[tex] A = \left(\begin{array}{cc}2&2\\-2&-2\end{array}\right) \\ \\ (I_{2}+A)(aA+I_2)= I_2 \\ \\ \left[\left(\begin{array}{cc}1&0\\0&1\end{array}\right) + \left(\begin{array}{cc}2&2\\-2&-2\end{array}\right)\right] \cdot \left[a\cdot \left(\begin{array}{cc}2&2\\-2&-2\end{array}\right) + \left(\begin{array}{cc}1&0\\0&1\end{array}\right)\right] =[/tex]
[tex] =\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\\ \\ \\ \Rightarrow \left(\begin{array}{cc}3&2\\-2&-1\end{array}\right) \cdot \left[ \left(\begin{array}{cc}2a&2a\\-2a&-2a\end{array}\right) + \left(\begin{array}{cc}1&0\\0&1\end{array}\right)\right] =\left(\begin{array}{cc}1&0\\0&1\end{array}\right)[/tex]
[tex] \Rightarrow \left(\begin{array}{cc}3&2\\-2&-1\end{array}\right) \cdot\left(\begin{array}{cc}2a+1&2a\\-2a&-2a+1\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\\ \\ \\ \Rightarrow \left(\begin{array}{cc}3(2a+1)+2\cdot (-2a)&3\cdot 2a+2\cdot (-2a+1)\\-2(2a+1)+(-1)\cdot (-2a)&(-2)\cdot (2a)+(-1)\cdot (-2a+1)\end{array}\right) \\ =\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\\ \\ \\ \text{E de ajuns sa alegem doar un element din matrice} \\ \text{fiindca toate verifica oricum.}\\ \text{Alegem primul element de sus:} [/tex]
[tex] \Rightarrow 6a+3-4a = 1 \Rightarrow 2a + 3 = 1 \Rightarrow 2a = 1 - 3 \Rightarrow \\ \\ \Rightarrow 2a = -2 \Rightarrow \boxed{a = -1} [/tex]
