Let A = begin{bmatrix} frac{1}{6} & frac{-1}{ | vedclass

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(B) Given $A = \begin{bmatrix} \frac{1}{6} & -\frac{1}{3} & -\frac{1}{6} \ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \ -\frac{1}{6} & \frac{1}{3} & \

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$\frac{1}{3}$

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(B) Given $A = \begin{bmatrix} \frac{1}{6} & -\frac{1}{3} & -\frac{1}{6} \ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \ -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix}$.First, we calculate $A^2 = A 	imes A$.$A^2 = \begin{bmatrix} \frac{1}{6} & -\frac{1}{3} & -\frac{1}{6} \ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \ -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix} \begin{bmatrix} \frac{1}{6} & -\frac{1}{3} & -\frac{1}{6} \ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \ -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix} = \frac{1}{36} \begin{bmatrix} 6 & -12 & -6 \ -12 & 24 & 12 \ -6 & 12 & 6 \end{bmatrix} = \begin{bmatrix} \frac{1}{6} & -\frac{1}{3} & -\frac{1}{6} \ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \ -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix} = A$.Since $A^2 = A$, it follows that $A^k = A$ for all $k \in N$.Therefore, $A^{2016l} = A$, $A^{2017m} = A$, and $A^{2018n} = A$.The given equation becomes $A + A + A = \frac{1}{\alpha} A$, which simplifies to $3A = \frac{1}{\alpha} A$.Comparing both sides, we get $\frac{1}{\alpha} = 3$, which implies $\alpha = \frac{1}{3}$.

Let $A = \begin{bmatrix} \frac{1}{6} & \frac{-1}{3} & \frac{-1}{6} \\ \frac{-1}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{-1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix}$. If $A^{2016l} + A^{2017m} + A^{2018n} = \frac{1}{\alpha} A$, for every $l, m, n \in N$, then the value of $\alpha$ is

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