Question

Let $a,b$ and $c$ be three real numbers satisfying

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 $

Answer 1

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

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 ${S_\infty } = \dfrac{a}{{1 - r}}$.