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 solve the problem, we need to calculate the determinant \( \Delta \) given by:

\[ \Delta = \begin{vmatrix} 1 + a^2 - b^2 & 2ab & -2b \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix} \]

where \( ab = 1 \).

### Step 1: Substitute \( b \) in terms of \( a \)

Since \( ab = 1 \), we can express \( b \) as \( b = \frac{1}{a} \).

### Step 2: Substitute \( b \) in the determinant

Now, we substitute \( b = \frac{1}{a} \) into the determinant:

\[ \Delta = \begin{vmatrix} 1 + a^2 - \left(\frac{1}{a}\right)^2 & 2a\left(\frac{1}{a}\right) & -2\left(\frac{1}{a}\right) \\ 2a\left(\frac{1}{a}\right) & 1 - a^2 + \left(\frac{1}{a}\right)^2 & 2a \\ 2\left(\frac{1}{a}\right) & -2a & 1 - a^2 - \left(\frac{1}{a}\right)^2 \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} 1 + a^2 - \frac{1}{a^2} & 2 & -\frac{2}{a} \\ 2 & 1 - a^2 + \frac{1}{a^2} & 2a \\ \frac{2}{a} & -2a & 1 - a^2 - \frac{1}{a^2} \end{vmatrix} \]

### Step 3: Simplify the determinant

Now we perform column operations to simplify the determinant.

1. **Column Operations**:
- \( C_1 \leftarrow C_1 - \frac{2}{a} C_3 \)
- \( C_2 \leftarrow C_2 + 2 C_3 \)

After performing these operations, we get:

\[ \Delta = \begin{vmatrix} 1 + a^2 - \frac{1}{a^2} + 2 & 0 & -\frac{2}{a} \\ 2 & 1 - a^2 + \frac{1}{a^2} + 2a & 2a \\ \frac{2}{a} & -2a & 1 - a^2 - \frac{1}{a^2} \end{vmatrix} \]

### Step 4: Calculate the determinant

Now we can expand the determinant. The determinant can be calculated using the rule of Sarrus or cofactor expansion. After simplification, we find that:

\[ \Delta = (1 + a^2 + \frac{1}{a^2})^3 \]

### Step 5: Find the minimum value of \( \Delta \)

To find the minimum value of \( \Delta \), we need to minimize \( 1 + a^2 + \frac{1}{a^2} \).

Using the AM-GM inequality:

\[ 1 + a^2 + \frac{1}{a^2} \geq 3 \]

The equality holds when \( a^2 = 1 \) (i.e., \( a = 1 \) or \( a = -1 \)). Thus, the minimum value of \( \Delta \) is:

\[ \Delta_{\text{min}} = 3^3 = 27 \]

### Final Answer

The minimum value of \( \Delta \) is \( \boxed{27} \).

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