Question

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
$A^{2016 l}+A^{2017 m}+A^{2018 n}=\frac{1}{\alpha} A$, for every
l, $m, n \in N$, then the value of $\alpha$ is

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

Answer: (B)

$\because A=\begin{bmatrix}\frac{1}{6} & -\frac{1}{3} & -\frac{1}{6} \\ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{6} & \frac{1}{3} & \frac{1}{6}\end{bmatrix}=6\begin{bmatrix}1 & -2 & -1 \\ -2 & 4 & 2 \\ -1 & 2 & 1\end{bmatrix}$
$A^{p}=A$, for every $p \in N$
So, $A^{2016 l}=A^{2017 m}=A^{2018 n}$
$=A$, for every $l, m, n \in N$
So, $A^{2016 l}+A^{2017 m}+A^{2018 n}=3 A=\frac{1}{\alpha} A $ (given)
$\Rightarrow \alpha=\frac{1}{3}$