If \({\rm{f}}\left( {\rm{x}} \right) = \left[ {\begin{array}{c} {\cos {\rm{x}}}&{ - \sin {\rm{x}}}&0\\ {\sin {\rm{x}}}&{\cos {\rm{x}}}&0\\ 0&0&1 \end{array}} \right]\) , then which of the following are correct? 1. f(θ) × f(ϕ) = f(θ + ϕ) 2. The value of the determinant of the matrix f(θ) × f(ϕ) is 1 3. The determinant of f(x) is an even function. Select the correct answer using the code given below:
1, 2 and 3
The question asks us to analyze the properties of a given matrix function \({\rm{f}}\left( {\rm{x}} \right) = \left[ {\begin{array}{c} {\cos {\rm{x}}}&{ - \sin {\rm{x}}}&0\\ {\sin {\rm{x}}}&{\cos {\rm{x}}}&0\\ 0&0&1 \end{array}} \right]\) and determine which of the provided statements are correct.
Analyzing Statement 1: Matrix Multiplication Property
Statement 1 says that \({\rm{f}}(\theta) \times {\rm{f}}(\phi) = {\rm{f}}(\theta + \phi)\).
Let's write down the matrices \({\rm{f}}(\theta)\) and \({\rm{f}}(\phi)\):
\({\rm{f}}(\theta) = \left[ {\begin{array}{c} {\cos \theta}&{ - \sin \theta}&0\\ {\sin \theta}&{\cos \theta}&0\\ 0&0&1 \end{array}} \right]\)
\({\rm{f}}(\phi) = \left[ {\begin{array}{c} {\cos \phi}&{ - \sin \phi}&0\\ {\sin \phi}&{\cos \phi}&0\\ 0&0&1 \end{array}} \right]\)
Now, let's perform the matrix multiplication \({\rm{f}}(\theta) \times {\rm{f}}(\phi)\):
| \({\cos \phi}\) | \(-\sin \phi\) | 0 | |
|---|---|---|---|
| \({\cos \theta}\) | \({\cos \theta \cos \phi - \sin \theta \sin \phi}\) | \({\cos \theta (-\sin \phi) + (-\sin \theta) \cos \phi}\) | \({\cos \theta \times 0 + (-\sin \theta) \times 0 + 0 \times 1}\) |
| \({\sin \theta}\) | \({\sin \theta \cos \phi + \cos \theta \sin \phi}\) | \({\sin \theta (-\sin \phi) + \cos \theta \cos \phi}\) | \({\sin \theta \times 0 + \cos \theta \times 0 + 0 \times 1}\) |
| 0 | \(0 \times \cos \phi + 0 \times \sin \phi + 1 \times 0\) | \(0 \times (-\sin \phi) + 0 \times \cos \phi + 1 \times 0\) | \(0 \times 0 + 0 \times 0 + 1 \times 1\) |
Simplifying the elements using trigonometric identities \(\cos(A+B) = \cos A \cos B - \sin A \sin B\) and \(\sin(A+B) = \sin A \cos B + \cos A \sin B\):
\({\rm{f}}(\theta) \times {\rm{f}}(\phi) = \left[ {\begin{array}{c} {\cos(\theta+\phi)}&{ -(\cos \theta \sin \phi + \sin \theta \cos \phi)}&0\\ {\sin(\theta+\phi)}&{\cos \theta \cos \phi - \sin \theta \sin \phi}&0\\ 0&0&1 \end{array}} \right]\)
\({\rm{f}}(\theta) \times {\rm{f}}(\phi) = \left[ {\begin{array}{c} {\cos(\theta+\phi)}&{ - \sin(\theta+\phi)}&0\\ {\sin(\theta+\phi)}&{\cos(\theta+\phi)}&0\\ 0&0&1 \end{array}} \right]\)
This result is exactly in the form of \({\rm{f}}(\theta + \phi)\), just with \(x\) replaced by \((\theta + \phi)\).
Thus, statement 1 is correct.
Analyzing Statement 2: Determinant Value
Statement 2 says that the value of the determinant of the matrix \({\rm{f}}(\theta) \times {\rm{f}}(\phi)\) is 1.
From statement 1, we found that \({\rm{f}}(\theta) \times {\rm{f}}(\phi) = {\rm{f}}(\theta + \phi)\).
So, we need to find the determinant of \({\rm{f}}(\theta + \phi)\). Let's calculate the determinant of the general matrix \({\rm{f}}(x)\).
\(\det({\rm{f}}(x)) = \det \left[ {\begin{array}{c} {\cos x}&{ - \sin x}&0\\ {\sin x}&{\cos x}&0\\ 0&0&1 \end{array}} \right]\)
Using the cofactor expansion along the third row (or third column, as they contain zeros):
\(\det({\rm{f}}(x)) = 0 \times (\text{cofactor of } 0) - 0 \times (\text{cofactor of } 0) + 1 \times \det \left[ {\begin{array}{c} {\cos x}&{ - \sin x}\\ {\sin x}&{\cos x} \end{array}} \right]\)
\(\det({\rm{f}}(x)) = 1 \times ((\cos x)(\cos x) - (-\sin x)(\sin x))\)
\(\det({\rm{f}}(x)) = \cos^2 x + \sin^2 x\)
Using the fundamental trigonometric identity \(\cos^2 x + \sin^2 x = 1\):
\(\det({\rm{f}}(x)) = 1\)
Since the determinant of \({\rm{f}}(x)\) is always 1, regardless of the value of \(x\), the determinant of \({\rm{f}}(\theta + \phi)\) is also 1.
Alternatively, using the property that for two matrices A and B, \(\det(AB) = \det(A)\det(B)\):
\(\det({\rm{f}}(\theta) \times {\rm{f}}(\phi)) = \det({\rm{f}}(\theta)) \times \det({\rm{f}}(\phi))\)
We know \(\det({\rm{f}}(\theta)) = 1\) and \(\det({\rm{f}}(\phi)) = 1\).
\(\det({\rm{f}}(\theta) \times {\rm{f}}(\phi)) = 1 \times 1 = 1\)
Thus, statement 2 is correct.
Analyzing Statement 3: Determinant as an Even Function
Statement 3 says that the determinant of \({\rm{f}}(x)\) is an even function.
We have already calculated the determinant of \({\rm{f}}(x)\) in the previous step:
\(\det({\rm{f}}(x)) = 1\)
Let \(g(x) = \det({\rm{f}}(x))\). So, \(g(x) = 1\).
A function \(g(x)\) is considered an even function if \(g(-x) = g(x)\) for all \(x\) in its domain.
Let's evaluate \(g(-x)\):
\(g(-x) = 1\)
Comparing \(g(-x)\) and \(g(x)\), we see that \(g(-x) = 1\) and \(g(x) = 1\).
Therefore, \(g(-x) = g(x)\) is true for all values of \(x\).
Thus, the determinant of \({\rm{f}}(x)\), which is the constant function \(g(x)=1\), is indeed an even function.
So, statement 3 is correct.
Summary of Correct Statements
Based on our analysis:
- Statement 1: \({\rm{f}}(\theta) \times {\rm{f}}(\phi) = {\rm{f}}(\theta + \phi)\) is Correct.
- Statement 2: The value of the determinant of the matrix \({\rm{f}}(\theta) \times {\rm{f}}(\phi)\) is 1 is Correct.
- Statement 3: The determinant of \({\rm{f}}(x)\) is an even function is Correct.
All three statements are correct.
Let's check the given options:
- 1 and 2 only
- 2 and 3 only
- 1 and 3 only
- 1, 2 and 3
The correct option is the one that includes all three statements.
The final answer is that statements 1, 2, and 3 are correct.
Revision Table: Matrix Properties and Determinants
| Concept | Description | Key Property/Formula |
|---|---|---|
| Matrix Multiplication | Process of multiplying two matrices. The number of columns in the first matrix must equal the number of rows in the second. | Resulting matrix element \(C_{ij} = \sum_k A_{ik}B_{kj}\) |
| Determinant of a Matrix | A scalar value that can be computed for a square matrix. It provides information about the matrix's properties (e.g., invertibility). | For a 2x2 matrix \(\left[ {\begin{array}{c} a&b\\ c&d \end{array}} \right]\), determinant is \(ad-bc\). For larger matrices, use cofactor expansion. |
| Determinant of a Product | The determinant of the product of two square matrices is the product of their determinants. | \(\det(AB) = \det(A)\det(B)\) |
| Even Function | A function \(g(x)\) is even if \(g(-x) = g(x)\) for all \(x\) in its domain. The graph is symmetric about the y-axis. | \(g(-x) = g(x)\) |
Additional Information: Special Matrices and Functions
The matrix \({\rm{f}}(x)\) given in the question is a type of rotation matrix in 3D space. Specifically, it represents a rotation around the z-axis by an angle \(x\).
- The top-left 2x2 block \(\left[ {\begin{array}{c} {\cos x}&{ - \sin x}\\ {\sin x}&{\cos x} \end{array}} \right]\) is the standard 2D rotation matrix.
- The third row and column \(\left[ {\begin{array}{c} 0\\ 0\\ 1 \end{array}} \right]\) and \(\left[ {\begin{array}{ccc} 0&0&1 \end{array}} \right]\) with the element '1' at (3,3) indicate that the z-axis is fixed during the transformation.
The property \({\rm{f}}(\theta) \times {\rm{f}}(\phi) = {\rm{f}}(\theta + \phi)\) makes sense geometrically: rotating by angle \(\phi\) and then by angle \(\theta\) is equivalent to a single rotation by angle \((\theta + \phi)\).
The determinant of a rotation matrix is always 1. This is consistent with our finding that \(\det({\rm{f}}(x)) = 1\). A determinant of 1 implies that the transformation preserves area (in 2D) or volume (in 3D) and orientation.
The concept of even and odd functions is important in calculus and analysis. Functions can be even, odd (\(g(-x) = -g(x)\)), or neither.
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