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 find the minimum value of the determinant of the matrix \( A = |a_{ij}| \) where \( a_{ij} \) is defined as follows:

\[ a_{ij} = \begin{cases} i^j - j^i + 2ijx & \text{if } i < j \\ 1 & \text{if } i > j \\ 0 & \text{if } i = j \end{cases} \]

We will construct the \( 3 \times 3 \) matrix \( A \) and then calculate its determinant.

### Step 1: Construct the Matrix \( A \)

The elements of the matrix \( A \) can be filled as follows:

- For \( i = 1 \):
- \( a_{11} = 0 \)
- \( a_{12} = 1^2 - 2^1 + 2 \cdot 1 \cdot 2 \cdot x = 1 - 2 + 4x = 4x - 1 \)
- \( a_{13} = 1^3 - 3^1 + 2 \cdot 1 \cdot 3 \cdot x = 1 - 3 + 6x = 6x - 2 \)

- For \( i = 2 \):
- \( a_{21} = 1 \)
- \( a_{22} = 0 \)
- \( a_{23} = 2^3 - 3^2 + 2 \cdot 2 \cdot 3 \cdot x = 8 - 9 + 12x = 12x - 1 \)

- For \( i = 3 \):
- \( a_{31} = 1 \)
- \( a_{32} = 1 \)
- \( a_{33} = 0 \)

Thus, the matrix \( A \) is:

\[ A = \begin{pmatrix} 0 & 4x - 1 & 6x - 2 \\ 1 & 0 & 12x - 1 \\ 1 & 1 & 0 \end{pmatrix} \]

### Step 2: Calculate the Determinant of Matrix \( A \)

To calculate the determinant \( |A| \), we can use the formula for the determinant of a \( 3 \times 3 \) 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 from matrix \( A \):

\[ |A| = 0 \cdot (0 \cdot 0 - (12x - 1) \cdot 1) - (4x - 1)(1 \cdot 0 - (12x - 1) \cdot 1) + (6x - 2)(1 \cdot 1 - 0 \cdot 1) \]

This simplifies to:

\[ |A| = 0 - (4x - 1)(-12x + 1) + (6x - 2)(1) \]

Calculating each term:

1. \( (4x - 1)(-12x + 1) = -48x^2 + 4x + 12x - 1 = -48x^2 + 16x - 1 \)
2. \( (6x - 2)(1) = 6x - 2 \)

Combining these:

\[ |A| = 48x^2 - 16x + 1 + 6x - 2 \]
\[ |A| = 48x^2 - 10x - 1 \]

### Step 3: Find the Minimum Value of the Determinant

To find the minimum value of the quadratic \( |A| = 48x^2 - 10x - 1 \), 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:

\[ \text{Minimum value of } |A| = -\frac{3504}{2304} \]