Key Concepts and Formulas:

This problem involves two main mathematical concepts:

1. Determinant of a $3 \times 3$ Matrix: For a matrix $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$, its determinant, denoted as $\det(A)$ or $|A|$, can be calculated using cofactor expansion. Expanding along the first row, for instance, we have: $\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$ This can be remembered by the sign pattern for cofactors: $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$.
2. Arithmetic Mean - Geometric Mean (AM-GM) Inequality: For any two non-negative real numbers $x$ and $y$, their arithmetic mean is greater than or equal to their geometric mean: $\frac{x+y}{2} \ge \sqrt{xy}$ Equality holds if and only if $x=y$. This inequality is particularly useful for finding the minimum value of expressions involving a term and its reciprocal (or a multiple of its reciprocal).

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Step-by-Step Solution:

Step 1: Understand the Given Matrix and the Objective

We are given the matrix $A$: $A = \begin{bmatrix} 2 & b & 1 \\ b & {{b^2} + 1} & b \\ 1 & b & 2 \end{bmatrix}$ We are also given that $b > 0$. Our goal is to find the minimum value of the expression $\frac{\det(A)}{b}$. This means we first need to calculate $\det(A)$, then divide it by $b$, and finally find the minimum value of the resulting expression.

Step 2: Calculate the Determinant of Matrix A

We will calculate $\det(A)$ by expanding along the first row ($R_1$) because it's a common and straightforward method. $\det(A) = 2 \cdot \det \begin{pmatrix} {{b^2} + 1} & b \\ b & 2 \end{pmatrix} - b \cdot \det \begin{pmatrix} b & b \\ 1 & 2 \end{pmatrix} + 1 \cdot \det \begin{pmatrix} b & {{b^2} + 1} \\ 1 & b \end{pmatrix}$

Let's calculate each $2 \times 2$ determinant:

• For the first term, associated with the element '2': $2 \cdot \left( ({{b^2} + 1})(2) - (b)(b) \right) = 2 \cdot \left( 2b^2 + 2 - b^2 \right) = 2 \cdot \left( b^2 + 2 \right)$ Explanation: We multiply the element '2' by the determinant of its minor matrix. The minor is obtained by removing the row and column containing '2'.

• For the second term, associated with the element 'b' (in the middle of $R_1$): $-b \cdot \left( (b)(2) - (b)(1) \right) = -b \cdot \left( 2b - b \right) = -b \cdot (b) = -b^2$ Explanation: We multiply the element 'b' by the determinant of its minor, but remember to apply the negative sign as per the cofactor expansion sign pattern for the second element in the first row.

• For the third term, associated with the element '1': $+1 \cdot \left( (b)(b) - ({{b^2} + 1})(1) \right) = 1 \cdot \left( b^2 - b^2 - 1 \right) = 1 \cdot \left( -1 \right) = -1$ Explanation: We multiply the element '1' by the determinant of its minor.

Now, we sum these three terms to get $\det(A)$: $\det(A) = 2(b^2 + 2) - b^2 - 1$ $\det(A) = 2b^2 + 4 - b^2 - 1$ $\det(A) = b^2 + 3$

Step 3: Form the Expression for Minimization

The problem asks for the minimum value of $\frac{\det(A)}{b}$. We have found $\det(A) = b^2 + 3$. So, the expression becomes: $\frac{\det(A)}{b} = \frac{b^2 + 3}{b}$ Now, we can simplify this algebraic expression: $\frac{b^2 + 3}{b} = \frac{b^2}{b} + \frac{3}{b} = b + \frac{3}{b}$ Let $f(b) = b + \frac{3}{b}$. We need to find the minimum value of $f(b)$ given that $b > 0$.

Step 4: Apply the AM-GM Inequality

We have the expression $f(b) = b + \frac{3}{b}$. Since $b > 0$, both terms $b$ and $\frac{3}{b}$ are positive. This is a perfect scenario to apply the AM-GM inequality.

Let $x = b$ and $y = \frac{3}{b}$. According to the AM-GM inequality: $\frac{x+y}{2} \ge \sqrt{xy}$ Substituting $x$ and $y$: $\frac{b + \frac{3}{b}}{2} \ge \sqrt{b \cdot \frac{3}{b}}$ $\frac{b + \frac{3}{b}}{2} \ge \sqrt{3}$

Now, multiply both sides by 2 to isolate the expression $b + \frac{3}{b}$: $b + \frac{3}{b} \ge 2\sqrt{3}$

This inequality tells us that the value of $b + \frac{3}{b}$ is always greater than or equal to $2\sqrt{3}$. Therefore, the minimum value of the expression is $2\sqrt{3}$.

Tip: The equality in AM-GM holds when $x=y$. In this case, $b = \frac{3}{b}$. This implies $b^2 = 3$, so $b = \sqrt{3}$ (since $b>0$). Since there exists a positive value of $b$ for which the equality holds, the minimum value is indeed achievable.

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Conclusion and Key Takeaway:

The minimum value of $\frac{\det(A)}{b}$ is $2\sqrt{3}$.

This problem beautifully illustrates how two distinct mathematical concepts – determinant calculation and the AM-GM inequality – can be combined to solve an optimization problem. The key steps were:

1. Accurate calculation of the determinant.
2. Algebraic simplification of the resulting expression.
3. Recognizing the applicability of the AM-GM inequality due to the structure of the simplified expression and the condition $b > 0$.

Common Mistakes to Avoid:

• Errors in determinant calculation, especially with signs during cofactor expansion.
• Forgetting the condition for AM-GM (non-negative terms). If $b$ could be negative, AM-GM would not be directly applicable in this way.
• Incorrectly simplifying algebraic expressions after determinant calculation.

The final answer is \boxed{\text{2\sqrt 3}}.