Key Concepts Utilized

This problem requires us to determine the range of the determinant of a given $3 \times 3$ matrix, where the matrix elements depend on a variable $\theta$ within a specified interval. The core concepts involved are:

1. Determinant of a $3 \times 3$ Matrix: For a matrix $M = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ its determinant, $\det(M)$ or $|M|$, can be calculated using cofactor expansion along the first row as: $\det(M) = a(ei - fh) - b(di - fg) + c(dh - eg)$ Alternatively, row/column operations can simplify the matrix before expansion.
2. Range of Trigonometric Functions: Understanding how $\sin\theta$ behaves in different quadrants and how its square, $\sin^2\theta$, affects its range.
3. Interval Notation: Correctly interpreting open and closed intervals for both $\theta$ and the resulting determinant.

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Step 1: Calculating the Determinant of Matrix A

First, let's find the determinant of the given matrix $A$: $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ { - \sin \theta } & 1 & {\sin \theta } \\ { - 1} & { - \sin \theta } & 1 \end{bmatrix}$

We can calculate the determinant using cofactor expansion along the first row: $\det(A) = 1 \cdot \left| {\begin{matrix} 1 & {\sin \theta } \\ { - \sin \theta } & 1 \end{matrix}} \right| - \sin \theta \cdot \left| {\begin{matrix} { - \sin \theta } & {\sin \theta } \\ { - 1} & 1 \end{matrix}} \right| + 1 \cdot \left| {\begin{matrix} { - \sin \theta } & 1 \\ { - 1} & { - \sin \theta } \end{matrix}} \right|$ Expanding the $2 \times 2$ determinants: $\det(A) = 1 \cdot ((1)(1) - (\sin \theta)(-\sin \theta)) - \sin \theta \cdot ((-\sin \theta)(1) - (\sin \theta)(-1)) + 1 \cdot ((-\sin \theta)(-\sin \theta) - (1)(-1))$ $\det(A) = 1 \cdot (1 + \sin^2 \theta) - \sin \theta \cdot (-\sin \theta + \sin \theta) + 1 \cdot (\sin^2 \theta + 1)$ $\det(A) = (1 + \sin^2 \theta) - \sin \theta \cdot (0) + (1 + \sin^2 \theta)$ $\det(A) = 1 + \sin^2 \theta + 1 + \sin^2 \theta$ $\det(A) = 2 + 2\sin^2 \theta$ This simplified expression for $\det(A)$ makes it easier to find its range.

Alternatively, using Row Operations: We can simplify the determinant calculation by applying row operations. Let $R_1, R_2, R_3$ denote the first, second, and third rows, respectively. Perform $R_1 \to R_1 + R_3$: $\det(A) = \left| {\begin{matrix} {1+(-1)} & {\sin \theta + (-\sin \theta)} & {1+1} \\ { - \sin \theta } & 1 & {\sin \theta } \\ { - 1} & { - \sin \theta } & 1 \end{matrix}} \right|$ $\det(A) = \left| {\begin{matrix} 0 & 0 & 2 \\ { - \sin \theta } & 1 & {\sin \theta } \\ { - 1} & { - \sin \theta } & 1 \end{matrix}} \right|$ Now, expand along the first row (which has two zeros): $\det(A) = 0 \cdot (\dots) - 0 \cdot (\dots) + 2 \cdot \left| {\begin{matrix} { - \sin \theta } & 1 \\ { - 1} & { - \sin \theta } \end{matrix}} \right|$ $\det(A) = 2 \cdot ((-\sin \theta)(-\sin \theta) - (1)(-1))$ $\det(A) = 2 \cdot (\sin^2 \theta + 1)$ $\det(A) = 2 + 2\sin^2 \theta$ Both methods yield the same result, confirming our determinant calculation.

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Step 2: Analyzing the Range of $\sin\theta$ for the Given Interval

The problem specifies that $\theta \in \left( \frac{3\pi}{4}, \frac{5\pi}{4} \right)$. Let's analyze the behavior of $\sin\theta$ in this interval:

• At $\theta = \frac{3\pi}{4}$ (which is $135^\circ$), $\sin\theta = \frac{1}{\sqrt{2}}$.
• As $\theta$ increases from $\frac{3\pi}{4}$ to $\pi$ ($180^\circ$), $\sin\theta$ decreases from $\frac{1}{\sqrt{2}}$ to $0$.
• As $\theta$ increases from $\pi$ to $\frac{5\pi}{4}$ ($225^\circ$), $\sin\theta$ decreases from $0$ to $-\frac{1}{\sqrt{2}}$.

Since the interval for $\theta$ is open, $\theta$ does not actually attain the values $\frac{3\pi}{4}$ or $\frac{5\pi}{4}$. Therefore, $\sin\theta$ does not attain the values $\frac{1}{\sqrt{2}}$ or $-\frac{1}{\sqrt{2}}$. However, it does pass through $0$ at $\theta=\pi$, which is within the given interval. So, the range of $\sin\theta$ for $\theta \in \left( \frac{3\pi}{4}, \frac{5\pi}{4} \right)$ is $\left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$.

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Step 3: Determining the Range of $\sin^2\theta$

Now we need to find the range of $\sin^2\theta$ based on the range of $\sin\theta$. We have $\sin\theta \in \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$. When squaring a quantity that ranges symmetrically around zero (like this), the minimum value of its square will be $0$, and the maximum value will be the square of the largest absolute value.

• The minimum value of $\sin^2\theta$ is $0$, which occurs when $\sin\theta = 0$ (at $\theta = \pi$). Since $\theta = \pi$ is in the interval $\left( \frac{3\pi}{4}, \frac{5\pi}{4} \right)$, this minimum value is included.
• The maximum value of $\sin^2\theta$ approaches $\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$. This value is approached as $\sin\theta$ approaches $\frac{1}{\sqrt{2}}$ or $-\frac{1}{\sqrt{2}}$, but it is never actually reached because the interval for $\theta$ is open. Therefore, the range of $\sin^2\theta$ is $\left[ 0, \frac{1}{2} \right)$.

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Step 4: Finding the Range of det(A)

We found $\det(A) = 2 + 2\sin^2 \theta$. Using the range of $\sin^2\theta \in \left[ 0, \frac{1}{2} \right)$:

• The minimum value of $\det(A)$ occurs when $\sin^2\theta = 0$: $\det(A)_{\min} = 2 + 2(0) = 2$ This minimum value is included in the range.
• The maximum value of $\det(A)$ occurs as $\sin^2\theta$ approaches $\frac{1}{2}$: $\det(A)_{\max} = 2 + 2\left(\frac{1}{2}\right) = 2 + 1 = 3$ This maximum value is not included in the range, as $\sin^2\theta$ never actually reaches $\frac{1}{2}$. So, the determinant $\det(A)$ lies in the interval $[2, 3)$.

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Step 5: Comparing with Options and Final Answer

Our derived interval for $\det(A)$ is $[2, 3)$. Now let's compare this with the given options: (A) $\left( {{3 \over 2},3} \right]$ which is $(1.5, 3]$ (B) $\left( {0,{3 \over 2}} \right]$ which is $(0, 1.5]$ (C) $\left[ {{5 \over 2},4} \right)$ which is $[2.5, 4)$ (D) $\left( {1,{5 \over 2}} \right]$ which is $(1, 2.5]$

We need to find the interval in which $\det(A)$ "lies". This means we are looking for an option that contains our calculated interval $[2, 3)$. Let's check each option:

• (A) $(1.5, 3]$: This interval includes all values from $1.5$ (exclusive) up to $3$ (inclusive). Our interval $[2, 3)$ is entirely contained within $(1.5, 3]$. Specifically, $1.5 < 2 \le \det(A) < 3 \le 3$. This option is a superset of our result.
• (B) $(0, 1.5]$: This interval does not contain $2$ or any values greater than $1.5$.
• (C) $[2.5, 4)$: This interval does not contain $2$ or any values between $2$ and $2.5$.
• (D) $(1, 2.5]$: This interval does not contain values approaching $3$.

Since our precise interval for $\det(A)$ is $[2, 3)$, and this interval is fully contained within option (A) $\left( {{3 \over 2},3} \right]$, option (A) is the correct choice.

The final answer is $\boxed{\text{A}}$.

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Tips for Success / Common Mistakes to Avoid

1. Determinant Calculation Accuracy: Be extremely careful with signs and calculations when expanding determinants. A common mistake is misplacing a negative sign. Using row/column operations can often simplify the process and reduce error.
2. Trigonometric Range: Precisely determine the range of $\sin\theta$ (or $\cos\theta$) for the given interval. Pay close attention to whether the interval is open or closed, as this affects whether endpoints are included.
3. Squaring a Range: When squaring a range that includes zero (e.g., $(-a, a)$), the squared range will be $[0, a^2)$. If the range does not include zero (e.g., $(a, b)$ where $a, b > 0$), then squaring simply gives $(a^2, b^2)$.
4. "Lies In" Interpretation: In multiple-choice questions, if your exact derived interval is not an option, look for the smallest option that contains your derived interval. This is a common test-taking strategy when options are not perfectly tight.

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Summary / Key Takeaway

This problem combined matrix determinant calculation with trigonometric range analysis. The key steps were:

1. Simplify the determinant expression to $2 + 2\sin^2\theta$.
2. Carefully determine the range of $\sin\theta$ within the given interval $\left( \frac{3\pi}{4}, \frac{5\pi}{4} \right)$, which was $\left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$.
3. From this, deduce the range of $\sin^2\theta$ as $\left[ 0, \frac{1}{2} \right)$.
4. Substitute this range into the determinant expression to find $\det(A) \in [2, 3)$.
5. Finally, identify the option that correctly encompasses this range, which was $\left( \frac{3}{2}, 3 \right]$.