Question

Va rog dau coroana....

Answer 1

$\\ I_2 = \begin{pmatrix}1&0\\ \:0&1\end{pmatrix} si \:\ A = \begin{pmatrix}1&-1\\ \:-2&2\end{pmatrix}\\\\ Ne \:\ cere \:\ sa \:\ demonstram \:\ ca \:\ X_{(a)} \cdot X_{(a)} = X_{(a+b+3ab)}\\ Acum \:\ vine: \\ X_{(a)} = I_2 + a \cdot A\\ X_{(b)} = I_2 + b \cdot A\\ X_{(a+b+3ab)} = I_2 + (a+b+3ab) \cdot A$

$\\ X_{(a)} = \begin{pmatrix}1&0\\ 0&1\end{pmatrix} + a \cdot \begin{pmatrix}1&-1\\ -2&2\end{pmatrix} => \begin{pmatrix}1+a&-a\\ -2a&1+a\cdot \:2\end{pmatrix}\\\\\\ X_{(b)} = \begin{pmatrix}1&0\\ 0&1\end{pmatrix} + b \cdot \begin{pmatrix}1&-1\\ -2&2\end{pmatrix} => \begin{pmatrix}1+b&-b\\ -2b&1+b\cdot \:2\end{pmatrix} \\\\\\ X_{(a+b+3ab)} = \begin{pmatrix}1+a+b+3ab&-a-3ab-b\\ -2\left(a+3ab+b\right)&1+\left(a+b+3ab\right)\cdot \:2\end{pmatrix}$

$\\ Acum: X_{(a)} \cdot X_{(b)} => \begin{pmatrix}1+a&-a\\ -2a&1+a\cdot \:2\end{pmatrix} \cdot \begin{pmatrix}1+b&-b\\ -2b&1+b\cdot \:2\end{pmatrix} =\\\\\\ \begin{pmatrix}1+b+a+3ab&-3ab-a-b\\ -2a-6ab-2b&6ab+2a+2b+1\end{pmatrix}\\ \\ \\ Acum \:\ demonstram \:\ ca \:\ relatia \:\ este \:\ adevarata:$

$\begin{pmatrix}1+b+a+3ab&-3ab-a-b\\ -2a-6ab-2b&6ab+2a+2b+1\end{pmatrix} = \\\\\\ \begin{pmatrix}1+a+b+3ab&-a-3ab-b\\ -2\left(a+3ab+b\right)&1+\left(a+b+3ab\right)\cdot \:2\end{pmatrix}\\\\\\ Care \:\ demonstreaza \:\ ca \:\ relatia \:\ este \:\ adevarata$