Let A = $\left[\begin{array}{ccc}1 & \sin θ & 1 \−\sin θ & 1 & \sin θ \−1 &−\sin θ & 1\end{array}\right]$, where 0 ≤ θ ≤ 2π, then:

Question

Let A = $\left[\begin{array}{ccc}1 & \sin &theta; & 1 \&minus;\sin &theta; & 1 & \sin &theta; \&minus;1 &&minus;\sin &theta; & 1\end{array}\right]$, where 0 &le; &theta;  &le; 2&pi;, then:

Prepp Team · Accepted Answer

Finding the Determinant of Matrix A

The question asks us to find the determinant of the given 3x3 matrix A and determine its range for a specific interval of θ.

The given matrix is:

$A = \left[\begin{array}{ccc}1 & \sin \theta & 1 \\-\sin \theta & 1 & \sin \theta \\-1 & -\sin \theta & 1\end{array}\right]$

We need to calculate the determinant of this matrix, denoted as Det A or |A|.

For a 3x3 matrix $\left[\begin{array}{ccc}a & b & c \\d & e & f \\g & h & i\end{array}\right]$, the determinant is calculated as: $a(ei - fh) - b(di - fg) + c(dh - eg)$

Calculating Det A

Applying the determinant formula to matrix A:

$\text{Det A} = 1 \cdot ((1)(1) - (\sin \theta)(-\sin \theta)) - \sin \theta \cdot ((-\sin \theta)(1) - (\sin \theta)(-1)) + 1 \cdot ((-\sin \theta)(-\sin \theta) - (1)(-1))$

Let's simplify each term:

First term: $1 \cdot (1 + \sin^2 \theta) = 1 + \sin^2 \theta$
Second term: $-\sin \theta \cdot (-\sin \theta + \sin \theta) = -\sin \theta \cdot (0) = 0$
Third term: $1 \cdot (\sin^2 \theta + 1) = \sin^2 \theta + 1$

Adding the terms together:

$\text{Det A} = (1 + \sin^2 \theta) + 0 + (\sin^2 \theta + 1)$

$\text{Det A} = 1 + \sin^2 \theta + \sin^2 \theta + 1$

$\text{Det A} = 2 + 2\sin^2 \theta$

Determining the Range of Det A

We are given that $0 \le \theta \le 2\pi$. We need to find the range of $\text{Det A} = 2 + 2\sin^2 \theta$ over this interval for $\theta$.

First, consider the range of $\sin \theta$ for $0 \le \theta \le 2\pi$. The sine function oscillates between -1 and 1. So, the range of $\sin \theta$ is $[-1, 1]$.

Next, consider the range of $\sin^2 \theta$. If $\sin \theta$ is in $[-1, 1]$, then $\sin^2 \theta$ will be between $0^2=0$ and $(\pm 1)^2=1$. So, the range of $\sin^2 \theta$ is $[0, 1]$.

Now, let's find the range of $2 + 2\sin^2 \theta$ using the range of $\sin^2 \theta$:

Minimum value:

When $\sin^2 \theta = 0$, $\text{Det A} = 2 + 2(0) = 2$.

Maximum value:

When $\sin^2 \theta = 1$, $\text{Det A} = 2 + 2(1) = 2 + 2 = 4$.

Therefore, the range of Det A for $0 \le \theta \le 2\pi$ is $[2, 4]$.

Analyzing the Options

Let's compare our result with the given options:

Det A = 0: This is incorrect, as the minimum value is 2.
Det A $\in$ [2, 4]: This matches our calculated range.
Det A belongs to (1, 4): This is incorrect, as our range includes 2 and 4, and the interval is closed, not open.
None of these: This is incorrect, as option 2 is correct.

Based on the calculation, the determinant of matrix A belongs to the interval [2, 4].

Revision Table: Matrix Determinant and Trigonometry

Concept	Description	Relevance to Problem
Matrix Determinant	A scalar value computed from the elements of a square matrix.	We calculated the determinant of the given 3x3 matrix A.
3x3 Determinant Formula	$\|A\| = a(ei - fh) - b(di - fg) + c(dh - eg)$ for $A = \left[\begin{smallmatrix}a & b & c \\d & e & f \\g & h & i\end{smallmatrix}\right]$.	This formula was used directly to find Det A.
Sine Function Range	For real θ, the range of $\sin \theta$ is [-1, 1]. For $0 \le \theta \le 2\pi$, the range is also [-1, 1].	Used to find the range of $\sin^2 \theta$.
$\sin^2 \theta$ Range	For $\sin \theta \in [-1, 1]$, the range of $\sin^2 \theta$ is [0, 1].	Crucial step in determining the range of Det A.

Additional Information: Matrix Properties and Ranges

Understanding determinants is fundamental in linear algebra. Determinants have various properties, such as $\det(AB) = \det(A)\det(B)$ and $\det(A^T) = \det(A)$. A matrix is invertible if and only if its determinant is non-zero.

Trigonometric functions often appear in matrix entries, especially in transformations related to rotation or scaling. When dealing with trigonometric functions in matrix determinants, understanding their range over the specified interval of the angle θ is essential to determine the possible values of the determinant.

For periodic functions like sine and cosine, their values repeat over intervals. The interval $0 \le \theta \le 2\pi$ covers one full period, ensuring that we capture the complete range of $\sin \theta$ and subsequently $\sin^2 \theta$.

The expression for the determinant, $2 + 2\sin^2 \theta$, shows how the determinant's value varies directly with $\sin^2 \theta$. Since $\sin^2 \theta$ is always non-negative, the determinant will always be greater than or equal to 2.