Q.If $A=\begin{bmatrix}1 & \omega\\\omega^2 & 1\end{bmatrix}$, $B=\begin{bmatrix}\omega^2 & 1\\1 & \omega\end{bmatrix}$, where $\omega$ is a complex cube root of unity, then show that $AB+BA+A-2B$ is a null matrix.

$A=\begin{bmatrix}1 & \omega\\\omega^2 & 1\end{bmatrix}$, $B=\begin{bmatrix}\omega^2 & 1\\1 & \omega\end{bmatrix}$, using $\omega^3=1$, $1+\omega+\omega^2=0$.

$AB$: entry(1,1)$=\omega^2+\omega=-1$; entry(1,2)$=1+\omega^2=-\omega$; entry(2,1)$=\omega^4+1=\omega+1=-\omega^2$; entry(2,2)$=\omega^2+\omega=-1$.

So $AB=\begin{bmatrix}-1 & -\omega\\-\omega^2 & -1\end{bmatrix}$.

$BA$: entry(1,1)$=\omega^2+\omega^2=2\omega^2$; entry(1,2)$=\omega^3+1=1+1=2$; entry(2,1)$=1+\omega^3=1+1=2$; entry(2,2)$=\omega+\omega=2\omega$.

So $BA=\begin{bmatrix}2\omega^2 & 2\\2 & 2\omega\end{bmatrix}$.

$AB+BA=\begin{bmatrix}-1+2\omega^2 & -\omega+2\\-\omega^2+2 & -1+2\omega\end{bmatrix}$. …

Compute AB and BA using $\omega^3=1$ and $1+\omega+\omega^2=0$, t …

• Forgetting $\omega^4=\omega^3\cdot\omega=\omega$ (reduce powers of $\omega$ modulo 3). …