444 Algebra
X^2 − 5 X+ 6 I 2 = 0 ,
which can be used to transform the original equation into
4 X− 12 I 2 =
(
− 2 − 2
− 2 − 2
)
with the solution
X=( 5
2 −
1
2
−^1252)
.
The case in which the second eigenvalue ofXis−1 is treated similarly and yields the
solution
X=
(
1 − 2
− 21
)
.
(Romanian competition, 2004, proposed by A. Buju)260.Because the trace of[A, B]is zero, the Cayley–Hamilton Theorem for this matrix
is[A, B]^2 +(det[A, B])I 2 =0, which shows that[A, B]^2 is a multiple of the identity.
The same argument applied to the matrices[C, D]and[A, B]+[C, D]shows that their
squares are also multiples of the identity.
We have
[A, B]·[C, D]+[C, D]·[A, B]=([A, B]+[C, D])^2 −[A, B]^2 −[C, D]^2.Hence[A, B]·[C, D]+[C, D]·[A, B]is also a multiple of the identity, and the problem
is solved.
(Romanian Mathematical Olympiad, 1981, proposed by C. Nast ̆ asescu) ̆
261.The Cayley–Hamilton Theorem gives
(AB−BA)^3 −c 1 (AB−BA)^2 +c 2 (AB−BA)−c 3 I 3 =O 3 ,wherec 1 =tr(AB−BA)=0, andc 3 =det(AB−BA). Taking the trace and using the
fact that the trace ofAB−BAis zero, we obtain tr((AB−BA)^3 )−3 det(AB−BA)=0,
and the equality is proved.
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
T. Andreescu)
262.LetC=AB−BA. We have
AB^2 +BA^2 =(AB−BA)B+B(AB−BA)=CB+BC= 2 BC.