Solution: Finding the Minimum and Maximum Values of a Determinant

This problem requires us to evaluate a given $3 \times 3$ determinant, simplify the resulting trigonometric expression, and then determine its minimum and maximum values using the properties of trigonometric functions.

Key Concepts Used:

1. Properties of Determinants:
   • The value of a determinant remains unchanged if we apply row operations of the type $R_i \to R_i + k R_j$ (or column operations $C_i \to C_i + k C_j$). In this case, we'll use $R_i \to R_i - R_j$.
   • Expansion of a $3 \times 3$ determinant: For a matrix $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$, its determinant can be expanded as $a(ei - fh) - b(di - fg) + c(dh - eg)$. Choosing a row or column with zeros simplifies this expansion.
2. Trigonometric Identities:
   • $\sin^2 x + \cos^2 x = 1$
3. Range of Trigonometric Functions:
   • For any real number $\theta$, the sine function has a range of $[-1, 1]$, i.e., $-1 \le \sin \theta \le 1$.

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Step 1: Simplify the Determinant using Row Operations

We are given the determinant: \Delta = \left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right| Our goal is to simplify this determinant by introducing zeros or simpler terms. Row operations are an excellent tool for this, as they do not change the value of the determinant.

Let's apply the operations $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$.

Applying $R_1 \to R_1 - R_2$:

• First element: $\cos^2 x - (1 + \cos^2 x) = \cos^2 x - 1 - \cos^2 x = -1$
• Second element: $(1 + \sin^2 x) - \sin^2 x = 1$
• Third element: $\sin 2x - \sin 2x = 0$ The new Row 1 becomes $\begin{pmatrix} -1 & 1 & 0 \end{pmatrix}$.

Applying $R_2 \to R_2 - R_3$:

• First element: $(1 + \cos^2 x) - \cos^2 x = 1$
• Second element: $\sin^2 x - \sin^2 x = 0$
• Third element: $\sin 2x - (1 + \sin 2x) = \sin 2x - 1 - \sin 2x = -1$ The new Row 2 becomes $\begin{pmatrix} 1 & 0 & -1 \end{pmatrix}$.

The determinant now transforms to: \Delta = \left| {\matrix{ { - 1} & 1 & 0 \cr 1 & 0 & { - 1} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right| Tip: Performing row/column operations is often the most efficient way to evaluate determinants, especially when elements are complex expressions. Look for opportunities to create zeros or common factors.

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Step 2: Expand the Simplified Determinant

Now we expand the determinant along the first row ($R_1$) because it contains a zero, which simplifies the calculation.

\Delta = (-1) \cdot \left| {\matrix{ 0 & { - 1} \cr {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right| - (1) \cdot \left| {\matrix{ 1 & { - 1} \cr {{{\cos }^2}x} & {1 + \sin 2x} \cr } } \right| + (0) \cdot \left| {\matrix{ 1 & 0 \cr {{{\cos }^2}x} & {{{\sin }^2}x} \cr } } \right|

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

• For the first term: $0 \cdot (1 + \sin 2x) - (-1) \cdot \sin^2 x = 0 - (-\sin^2 x) = \sin^2 x$
• For the second term: $1 \cdot (1 + \sin 2x) - (-1) \cdot \cos^2 x = 1 + \sin 2x + \cos^2 x$
• The third term is $0 \cdot (\dots)$, so it's $0$.

Substitute these back into the expansion: $\Delta = (-1) \cdot (\sin^2 x) - (1) \cdot (1 + \sin 2x + \cos^2 x) + 0$ $\Delta = -\sin^2 x - (1 + \sin 2x + \cos^2 x)$ Now, we use the trigonometric identity $\sin^2 x + \cos^2 x = 1$: $\Delta = -\sin^2 x - (1 + \sin 2x + \cos^2 x)$ $\Delta = -\sin^2 x - 1 - \sin 2x - \cos^2 x$ Rearrange the terms: $\Delta = -(\sin^2 x + \cos^2 x) - 1 - \sin 2x$ Substitute $\sin^2 x + \cos^2 x = 1$: $\Delta = -(1) - 1 - \sin 2x$ $\Delta = -2 - \sin 2x$ So, the determinant simplifies to the expression $-2 - \sin 2x$.

Common Mistake: Be very careful with the signs during determinant expansion. The alternating signs for cofactors ($+,-,+$ for $R_1$ expansion) are crucial.

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Step 3: Determine the Minimum and Maximum Values

We need to find the minimum ($m$) and maximum ($M$) values of the expression $\Delta = -2 - \sin 2x$.

We know that for any real angle $\theta$, the value of $\sin \theta$ lies in the interval $[-1, 1]$. Therefore, for $2x$: $-1 \le \sin 2x \le 1$

To find the minimum value ($m$) of $\Delta = -2 - \sin 2x$: We need to make the term $-\sin 2x$ as small as possible. This happens when $\sin 2x$ is at its maximum value, which is $1$. So, when $\sin 2x = 1$: $m = -2 - (1)$ $m = -3$

To find the maximum value ($M$) of $\Delta = -2 - \sin 2x$: We need to make the term $-\sin 2x$ as large as possible. This happens when $\sin 2x$ is at its minimum value, which is $-1$. So, when $\sin 2x = -1$: $M = -2 - (-1)$ $M = -2 + 1$ $M = -1$

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Step 4: Form the Ordered Pair

The minimum value is $m = -3$ and the maximum value is $M = -1$. Therefore, the ordered pair $(m, M)$ is $(-3, -1)$.

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Summary and Key Takeaway:

This problem demonstrates a common strategy for solving determinant problems involving trigonometric functions:

1. Simplify the determinant first using row/column operations to reduce complex expressions and introduce zeros. This makes the expansion much easier and less prone to errors.
2. Expand the simplified determinant carefully, paying attention to signs.
3. Utilize trigonometric identities to further simplify the resulting expression.
4. Determine the range of the final trigonometric expression to find its minimum and maximum values. Remember the fundamental range of $\sin \theta$ and $\cos \theta$ is $[-1, 1]$.

The final answer is $\boxed{\text{(–3, –1)}}$.