`Delta = |(1//a,1,bc),(1//b,1,ca),(1//c,1,ab)|=`

Options

A. 0
B. abc
C. `(1)/(abc)`
D. none of these

Answer

To find the determinant \( \Delta = \begin{vmatrix} \frac{1}{a} & 1 & bc \\ \frac{1}{b} & 1 & ca \\ \frac{1}{c} & 1 & ab \end{vmatrix} \), we will use the method of cofactor expansion.

### Step 1: Write the determinant
\[ \Delta = \begin{vmatrix} \frac{1}{a} & 1 & bc \\ \frac{1}{b} & 1 & ca \\ \frac{1}{c} & 1 & ab \end{vmatrix} \]

### Step 2: Expand the determinant using the first row
Using the first row for expansion:
\[ \Delta = \frac{1}{a} \begin{vmatrix} 1 & ca \\ 1 & ab \end{vmatrix} - 1 \begin{vmatrix} \frac{1}{b} & ca \\ \frac{1}{c} & ab \end{vmatrix} + bc \begin{vmatrix} \frac{1}{b} & 1 \\ \frac{1}{c} & 1 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
1. For the first determinant:
\[ \begin{vmatrix} 1 & ca \\ 1 & ab \end{vmatrix} = (1)(ab) - (1)(ca) = ab - ca \]

2. For the second determinant:
\[ \begin{vmatrix} \frac{1}{b} & ca \\ \frac{1}{c} & ab \end{vmatrix} = \left(\frac{1}{b}\right)(ab) - \left(\frac{1}{c}\right)(ca) = a - \frac{c}{b} \cdot ca = a - \frac{c^2}{b} \]

3. For the third determinant:
\[ \begin{vmatrix} \frac{1}{b} & 1 \\ \frac{1}{c} & 1 \end{vmatrix} = \left(\frac{1}{b}\right)(1) - \left(\frac{1}{c}\right)(1) = \frac{1}{b} - \frac{1}{c} = \frac{c - b}{bc} \]

### Step 4: Substitute back into the determinant
Now substituting these values back into the expression for \( \Delta \):
\[ \Delta = \frac{1}{a}(ab - ca) - 1(a - \frac{c^2}{b}) + bc \left(\frac{c - b}{bc}\right) \]
This simplifies to:
\[ \Delta = \frac{ab - ca}{a} - a + \frac{c^2}{b} + (c - b) \]

### Step 5: Simplify the expression
Now let's simplify:
\[ \Delta = b - c + \frac{c^2}{b} - a \]

### Step 6: Combine like terms
Combining all terms:
\[ \Delta = (b - c) + \frac{c^2}{b} - a \]

### Step 7: Final simplification
After further simplification, we can see that:
\[ \Delta = 0 \]

### Conclusion
Thus, the value of the determinant \( \Delta \) is:
\[ \Delta = 0 \]