Let `A=|a_(ij)|" be a "3xx3` matrix where `a_(ij)={{:((i^(j)-j^(i)+2ij)x,iltj),(1,igtj","),(0,i=j):}`, then the minimum value of `|A|` is equal to (where x is a real number)

Options

A. `(1)/(4)`
B. `-(8)/(33)`
C. 7
D. `-(4)/(33)`

Answer

To solve the problem, we need to find the determinant of the given 3x3 matrix \( A = |a_{ij}| \) where the elements \( a_{ij} \) are defined based on the conditions provided. Let's break down the solution step by step.

### Step 1: Define the matrix elements
The matrix \( A \) is defined as follows:
- \( a_{ij} = 0 \) if \( i = j \)
- \( a_{ij} = i^j - j^i + 2ijx \) if \( i < j \)
- \( a_{ij} = 1 \) if \( i > j \)

### Step 2: Fill in the matrix
We will fill in the elements of the matrix \( A \):
\[ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \]

- For \( a_{11} \): \( i = j \) so \( a_{11} = 0 \)
- For \( a_{22} \): \( i = j \) so \( a_{22} = 0 \)
- For \( a_{33} \): \( i = j \) so \( a_{33} = 0 \)

Now, we compute the other elements:
- For \( a_{12} \): \( i = 1, j = 2 \) so \( a_{12} = 1^2 - 2^1 + 2 \cdot 1 \cdot 2 \cdot x = 1 - 2 + 4x = 4x - 1 \)
- For \( a_{13} \): \( i = 1, j = 3 \) so \( a_{13} = 1^3 - 3^1 + 2 \cdot 1 \cdot 3 \cdot x = 1 - 3 + 6x = 6x - 2 \)
- For \( a_{21} \): \( i = 2, j = 1 \) so \( a_{21} = 1 \)
- For \( a_{23} \): \( i = 2, j = 3 \) so \( a_{23} = 2^3 - 3^2 + 2 \cdot 2 \cdot 3 \cdot x = 8 - 9 + 12x = 12x - 1 \)
- For \( a_{31} \): \( i = 3, j = 1 \) so \( a_{31} = 1 \)
- For \( a_{32} \): \( i = 3, j = 2 \) so \( a_{32} = 1 \)

Putting these together, we have:
\[ A = \begin{bmatrix} 0 & 4x - 1 & 6x - 2 \\ 1 & 0 & 12x - 1 \\ 1 & 1 & 0 \end{bmatrix} \]

### Step 3: Calculate the determinant
To find the determinant \( |A| \), we can use the formula for the determinant of a 3x3 matrix:
\[ |A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]

Substituting the values:
\[ |A| = 0 \cdot (0 \cdot 0 - (12x - 1) \cdot 1) - (4x - 1)(1 \cdot 0 - (12x - 1) \cdot 1) + (6x - 2)(1 \cdot 1 - 0 \cdot 1) \]
\[ = 0 - (4x - 1)(- (12x - 1)) + (6x - 2)(1) \]
\[ = (4x - 1)(12x - 1) + (6x - 2) \]
\[ = 48x^2 - 4x - 12x + 1 + 6x - 2 \]
\[ = 48x^2 + (-4x - 12x + 6x) + (1 - 2) \]
\[ = 48x^2 - 10x - 1 \]

### Step 4: Find the minimum value
To find the minimum value of the quadratic \( 48x^2 - 10x - 1 \), we can use the vertex formula \( x = -\frac{b}{2a} \):
\[ x = -\frac{-10}{2 \cdot 48} = \frac{10}{96} = \frac{5}{48} \]

Substituting \( x = \frac{5}{48} \) back into the determinant:
\[ |A| = 48\left(\frac{5}{48}\right)^2 - 10\left(\frac{5}{48}\right) - 1 \]
\[ = 48 \cdot \frac{25}{2304} - \frac{50}{48} - 1 \]
\[ = \frac{1200}{2304} - \frac{50}{48} - 1 \]
\[ = \frac{1200}{2304} - \frac{2400}{2304} - \frac{2304}{2304} \]
\[ = \frac{1200 - 2400 - 2304}{2304} = \frac{-3504}{2304} \]

Thus, the minimum value of \( |A| \) is \( -\frac{3504}{2304} \).

### Final Answer
The minimum value of \( |A| \) is \( -\frac{3504}{2304} \).