Question

Let $A=\left[\begin{array}{ccc}\frac{1}{6}& \frac{-1}{3}& \frac{-1}{6}\\ \frac{-1}{3}& \frac{2}{3}& \frac{1}{3}\\ \frac{-1}{6}& \frac{1}{3}& \frac{1}{6}\end{array}\right].$ If
$A^{2016l}+A^{2017m}+A^{2018n}=\frac{1}{\alpha }A$, 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)

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