Let $a,\ b$ and $c$ be three real numbers satisfying $\left[ \begin{matrix} a & b & c \end{matrix} \right] \begin{bmatrix} 1 & 9 & 7 \ 8 & 2 & 7 \ 7 & 3 & 7 \end{bmatrix}=\left[ \begin{matrix} 0 & 0 & 0 \end{matrix} \right]$ ........ $(E)$ Let $b=6$, with $a$ and $c$ satisfying (E). lf $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2}+bx+c=0$, then $\displaystyle \sum_{n=0}^{\infty} \left (\displaystyle \frac{1}{\alpha}+\displaystyle \frac{1}{\beta}\right )^{n} $ is:

Expanding the matrix multiplication yields three linear equations: a+8b+7c=0, 9a+2b+3c=0, and 7a+7b+7c=0; the third simplifies to a+b+c=0. Substituting b=6 gives a+c=-6, and using this in the second equation yields 9a+12+3(-6-a)=0, which solves to a=1 and subsequently c=-7. The quadratic equation becomes x^2+6x-7=0, and by Vieta's formulas the sum of the roots is -6 and the product is -7, so (1/alpha + 1/beta) = (alpha+beta)/(alpha beta) = -6/-7 = 6/7. Evaluating the infinite geometric series with first term 1 and common ratio 6/7 gives the sum 1 / (1 - 6/7) = 7.

• Thanks
• Wrong explanation