If `a^(-1)+b^(-1)+c^(-1)=0` such that `|{:(1+a," "1," "1),(" "1,1+b," "1),(" "1," "1,1+c):}|=lambda`, then what is `lambda` equal to ?

Options

A. `-abc`
B. abc
C. 0
D. 1

Explanation

To find the value of \(\lambda\) for the given determinant, we need to follow these steps:

1. **Given Condition**:
\[ a^{-1} + b^{-1} + c^{-1} = 0 \]
This can be rewritten as:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \]
Taking the common denominator, we get:
\[ \frac{bc + ac + ab}{abc} = 0 \]
Hence:
\[ bc + ac + ab = 0 \]

2. **Matrix Determinant**:
We need to find the determinant of the matrix:
\[ \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} \]

3. **Expanding the Determinant**:
Expand along the first row:
\[ \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} = (1 + a) \begin{vmatrix} 1 + b & 1 \\ 1 & 1 + c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & 1 + c \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 + b \\ 1 & 1 \end{vmatrix} \]

4. **Calculating 2x2 Determinants**:
\[ \begin{vmatrix} 1 + b & 1 \\ 1 & 1 + c \end{vmatrix} = (1 + b)(1 + c) - 1 \cdot 1 = 1 + b + c + bc - 1 = b + c + bc \]
\[ \begin{vmatrix} 1 & 1 \\ 1 & 1 + c \end{vmatrix} = 1(1 + c) - 1 \cdot 1 = 1 + c - 1 = c \]
\[ \begin{vmatrix} 1 & 1 + b \\ 1 & 1 \end{vmatrix} = 1 \cdot 1 - 1 \cdot (1 + b) = 1 - (1 + b) = -b \]

5. **Substituting Back**:
\[ (1 + a)(b + c + bc) - c - b \]
Expanding:
\[ (1 + a)(b + c + bc) = b + c + bc + ab + ac + abc \]
So:
\[ \lambda = b + c + bc + ab + ac + abc - c - b \]
Simplifying:
\[ \lambda = bc + ab + ac + abc \]

6. **Using the Given Condition**:
From the given condition \(ab + bc + ac = 0\), we substitute:
\[ \lambda = abc \]

Therefore, the value of \(\lambda\) is:
\[ \boxed{abc} \]