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# 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) $\infty$

Last updated date: 18th Sep 2024
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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.

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 = $$\dfrac{{ - b}}{a} = \dfrac{{ - 6}}{1} = - 6 Multiplication of roots, \alpha \beta = \dfrac{c}{a} = \dfrac{{ - 7}}{1} = - 7 Now, \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}} =\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{{\beta + \alpha }}{{\alpha \beta }}} \right)}^n}} =\sum\limits_{n = 0}^\infty {{{\left( {\dfrac{{ - 6}}{{ - 7}}} \right)}^n}} = \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{6}{7}} \right)}^n}} On expand it, we get- = {\left( {\dfrac{6}{7}} \right)^0} + {\left( {\dfrac{6}{7}} \right)^1} + {\left( {\dfrac{6}{7}} \right)^2} + {\left( {\dfrac{6}{7}} \right)^3} + ........................ + {\left( {\dfrac{6}{7}} \right)^n} = 1 + {\left( {\dfrac{6}{7}} \right)^1} + {\left( {\dfrac{6}{7}} \right)^2} + {\left( {\dfrac{6}{7}} \right)^3} + ........................ + {\left( {\dfrac{6}{7}} \right)^n} This is an infinite Geometric Progression, whose sum of infinite terms is given by {S_\infty } = \dfrac{a}{{1 - r}} Where a is the first term of G.P. and r is the common ratio of G.P. Here we have, a = 1and r = \dfrac{6}{7} \therefore \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right)}^n}}$$ = \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 ${S_\infty } = \dfrac{a}{{1 - r}}$.