Mathematics - Matrices Question with Solution | TestHub

$$A^3+B^3+C^3-3ABC=(A+B+C)(A+\omega B+{\omega }^{2}C)(A+{\omega }^{2}B+\omega C)$$

$$\det (A^3+B^3+C^3)=\det (A+B+C)\det (A+\omega B+{\omega }^{2}C)\det (A+{\omega }^{2}B+\omega C)$$

Since $A$, $B$, and $C$ are real matrices, $\det (A+{\omega }^{2}B+\omega C)$ is the complex conjugate of $\det (A+\omega B+{\omega }^{2}C)$.

Therefore,

$$\det (A^3+B^3+C^3)=\det (A+B+C)|\det (A+\omega B+{\omega }^{2}C){|}^2$$

Since $|\det (A+\omega B+{\omega }^{2}C){|}^2\ge 0$, if $\det (A+B+C)\ge 0$, then $\det (A^3+B^3+C^3)\ge 0$.

Thus, $\det (A^3+B^3+C^3)\ge 0$.