}&b_{13} \\ b_{21} & b_{22}&b_{23} \\ b_{31} & b_{32}&b_{33} \end{pmatrix}, b_{ij}\in Q,\\ detB\neq0$

$A*B=X^{-1}ABX$

$X^{-1},A,B,X$ 3 $\times$ 3 matrices

$A*B$ 3$\times$ 3 matrices with elemets in Q

$det(A*B)=det(X^{-1}ABX)=\\ =det(X^{-1})det(A)det(B)det(X)\neq0\\ A*B\in G$

So $*$ a binary operation.

Prove that G is a group.

1. $A*B\in G$

2.$(A*B)*C=A*(B*C)$

3.$I=\begin{pmatrix} 1& 0&0 \\ 0& 1&0\\ 0&0&1 \end{pmatrix}\in G,1\in Q, det(I)=1\neq0$

4.$A*A^{-1}=I$

Find $A^{-1}:$

$A*A^{-1}=X^{-1}AA^{-1}X=X^{-1}IX=X^{-1}X=I\\ det(A^{-1})\neq0$

So G is a group with respect to binary operation