Resolvi assim:
(x,y)= y(1,1) + (x-y)(1,0)
Substituindo, valores de A:
(1,0)= (1,0)
(0,1)=(0,1)
Mas no gabarito a matriz é:
![A=\left[\begin{array}{rr}
0&1\\
1&-1
\end{array}\right],\quad A=\left[\begin{array}{rr}
0&1\\
1&-1
\end{array}\right],\quad](/latexrender/pictures/db02409b98154dab4b36ec36803a3d31.png)
Podem me ajudar?
![A=\left[\begin{array}{rr}
0&1\\
1&-1
\end{array}\right],\quad A=\left[\begin{array}{rr}
0&1\\
1&-1
\end{array}\right],\quad](/latexrender/pictures/db02409b98154dab4b36ec36803a3d31.png)

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![\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}} \frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}](/latexrender/pictures/981987c7bcdf9f8f498ca4605785636a.png)
(dica : igualar a expressão a
e elevar ao quadrado os dois lados)