Problems and Solutions #038
Problem #038$3$차 정사각행렬 $A$, $B$가 $A^2 = AB + BA$를 만족할 때, $\det(AB - BA) = 0$임을 보여라. $A^2 = AB + BA$이므로 $AB = A^2 - BA$, $BA = A^2 - AB$가 성립한다. 따라서\[ \begin{align*} \det(AB - BA) &= \det(A^2 - 2BA) = \det((A - 2B)A) \\[5px] &= \det(A - 2B) \det(A) = \det(A) \det(A - 2B) \\[5px] &= \det(A(A - 2B)) = \det(A^2 - 2AB) \\[5px] &= \det(BA - AB) = \det(-(AB - BA)) \\[5px] &= (-1)^3 \det(AB - BA) = -\det..