Let `A=[(2,b,1),(b,b^(2)+1,b),(1,b,2)]` where `b gt 0`. Then the minimum value of `("det.(A)")/(b)` is

Options

A. `sqrt(3)`
B. `-sqrt(3)`
C. `-2sqrt(3)`
D. `2sqrt(3)`

Answer

To solve the problem, we need to find the minimum value of \(\frac{\text{det}(A)}{b}\) for the matrix

\[ A = \begin{pmatrix} 2 & b & 1 \\ b & b^2 + 1 & b \\ 1 & b & 2 \end{pmatrix} \]

where \(b > 0\).

### Step 1: Calculate the Determinant of Matrix A

To find the determinant of matrix \(A\), we can use the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

\[ A = \begin{pmatrix} 2 & b & 1 \\ b & b^2 + 1 & b \\ 1 & b & 2 \end{pmatrix} \]

We can assign:

- \(a = 2\), \(b = b\), \(c = 1\)
- \(d = b\), \(e = b^2 + 1\), \(f = b\)
- \(g = 1\), \(h = b\), \(i = 2\)

Thus, the determinant can be calculated as follows:

\[ \text{det}(A) = 2((b^2 + 1) \cdot 2 - b \cdot b) - b(b \cdot 2 - b \cdot 1) + 1(b \cdot b - (b^2 + 1) \cdot 1) \]

Calculating each term:

1. \(2((b^2 + 1) \cdot 2 - b^2) = 2(2b^2 + 2 - b^2) = 2(b^2 + 2) = 2b^2 + 4\)
2. \(-b(2b - b) = -b(b) = -b^2\)
3. \(1(b^2 - (b^2 + 1)) = 1(b^2 - b^2 - 1) = -1\)

Combining these, we get:

\[ \text{det}(A) = (2b^2 + 4 - b^2 - 1) = (b^2 + 3) \]

### Step 2: Express the Function to Minimize

Now we need to find the minimum value of:

\[ \frac{\text{det}(A)}{b} = \frac{b^2 + 3}{b} = b + \frac{3}{b} \]

### Step 3: Find the Critical Points

To find the minimum value of \(f(b) = b + \frac{3}{b}\), we take the derivative and set it to zero:

\[ f'(b) = 1 - \frac{3}{b^2} \]

Setting \(f'(b) = 0\):

\[ 1 - \frac{3}{b^2} = 0 \implies \frac{3}{b^2} = 1 \implies b^2 = 3 \implies b = \sqrt{3} \]

### Step 4: Verify Minimum Value

To confirm that this critical point is a minimum, we can check the second derivative:

\[ f''(b) = \frac{6}{b^3} \]

Since \(b > 0\), \(f''(b) > 0\), indicating that \(f(b)\) is concave up at \(b = \sqrt{3}\).

### Step 5: Calculate the Minimum Value

Now we substitute \(b = \sqrt{3}\) back into \(f(b)\):

\[ f(\sqrt{3}) = \sqrt{3} + \frac{3}{\sqrt{3}} = \sqrt{3} + \sqrt{3} = 2\sqrt{3} \]

### Final Answer

Thus, the minimum value of \(\frac{\text{det}(A)}{b}\) is:

\[ \boxed{2\sqrt{3}} \]