If a,b,c are negative distinct real numbers then the determinant `|(a,b,c),(b,c,a),(c,a,b)|` is

Options

A. `lt 0`
B. `le 0`
C. `gt0`
D. `ge0`

Answer

To find the determinant \( D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \), where \( a, b, c \) are distinct negative real numbers, we can follow these steps:

### Step 1: Apply Column Operations
We can simplify the determinant by performing column operations. Let's add all three columns together:

\[ C_1 \to C_1 + C_2 + C_3 \]

This gives us:

\[ D = \begin{vmatrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \end{vmatrix} \]

### Step 2: Simplify the Determinant
Now, we can factor out \( a + b + c \) from the first column:

\[ D = (a + b + c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} \]

### Step 3: Expand the 3x3 Determinant
Next, we can expand the 3x3 determinant:

\[ D = (a + b + c) \left( 1 \cdot \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \cdot \begin{vmatrix} 1 & a \\ 1 & b \end{vmatrix} + c \cdot \begin{vmatrix} 1 & c \\ 1 & a \end{vmatrix} \right) \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} c & a \\ a & b \end{vmatrix} = cb - a^2 \)
2. \( \begin{vmatrix} 1 & a \\ 1 & b \end{vmatrix} = b - a \)
3. \( \begin{vmatrix} 1 & c \\ 1 & a \end{vmatrix} = a - c \)

Substituting these back into the determinant expression:

\[ D = (a + b + c) \left( cb - a^2 - b(b - a) + c(a - c) \right) \]

### Step 4: Combine Like Terms
Now we simplify the expression:

\[ D = (a + b + c) \left( cb - a^2 - b^2 + ab + ca - c^2 \right) \]

### Step 5: Analyze the Sign of the Determinant
Since \( a, b, c \) are distinct negative real numbers, \( a + b + c < 0 \). The expression \( cb - a^2 - b^2 + ab + ca - c^2 \) is a quadratic in terms of \( a, b, c \) and will yield a positive value when evaluated due to the distinctness and negativity of \( a, b, c \).

Thus, the overall determinant \( D \) will be negative because it is the product of a negative number \( (a + b + c) \) and a positive quantity.

### Final Result
Therefore, the determinant \( D \) is negative:

\[ D < 0 \]