Paragraph:
Let \(a\), b and \(c\) be three real numbers satisfying \(\left[\begin{array}{lll}a & b & c\end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]\)Question:
Let \(b=6\), with \(a\) and \(c\) satisfying Eq. (E). If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(a x^2+b x+c=0\), then \(\sum_{n=0}^{\infty}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^n\) is

\[
\begin{aligned}
& \text { Given, }[a b c]_{1 \times 3}\left[\begin{array}{lll}
1 & 9 & 7 \\
8 & 2 & 7 \\
7 & 3 & 7
\end{array}\right]_{3 \times 3}=\left[\begin{array}{lll}
0 & 0 & 0
\end{array}\right] \\
& \Rightarrow \quad\left[\begin{array}{c}
a+8 b+7 c \\
9 a+2 b+3 c \\
7 a+7 b+7 c
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] \\
& \Rightarrow \quad a+8 b+7 c=0 \\
& \Rightarrow 9 a+2 b+3 c=0 \\
& \Rightarrow \quad a+b+c=0 \\
&
\end{aligned}
\]
On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get
\[
7 a+c=0
\]
Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get
\[
\begin{array}{ll}
& 6 a-b=0 \\
\therefore \quad & b=6 a \text { and } c=-7 a
\end{array}
\]
If \(b=6, a=1\) and \(c=-7\)
\[
\begin{array}{ll}
\therefore & a x^2+b x+c=0 \\
\Rightarrow & x^2+6 x-7=0 \\
\Rightarrow & (x+7)(x-1)=0 \\
\therefore & x=1,-7 \\
\Rightarrow & \quad \sum_{n=0}^{\infty}\left(\frac{1}{1}-\frac{1}{7}\right)^n \Rightarrow \sum_{n=0}^{\infty}\left(\frac{6}{7}\right)^n \\
\Rightarrow & 1+\frac{6}{7}+\left(\frac{6}{7}\right)^n+\ldots \infty \\
& =\frac{1}{1-\frac{6}{7}}=\frac{1}{1 / 7}=7
\end{array}
\]