Question

Solve the question in attachment.

Options are ;
(a) 1/6
(b) 1/3
(c) 1/2
(d) 2/3 ​

Answer 1

QUESTION:

$\displaystyle Let\:A=\left[\begin{array}{c c c} \frac{1}{6}& - \frac{1}{3} & - \frac{1}{6} \\ \\ - \frac{1}{3}& \frac{2}{3}&\frac{1}{3} \\ \\ - \frac{1}{6}& \frac{1}{3} & \frac{1}{6}\end{array} \right]\\ If \: \: {A}^{2016 \: l} +{A}^{2017 \: m} + {A}^{2018\: n} = \frac{1}{ \alpha }A\\ for \: \: every \: l, \: m, \: n \: \in \: N \: ,then \: \: find \: \: the \\ value \: \: of \: \: \alpha$

ANSWER:

$\textsf{\bf{The value of \alpha = \frac{1}{3} }}$

GIVEN:

${A}^{2016 \: l} +{A}^{2017 \: m} + {A}^{2018\: n} = \frac{1}{ \alpha }A\\ for \: \: every \: l, \: m, \: n \: \in \: N$

TO FIND:

$\textsf{The value of \alpha }$

EXPLANATION:

$\displaystyle A = \left[\begin{array}{c c c} \frac{1}{6}& - \frac{1}{3} & - \frac{1}{6} \\ \\ - \frac{1}{3}& \frac{2}{3}&\frac{1}{3} \\ \\ - \frac{1}{6}& \frac{1}{3} & \frac{1}{6} \end{array} \right] \\ \\ First \: \: \: find \: \: {A}^{2} \\ \\ {A}^{2} = \left[\begin{array}{c c c} \frac{1}{6}& - \frac{1}{3} & - \frac{1}{6} \\ \\ - \frac{1}{3}& \frac{2}{3}&\frac{1}{3} \\ \\ - \frac{1}{6}& \frac{1}{3} & \frac{1}{6} \end{array} \right] \left[\begin{array}{c c c} \frac{1}{6}& - \frac{1}{3} & - \frac{1}{6} \\ \\ - \frac{1}{3}& \frac{2}{3}&\frac{1}{3} \\ \\ - \frac{1}{6}& \frac{1}{3} & \frac{1}{6} \end{array} \right] \\ \\ Let \: \: {A}^{2} = \left[\begin{array}{c c c} a_{11}& a_{12}& a_{13}\\ \\ a_{21} & a_{22} &a_{23} \\ \\ a_{31} & a_{32} &a_{33} \end{array} \right]$

$a_{11} = \frac{1}{6} \times \frac{1}{6} + \frac{ - 1}{3} \times \frac{ - 1}{3} + \frac{ - 1}{6} \times \frac{ - 1}{6} \\ \\ a_{11} = \frac{1}{36} + \frac{1}{9} + \frac{1}{36} \\ \\ a_{11} = \frac{1 + 4 + 1}{36} \implies \frac{6}{36} = \frac{1}{6} \\ \\ a_{12} = \frac{ 1}{6} \times \frac{ - 1}{3} + \frac{ - 1}{3} \times \frac{2}{3} + \frac{ - 1}{6} \times \frac{ 1}{3} \\ \\ a_{12} = \frac{ - 1}{18} - \frac{2}{9} - \frac{1}{18} \\ \\ a_{12} = \frac{ - 1 - 4- 1}{18} \implies \frac{ - 6}{18} = - \frac{ 1}{3} \\ \\ a_{13} = \frac{1}{6} \times \frac{ - 1}{6} + \frac{ - 1}{3} \times \frac{1}{3} + \frac{ - 1}{6} \times \frac{1}{6} \\ \\ a_{13} = \frac{ - 1}{36} - \frac{1}{9} - \frac{1}{36} \\ \\ a_{13} = \frac{ - 1 - 4 - 1}{36} \implies \frac{ - 6}{36} = - \frac{1}{6}$

$a_{21} = \frac{ - 1}{3} \times \frac{1}{6} + \frac{ 2}{3} \times \frac{ - 1}{3} + \frac{ 1}{3} \times \frac{ - 1}{6} \\ \\a_{21} = - \frac{1}{18} - \frac{2}{9} - \frac{1}{18} \\ \\ a_{21} = \frac{ - 1 - 4 - 1}{18} \implies - \frac{6}{18} = - \frac{1}{3} \\ \\ a_{22} = \frac{ - 1}{3} \times \frac{ - 1}{3} + \frac{ 2}{3} \times \frac{2}{3} + \frac{ 1}{3} \times \frac{ 1}{3} \\ \\ a_{22} = \frac{1}{9} + \frac{4}{9} + \frac{1}{9} \\ \\ a_{22} \implies \frac{6}{9} = \frac{2}{3} \\ \\ a_{23} = \frac{ - 1}{3} \times \frac{ - 1}{6} + \frac{ 2}{3} \times \frac{1}{3} + \frac{ 1}{3} \times \frac{1}{6} \\ \\ a_{23} = \frac{1}{18} + \frac{2}{9} + \frac{1}{18} \\ \\ a_{23} = \frac{1 + 4 + 1}{18} \implies \frac{6}{18} = \frac{1}{3}$

$a_{31} = \frac{ - 1}{6} \times \frac{1}{6} + \frac{ 1}{3} \times \frac{ - 1}{3} + \frac{ 1}{6} \times \frac{ - 1}{6} \\ \\ a_{31} = - \frac{1}{36} - \frac{1}{9} - \frac{1}{36} \\ \\ a_{31} = \frac{ - 1 - 4 - 1}{36} \implies - \frac{6}{36} = - \frac{1}{6} \\ \\ a_{32} = \frac{ - 1}{6} \times \frac{ - 1}{3} + \frac{1}{3} \times \frac{2}{3} + \frac{ 1}{6} \times \frac{ 1}{3} \\ \\ a_{32} = \frac{1}{18} + \frac{2}{9} + \frac{1}{18} \\ \\ a_{33} = \frac{1 + 4 + 1}{18} \implies \frac{6}{18} = \frac{1}{3} \\ \\ a_{33} = \frac{ - 1}{6} \times \frac{ - 1}{6} + \frac{ 1}{3} \times \frac{1}{3} + \frac{ 1}{6} \times \frac{1}{6} \\ \\ a_{33} = \frac{1}{36} + \frac{1}{9} + \frac{1}{36} \\ \\ a_{33} = \frac{1 + 4 + 1}{36} \implies \frac{6}{36} = \frac{1}{6}$

$SUBSTITUTE \: \: THE \: \: VALUES \: \: OF \\ a_{11}, \: \: a_{12}, \: \: a_{13}, \: \: a_{21}, \: \: a_{22}, \: \: a_{23}, \\ a_{31}, \: \: a_{32}, \: \: a_{33} \: \: in \: \: A^{2}$

$A^{2} = \left[\begin{array}{c c c} \frac{1}{6}& - \frac{1}{3} & - \frac{1}{6} \\ \\ - \frac{1}{3}& \frac{2}{3}&\frac{1}{3} \\ \\ - \frac{1}{6}& \frac{1}{3} & \frac{1}{6} \end{array} \right] \\ \\ A^{2} = A \\ \\ \textsf{Multiply by A on both sides} \\ A^{3} = A^{2}$

Hence multiplying A n times(Aⁿ) also gives A

Also given l, m, n are natural numbers.

${A}^{2016 \: l} +{A}^{2017 \: m} + {A}^{2018\: n} = \frac{1}{ \alpha }A \\ \\ {A}^{2016 \: l} = {A}^{2017 \: m} = {A}^{2018\: n} = A \\ \\ 3A = \frac{1}{ \alpha}A \\ \\ \frac{1}{ \alpha} = 3 \\ \\ \alpha = \frac{1}{3}$

$\textsf{\bf{Hence the value of \alpha = \frac{1}{3} }}$

Answer 2

ANSWER :- 1/3

SOLUTION :- GIVEN IN THE ATTACHMENT

HOPE IT HELPS :)