If `a=x/(y-z),b=y/(z-x) and c = z/(x-y)` where x,y,z are not all zero , then `ab + bc + ca = `

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the problem, we need to find the value of \( ab + bc + ca \) given the definitions of \( a \), \( b \), and \( c \):

1. **Given Definitions:**
\[ a = \frac{x}{y - z}, \quad b = \frac{y}{z - x}, \quad c = \frac{z}{x - y} \]

2. **Rearranging the Definitions:**
We can rewrite these equations to express \( x \), \( y \), and \( z \) in terms of \( a \), \( b \), and \( c \):
\[ x = a(y - z) \quad (1) \]
\[ y = b(z - x) \quad (2) \]
\[ z = c(x - y) \quad (3) \]

3. **Substituting Values:**
From equation (1), we can express \( y \) and \( z \) in terms of \( x \):
\[ y = \frac{x}{a} + z \quad \text{(from } x = a(y - z) \text{)} \]
From equation (2), we can express \( z \):
\[ z = \frac{y}{b} + x \quad \text{(from } y = b(z - x) \text{)} \]
From equation (3), we can express \( x \):
\[ x = \frac{z}{c} + y \quad \text{(from } z = c(x - y) \text{)} \]

4. **Setting Up the Determinant:**
We can set up a system of equations to find a non-trivial solution (where \( x, y, z \) are not all zero):
\[ \begin{vmatrix} 1 & -a & a \\ b & 1 & -b \\ -c & c & 1 \end{vmatrix} = 0 \]

5. **Calculating the Determinant:**
We can calculate the determinant using the formula for a 3x3 matrix:
\[ D = 1 \cdot (1 \cdot 1 - (-b)(c)) - (-a) \cdot (b \cdot 1 - (-b)(-c)) + a \cdot (b \cdot (-c) - 1 \cdot (-b)) \]
Simplifying this gives:
\[ D = 1 + bc + ab - ac + ab + ac - bc \]
Thus:
\[ D = 1 + ab + bc + ac \]

6. **Setting the Determinant to Zero:**
Since we require a non-trivial solution, we set the determinant to zero:
\[ 1 + ab + bc + ac = 0 \]

7. **Final Result:**
Rearranging gives us:
\[ ab + bc + ac = -1 \]

Thus, the final answer is:
\[ \boxed{-1} \]