This solution demonstrates how to effectively simplify and evaluate a determinant involving trigonometric functions and then find a parameter's value based on the determinant's minimum value.

Key Concepts Used:

1. Properties of Determinants:
   • The value of a determinant remains unchanged if we apply elementary row operations of the type $R_i \to R_i + k R_j$ (or column operations $C_i \to C_i + k C_j$). This property is crucial for simplifying complex determinants by creating zeros.
2. Expansion of a Determinant:
   • A $3 \times 3$ determinant can be expanded along any row or column using cofactor expansion. If a row or column contains multiple zeros, expanding along that row/column significantly reduces computation.
3. Range of Trigonometric Functions:
   • For $\theta \in [0, 2\pi]$, the range of $\sin\theta$ is $[-1, 1]$.
   • Consequently, the range of $\sin^2\theta$ is $[0, 1]$, as squaring a number between $-1$ and $1$ results in a number between $0$ and $1$.
4. Minimizing an Expression:
   • To find the minimum value of an expression of the form $K - f(x)$ (where $K$ is a constant), we need to maximize the function $f(x)$ that is being subtracted.

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Step 1: Express the Given Determinant

We are given the matrix $A$ as: A = \left[ {\matrix{ { - 2} & {4 + d} & {\left( {\sin \theta } \right) - 2} \cr 1 & {\left( {\sin \theta } \right) + 2} & d \cr 5 & {\left( {2\sin \theta } \right) - d} & {\left( { - \sin \theta } \right) + 2 + 2d} \cr } } \right] Our goal is to find $\det(A)$. Directly expanding this $3 \times 3$ determinant would be cumbersome due to the presence of both $d$ and $\sin\theta$ terms.

Step 2: Strategic Row Operation for Simplification

• Why this step? To simplify the calculation of the determinant, we aim to create as many zeros as possible in a single row or column. This allows us to expand the determinant along that row/column, reducing a $3 \times 3$ determinant to a simpler $2 \times 2$ determinant.

• How to choose the operation? We observe the terms in the matrix, especially those involving $\sin\theta$ and $d$. Notice how $d$ appears with coefficients $1, -1, 2$ across rows, and $\sin\theta$ appears with coefficients $1, 1, 2$. We look for a combination that can cancel these variable terms. Consider the row operation $R_1 \to R_1 + R_3 - 2R_2$. Let's examine how each element of the first row transforms:

  • New $R_{11}$ (first element of the first row): $(-2) + (5) - 2(1) = -2 + 5 - 2 = 1$
  • New $R_{12}$ (second element of the first row): $(4+d) + (2\sin\theta - d) - 2(\sin\theta + 2)$ $= 4+d+2\sin\theta-d-2\sin\theta-4 = 0$
  • New $R_{13}$ (third element of the first row): $(\sin\theta - 2) + (-\sin\theta + 2 + 2d) - 2(d)$ $= \sin\theta - 2 - \sin\theta + 2 + 2d - 2d = 0$

• Applying the operation: This specific operation is highly effective as it creates two zeros in the first row. The value of the determinant remains unchanged. \det(A) = \left| {\matrix{ { - 2 + 5 - 2(1)} & {(4 + d) + (2\sin\theta - d) - 2(\sin\theta + 2)} & {(\sin\theta - 2) + (-\sin\theta + 2 + 2d) - 2d} \cr 1 & {\sin\theta + 2} & d \cr 5 & {2\sin\theta - d} & {-\sin\theta + 2 + 2d} \cr } } \right| \det(A) = \left| {\matrix{ 1 & 0 & 0 \cr 1 & {\sin\theta + 2} & d \cr 5 & {2\sin\theta - d} & {-\sin\theta + 2 + 2d} \cr } } \right| Tip: When simplifying determinants, always look for operations that can make elements zero. Sometimes, trial and error or observing patterns (like terms that can cancel out) can help in finding the right operation.

Step 3: Expanding the Simplified Determinant

• Why this step? With two zeros in the first row, expanding the determinant along this row becomes very straightforward, as only one term will contribute to the sum. \det(A) = 1 \cdot \left| {\matrix{ {\sin\theta + 2} & d \cr {2\sin\theta - d} & {-\sin\theta + 2 + 2d} \cr } } \right| - 0 \cdot (\text{cofactor}) + 0 \cdot (\text{cofactor}) $\det(A) = (\sin\theta + 2)(-\sin\theta + 2 + 2d) - d(2\sin\theta - d)$

Step 4: Algebraic Simplification of det(A)

• Why this step? To obtain a clear expression for $\det(A)$ in terms of $d$ and $\sin\theta$, which will be necessary for finding its minimum value.
• Let's expand and simplify the terms:
  • First term: $(\sin\theta + 2)(-\sin\theta + 2 + 2d)$ $= -\sin^2\theta + 2\sin\theta + 2d\sin\theta - 2\sin\theta + 4 + 4d$ $= -\sin^2\theta + 2d\sin\theta + 4 + 4d$
  • Second term: $-d(2\sin\theta - d) = -2d\sin\theta + d^2$
• Now, combine both terms to get the full expression for $\det(A)$: $\det(A) = (-\sin^2\theta + 2d\sin\theta + 4 + 4d) + (-2d\sin\theta + d^2)$ $\det(A) = -\sin^2\theta + 2d\sin\theta + 4 + 4d - 2d\sin\theta + d^2$ $\det(A) = d^2 + 4d + 4 - \sin^2\theta$ We can recognize that $d^2 + 4d + 4$ is a perfect square, $(d+2)^2$. So, the simplified determinant is: $\det(A) = (d+2)^2 - \sin^2\theta$

Step 5: Determining the Minimum Value of det(A)

• Why this step? The problem asks for a value of $d$ given the minimum value of $\det(A)$. We need to express this minimum value in terms of $d$.
• We have $\det(A) = (d+2)^2 - \sin^2\theta$.
• The term $(d+2)^2$ is a constant with respect to $\theta$.
• To minimize the entire expression, we need to subtract the largest possible value from $(d+2)^2$. This means we need to maximize $\sin^2\theta$.
• We are given $\theta \in [0, 2\pi]$.
  • The range of $\sin\theta$ is $[-1, 1]$.
  • Therefore, the range of $\sin^2\theta$ is $[0, 1]$ (since squaring makes negative values positive, and the maximum value of $\sin\theta$ is $1$).
• The maximum value of $\sin^2\theta$ is $1$.
• Substituting this maximum value into the determinant expression yields the minimum value of $\det(A)$: $\min(\det(A)) = (d+2)^2 - 1$ Common Mistake: A frequent error is to incorrectly identify the range or maximum value of $\sin^2\theta$. Remember that $\sin^2\theta$ is always non-negative.

Step 6: Solving for 'd' using the Given Minimum Value

• Why this step? The problem states that the minimum value of $\det(A)$ is $8$. We use this information to form an equation and solve for $d$.
• We equate our derived minimum value to $8$: $(d+2)^2 - 1 = 8$
• Solve for $d$: $(d+2)^2 = 8 + 1$ $(d+2)^2 = 9$ Taking the square root of both sides gives two possibilities: $d+2 = \pm\sqrt{9}$ $d+2 = \pm 3$
• This yields two possible values for $d$:
  1. $d+2 = 3 \implies d = 3-2 \implies d = 1$
  2. $d+2 = -3 \implies d = -3-2 \implies d = -5$

Step 7: Final Conclusion

The possible values for $d$ are $1$ and $-5$. We check these against the given options: (A) $-5$ (B) $2(\sqrt{2} + 2)$ (C) $2(\sqrt{2} + 1)$ (D) $1$

Option (A) matches one of our calculated values for $d$.

The final answer is $\boxed{\text{-5}}$.

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Summary and Key Takeaway:

This problem effectively tests your understanding of determinant properties, algebraic manipulation, and the range of trigonometric functions. The key takeaway is that strategic simplification using row/column operations can drastically reduce the complexity of determinant calculations. Furthermore, careful analysis of the range of trigonometric expressions is essential for correctly determining minimum or maximum values. Recognizing perfect squares in algebraic expressions also aids in simplification.