Matrices & Determinants
Real matrices with A² + B² = AB — invertibility condition on n
MJAT_TS4_P1
Grade 12
Question:
Let $A$ and $B$ be real $n\times n$ matrices such that $A^2+B^2=AB$. If $BA-AB$ is an invertible matrix, then possible value(s) of $n$ is/are:
Step-by-Step Solution
Key Concept: Let $P=A+B\omega$ where $\omega=e^{2\pi i/3}=(-1+i\sqrt{3})/2$. Then $\overline{PP}=(A+\omega B)(A+\bar{\omega}B)=A^2+B^2+(\omega+\bar{\omega})BA-AB\cdot\bar{\omega}\omega=AB-BA+\omega(BA-AB)$... The key is $\det(PP)=\det(P)\overline{\det(P)}=|\det(P)|^2\geq 0$. But $\det(PP)=\det(\omega(BA-AB))=\omega^n\det(BA-AB)$. For this to be real and non-negative with $\det(BA-AB)\neq 0$: $\omega^n$ must be real and positive, which requires $n\equiv 0\pmod{3}$.
Condition: $3\mid n$. Options A (30) and D (33) satisfy this. Answer: A, D.
Correct Answer: AD