Let \[a,b\] and $c$ be three real numbers satisfying
\[\left[ {\begin{array}{*{20}{c}}
  a&b&c
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&9&7 \\
  8&2&7 \\
  7&3&7
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0&0
\end{array}} \right]......\left( E \right)\]
Let $b = 6$, with $a$ and $c$ satisfying $\left( E \right).$ If $\alpha $ and $\beta $ are the roots of the quadratic equation $a{x^2} + bx + c = 0$, then $\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}} $ is:
(A) 6
(B) 7
(C) $\dfrac{6}{7}$
(D) $\infty $

Hint: The multiplication of two matrices is possible if the no. of columns in matrix A is equal to the no. of rows in matrix B. Here we multiplied the two given matrix and form the equations by comparing the values of both sides.

Complete step-by-step answer:
Since, \[a,b\] and $c$ be three real numbers satisfies
\[\left[ {\begin{array}{*{20}{c}}
  a&b&c
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&9&7 \\
  8&2&7 \\
  7&3&7
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0&0
\end{array}} \right]\]
So, we get the equations
$
  a + 8b + 7c = 0 \\
  9a + 2b + 3c = 0 \\
  7a + 7b + 7c = 0 \Rightarrow a + b + c = 0 \\
 $
Since, $b = 6$, so the equations become
$
  a + 8\left( 6 \right) + 7c = 0 \Rightarrow a + 7c = - 48....(1) \\
  9a + 2\left( 6 \right) + 3c = 0 \Rightarrow 9a + 3c = - 12....(2) \\
  a + 6 + c = 0 \Rightarrow a + c = - 6....(3) \\
 $
On subtracting equation (3) from (1), we get
$a + 7c - \left( {a + c} \right) = - 48 - \left( { - 6} \right)$
$
   \Rightarrow a + 7c - a - c = - 48 + 6 \\
   \Rightarrow 6c = - 42 \\
   \Rightarrow c = - 7 \\
 $
 Substitute the value of $c$ in equation (3), we get
$
  a + \left( { - 7} \right) = - 6 \\
   \Rightarrow a - 7 = - 6 \\
   \Rightarrow a = - 6 + 7 \\
   \Rightarrow a = 1 \\
 $
So, we have $a = 1,b = 6,c = - 7$
Given quadratic equation is $a{x^2} + bx + c = 0$. After putting the values of \[a,b\] and $c$, it becomes ${x^2} + 6x - 7 = 0$.
Since, $\alpha $ and $\beta $ are the roots of this equation, So
Sum of roots, $\alpha + \beta = $$\dfrac{{ - b}}{a} = \dfrac{{ - 6}}{1} = - 6$
Multiplication of roots, $\alpha \beta = \dfrac{c}{a} = \dfrac{{ - 7}}{1} = - 7$
Now, $\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}} $
=$\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{{\beta + \alpha }}{{\alpha \beta }}} \right)}^n}} $
=$\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{{ - 6}}{{ - 7}}} \right)}^n}} $
$ = \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{6}{7}} \right)}^n}} $
On expand it, we get-
$ = {\left( {\dfrac{6}{7}} \right)^0} + {\left( {\dfrac{6}{7}} \right)^1} + {\left( {\dfrac{6}{7}} \right)^2} + {\left( {\dfrac{6}{7}} \right)^3} + ........................ + {\left( {\dfrac{6}{7}} \right)^n}$
$ = 1 + {\left( {\dfrac{6}{7}} \right)^1} + {\left( {\dfrac{6}{7}} \right)^2} + {\left( {\dfrac{6}{7}} \right)^3} + ........................ + {\left( {\dfrac{6}{7}} \right)^n}$
This is an infinite Geometric Progression, whose sum of infinite terms is given by
${S_\infty } = \dfrac{a}{{1 - r}}$
Where $a$ is the first term of G.P. and $r$ is the common ratio of G.P.
Here we have, $a = 1$and $r = \dfrac{6}{7}$
$\therefore $ $\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}} $$ = \dfrac{1}{{1 - \dfrac{6}{7}}}$
$ \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}} = \dfrac{1}{{\dfrac{{7 - 6}}{7}}}$
$ \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}} = \dfrac{1}{{\dfrac{1}{7}}}$
$ \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}} = 7$

Hence, option (B) is the correct answer.

Note: If $\alpha $ and $\beta $ are the roots of the quadratic equation $a{x^2} + bx + c = 0$, then sum of roots, $\alpha + \beta = \dfrac{{ - b}}{a}$ and multiplication of roots, $\alpha \beta = \dfrac{c}{a}$. Also, the sum of infinite terms of an G.P. is ${S_\infty } = \dfrac{a}{{1 - r}}$.