Let a,b, and c be three real numbers satistying `[a,b,c][(1,9,7),(8,2,7),(7,3,7)]=[0,0,0]`Let b=6, with a and c satisfying (E). If alpha and beta are the roots of the quadratic equation `ax^2+bx+c=0 then sum_(n=0)^oo (1/alpha+1/beta)^n` is (A) 6 (B) 7 (C) `6/7` (D) oo

Options

A. 6
B. 7
C. `6/7`
D. `oo`

Explanation

To solve the problem step by step, we will follow the instructions given in the video transcript and derive the values of \(a\), \(b\), and \(c\) from the matrix equation, then find the roots of the quadratic equation, and finally compute the infinite series.

### Step 1: Set up the matrix equation
We have the equation:
\[ [a, b, c] \begin{pmatrix} 1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7 \end{pmatrix} = [0, 0, 0] \]
This implies that the linear combinations of the rows of the matrix with coefficients \(a\), \(b\), and \(c\) must equal zero.

### Step 2: Write the equations from the matrix multiplication
From the matrix multiplication, we can derive the following equations:
1. \( a + 8b + 7c = 0 \)
2. \( 9a + 2b + 3c = 0 \)
3. \( 7a + 7b + 7c = 0 \)

### Step 3: Substitute \(b = 6\)
Given that \(b = 6\), we can substitute this value into the equations:
1. \( a + 8(6) + 7c = 0 \) → \( a + 48 + 7c = 0 \) → \( a + 7c = -48 \) (Equation 1)
2. \( 9a + 2(6) + 3c = 0 \) → \( 9a + 12 + 3c = 0 \) → \( 9a + 3c = -12 \) (Equation 2)
3. \( 7a + 7(6) + 7c = 0 \) → \( 7a + 42 + 7c = 0 \) → \( 7a + 7c = -42 \) (Equation 3)

### Step 4: Simplify the equations
From Equation 1:
\[ a + 7c = -48 \quad \text{(1)} \]
From Equation 2:
\[ 9a + 3c = -12 \quad \text{(2)} \]
From Equation 3:
\[ 7a + 7c = -42 \quad \text{(3)} \]

### Step 5: Solve for \(c\) using Equations (1) and (3)
From Equation (3):
\[ 7a + 7c = -42 \implies a + c = -6 \quad \text{(4)} \]

### Step 6: Substitute Equation (4) into Equation (1)
Substituting \(c = -6 - a\) into Equation (1):
\[ a + 7(-6 - a) = -48 \implies a - 42 - 7a = -48 \implies -6a = -6 \implies a = 1 \]

### Step 7: Find \(c\)
Substituting \(a = 1\) back into Equation (4):
\[ 1 + c = -6 \implies c = -7 \]

### Step 8: Summary of values
We have found:
- \(a = 1\)
- \(b = 6\)
- \(c = -7\)

### Step 9: Form the quadratic equation
The quadratic equation is:
\[ 1x^2 + 6x - 7 = 0 \]

### Step 10: Find the roots \(\alpha\) and \(\beta\)
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4(1)(-7)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 28}}{2} = \frac{-6 \pm \sqrt{64}}{2} = \frac{-6 \pm 8}{2} \]
Thus, the roots are:
\[ \alpha = 1, \quad \beta = -7 \]

### Step 11: Calculate the infinite series
We need to compute:
\[ \sum_{n=0}^{\infty} \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)^n \]
Calculating \(\frac{1}{\alpha} + \frac{1}{\beta}\):
\[ \frac{1}{1} + \frac{1}{-7} = 1 - \frac{1}{7} = \frac{6}{7} \]
This series is a geometric series with first term \(1\) and common ratio \(\frac{6}{7}\):
\[ \sum_{n=0}^{\infty} \left(\frac{6}{7}\right)^n = \frac{1}{1 - \frac{6}{7}} = \frac{1}{\frac{1}{7}} = 7 \]

### Final Answer
Thus, the value of the infinite series is \(7\).