Question

Ajutorrr!Legi de compozitie!!!
E2/e, cel cu matricea.

Answer 1

$A\in M\Rightarrow A=\left(\begin{array}{cc}1&a\\0&1\end{array}\right),a\in\mathbb{R}\\B\in M\Rightarrow B=\left(\begin{array}{cc}1&b\\0&1\end{array}\right),b\in\mathbb{R}\\A\circ B=B\circ A,\forall A,B\in M\\A\circ B=AB-A-B+2I_2\\AB=\left(\begin{array}{cc}1&a\\0&1\end{array}\right)\cdot\left(\begin{array}{cc}1&b\\0&1\end{array}\right)=\left(\begin{array}{cc}1&a+b\\0&1\end{array}\right),a,b\in\mathbb{R}$

$A\circ B=\left(\begin{array}{cc}1&a+b\\0&1\end{array}\right)-\left(\begin{array}{cc}1&a\\0&1\end{array}\right)-\left(\begin{array}{cc}1&b\\0&1\end{array}\right)+2\left(\begin{array}{cc}1&0\\0&1\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)=I_2$

$B\circ A=BA-B-A+2I_2\\BA=\left(\begin{array}{cc}1&b\\0&1\end{array}\right)\left(\begin{array}{cc}1&a\\0&1\end{array}\right)=\left(\begin{array}{cc}1&a+b\\0&1\end{array}\right), a,b\in\mathbb{R}\\B\circ A=\left(\begin{array}{cc}1&a+b\\0&1\end{array}\right)-\left(\begin{array}{cc}1&b\\0&1\end{array}\right)-\left(\begin{array}{cc}1&a\\0&1\end{array}\right)+2\left(\begin{array}{cc}1&0\\0&1\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)=I_2=A\circ B\Rightarrow \circ\text{ comutativ\u a},\forall A,B\in M$