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

Options

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

Explanation

To find the maximum value of \(\frac{\text{det}(A)}{b}\) where \(A = \begin{pmatrix} 2 & b & 1 \\ b & b^2 + 1 & b \\ 1 & b & 2 \end{pmatrix}\) and \(b > 0\), we will follow these steps:

### Step 1: Calculate the determinant of matrix \(A\)

The determinant of a \(3 \times 3\) matrix can be calculated using the formula:

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

For our matrix \(A\):

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

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

Now we can compute the determinant:

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

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. \(1(b^2 - (b^2 + 1)) = b^2 - b^2 - 1 = -1\)

Now, combine these results:

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

### Step 2: Compute \(\frac{\text{det}(A)}{b}\)

Now we need to find:

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

### Step 3: Find the maximum value of \(b + \frac{3}{b}\)

To find the maximum value of \(b + \frac{3}{b}\), we can apply the AM-GM inequality. According to AM-GM:

\[ \frac{b + \frac{3}{b}}{2} \geq \sqrt{b \cdot \frac{3}{b}} = \sqrt{3} \]

Thus,

\[ b + \frac{3}{b} \geq 2\sqrt{3} \]

The equality holds when \(b = \frac{3}{b}\), which gives \(b^2 = 3\) or \(b = \sqrt{3}\).

### Step 4: Conclusion

The maximum value of \(\frac{\text{det}(A)}{b}\) is:

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