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Fie matricele \( A, B\in M_{2}(\mathbb{Z}) \) astfel incat \( AB=BA \) si \( \det B=1 \). Sa se arate ca daca \( \det(A^{3}+B^{3})=1 \), atunci \( A^{2}=O_{2} \).

\( \det(A^3+B^3)=\det(A+B)\cdot \det(A^2-AB+B^2)=1 \)
Cum matricele au toate elementele intregi atunci si determinanti lor vor fi intregi. Deci \( \det(A^2-AB+B^2)\in \lbrace -1,1 \rbrace \).
Dar
\( \det(A^2-AB+B^2)=\det((A-\frac{1}{2}B)^2+\frac{3}{4}B^2)=\det(A-\frac{1}{2}B+\frac{\sqrt 3}{2}iB)\cdot \det(A-\frac{1}{2}B-\frac{\sqrt 3}{2}iB)=det {\det A-\frac{1}{2}B+\frac{\sqrt 3}{2}iB} \geq 0 \).

Deci \( \det(A^2-AB+B^2)=1 \) si \( \det(A+B)=1 \)

Pentru matricele de ordinul 2 avem relatia
\( \det(X+Y)+\det(X-Y)=2(\det(X)+\det(Y)) \)

Pt. \( X=A,\ Y=B \)
\( \det(A+B)+\det(A-B)=2(\det(A)+\det(B)) \)
\( \det(A-B)=2\det(A)+1 \).

Pt. \( X=A^2+B^2,\ Y=2AB \)
\( \det(A+B)^2+\det(A-B)^2=2(\det(A^2+B^2)+\det(2AB)) \)
\( 1+(2\det(A)+1)^2=2\det(A^2+B^2)+8\det(A) \)
\( \det(A^2+B^2)=2\det^2(A)-2\det(A)+1 \).

Pt. \( X=A^2+B^2-AB,\ Y=AB \)
\( \det(A^2+B^2)+\det(A-B)^2=2\det(A^2+B^2-AB)+2\det(A) \)
\( 2\det^2(A)-2\det(A)+1+(2\det(A)+1)^2=2+2\det(A) \)
\( 6\det^2(A)=0 \)
\( \det(A)=0 \).

Din Teorema Cayley-Hamilton
\( A^2-\tr(A)A=O_2\ (1) \)
\( B^2-\tr(B)B+I_2=O_2 \)
\( (A+B)^2-\tr(A+B)(A+B)+I_2=O_2 \)

\( A^2+2AB+B^2-\tr(A)A-\tr(A)B-\tr(B)B-\tr(B)A+I_2=O_2 \)
\( 2AB-\tr(A)B-\tr(B)A=O_2 \)
\( A(2B-\tr(B))=\tr(A)B \)
\( \det(A(2B-\tr(B))=\det(\tr(A)B) \)
\( 0=\tr^2(A) \)
\( \tr(A)=0\ (2) \)

Din relatiile (1) si (2) obtinem ca \( A^2=O_2 \).