Let `ab=1,Delta=|{:(1+a^2-b^2, 2ab,-2b),(2ab,1-a^2+b^2, 2a),(2b,-2a,1-a^2-b^2):}|` then the minimum value of `Delta` is :

Options

A. 3
B. 9
C. 27
D. 81

Answer

To find the minimum value of the determinant \( \Delta \) given that \( ab = 1 \), we will follow these steps:

### Step 1: Write the Determinant
Given the determinant:
\[ \Delta = \begin{vmatrix} 1 + a^2 - b^2 & 2ab & -2b \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix} \]

### Step 2: Apply Row Operations
We will perform row operations to simplify the determinant. Let's apply the operation \( R_1 \to R_1 + bR_3 \):
\[ R_1 = (1 + a^2 - b^2 + 2b^2, 2ab - 2ab, 2a - 2b + b(1 - a^2 - b^2)) \]
This results in:
\[ \Delta = \begin{vmatrix} 1 + a^2 + b^2 & 0 & 2a - 2b + b(1 - a^2 - b^2) \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix} \]

### Step 3: Simplify the Rows
Next, we simplify the second row by performing \( R_2 \to R_2 - aR_3 \):
\[ R_2 = (2ab - 2ab, 1 - a^2 + b^2 + 2a, 2a - a(1 - a^2 - b^2)) \]
This results in:
\[ \Delta = \begin{vmatrix} 1 + a^2 + b^2 & 0 & 2a - 2b + b(1 - a^2 - b^2) \\ 0 & 1 + a^2 + b^2 & 0 \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix} \]

### Step 4: Factor Out Common Terms
Notice that \( 1 + a^2 + b^2 \) is common in the second row:
\[ \Delta = (1 + a^2 + b^2) \begin{vmatrix} 1 + a^2 + b^2 & 0 & 2a - 2b + b(1 - a^2 - b^2) \\ 0 & 1 & 0 \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix} \]

### Step 5: Expand the Determinant
Now, we can expand the determinant along the first row:
\[ \Delta = (1 + a^2 + b^2) \cdot (1) \cdot \begin{vmatrix} 1 - a^2 - b^2 & 0 \\ 2b & -2a \end{vmatrix} \]
Calculating this determinant gives:
\[ \Delta = (1 + a^2 + b^2)(-2a(1 - a^2 - b^2)) \]

### Step 6: Substitute \( ab = 1 \)
Since \( ab = 1 \), we can express \( b \) in terms of \( a \):
\[ b = \frac{1}{a} \]
Now substitute \( b \) into the expression for \( \Delta \):
\[ \Delta = (1 + a^2 + \frac{1}{a^2})(-2a(1 - a^2 - \frac{1}{a^2})) \]

### Step 7: Find the Minimum Value
To minimize \( \Delta \), we need to analyze the expression:
\[ 1 + a^2 + \frac{1}{a^2} \geq 3 \quad \text{(by AM-GM inequality)} \]
Thus, the minimum occurs when \( a = 1 \) (and hence \( b = 1 \)):
\[ \Delta_{\text{min}} = 3^3 = 27 \]

### Final Answer
The minimum value of \( \Delta \) is:
\[ \boxed{27} \]