Chapter 8: Problem 32
$$ \text { Find the inverse of } \mathbf{A}=\left(\begin{array}{cc} \sin \theta & \cos \theta \\ -\cos \theta & \sin \theta \end{array}\right) $$
Short Answer
Expert verified
The inverse of \( \mathbf{A} \) is \( \begin{pmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{pmatrix} \).
Step by step solution
01
Understand the Problem
The problem requires finding the inverse of a 2x2 matrix \( \mathbf{A} \). An inverse of a matrix \( \mathbf{A} \), denoted as \( \mathbf{A}^{-1} \), is a matrix that, when multiplied by \( \mathbf{A} \), results in the identity matrix. Thus, we need to determine \( \mathbf{A}^{-1} \) such that \( \mathbf{A} \mathbf{A}^{-1} = \mathbf{I} \).
02
Calculate the Determinant
For a 2x2 matrix \( \mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is calculated using the formula \( \text{det}(\mathbf{A}) = ad - bc \). For our matrix, \( a = \sin \theta \), \( b = \cos \theta \), \( c = -\cos \theta \), and \( d = \sin \theta \). Substitute these into the determinant formula: \( \text{det}(\mathbf{A}) = (\sin \theta)(\sin \theta) - (-\cos \theta)(\cos \theta) = \sin^2 \theta + \cos^2 \theta \).
03
Use the Pythagorean Identity
Recall that the Pythagorean identity states \( \sin^2 \theta + \cos^2 \theta = 1 \). Hence, the determinant of \( \mathbf{A} \) is \( \text{det}(\mathbf{A}) = 1 \). Since the determinant is non-zero, \( \mathbf{A} \) has an inverse.
04
Apply the Inverse Formula for 2x2 Matrices
The inverse of a 2x2 matrix \( \mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is given by \( \mathbf{A}^{-1} = \frac{1}{\text{det}(\mathbf{A})} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \). For our matrix, \( d = \sin \theta \), \( -b = -\cos \theta \), \( -c = \cos \theta \), and \( a = \sin \theta \). Substitute these values: \( \mathbf{A}^{-1} = \begin{pmatrix} \sin \theta & -\cos \theta \ \cos \theta & \sin \theta \end{pmatrix} \).
05
Confirm with the Identity Matrix
Multiply \( \mathbf{A} \) and \( \mathbf{A}^{-1} \) to confirm they produce the identity matrix. We calculate \( \mathbf{A} \mathbf{A}^{-1} = \begin{pmatrix} \sin \theta & \cos \theta \ -\cos \theta & \sin \theta \end{pmatrix} \begin{pmatrix} \sin \theta & -\cos \theta \ \cos \theta & \sin \theta \end{pmatrix} \). The result is the identity matrix \( \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \), confirming that \( \mathbf{A}^{-1} \) is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
2x2 matrix
Matrices are rectangular arrays of numbers set up in rows and columns. A 2x2 matrix is one of the simplest forms with two rows and two columns. It looks like this:
In this context, we deal with the matrix \( \mathbf{A}=\left(\begin{array}{cc} \sin \theta & \cos \theta \ -\cos \theta & \sin \theta \end{array}\right) \), which represents a rotation in the 2D coordinate plane.
The ability to find the inverse of a 2x2 matrix is important because the inverse matrix reverses the effect of the original transformation. This means if the matrix represents a rotation, its inverse will rotate objects back to their original position.
- Examples of 2x2 matrices are: - \( \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix} \) - \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \)
In this context, we deal with the matrix \( \mathbf{A}=\left(\begin{array}{cc} \sin \theta & \cos \theta \ -\cos \theta & \sin \theta \end{array}\right) \), which represents a rotation in the 2D coordinate plane.
The ability to find the inverse of a 2x2 matrix is important because the inverse matrix reverses the effect of the original transformation. This means if the matrix represents a rotation, its inverse will rotate objects back to their original position.
Determinant
The determinant is a special number that can be calculated from a square matrix. For 2x2 matrices, it helps determine if the matrix has an inverse.
Plug the values into the formula:
If the determinant is not zero, the 2x2 matrix has an inverse, which means the matrix transformation is reversible.
- The formula for the determinant of a 2x2 matrix \( \mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is \( \text{det}(\mathbf{A}) = ad - bc \).
Plug the values into the formula:
- \( a = \sin \theta \)
- \( b = \cos \theta \)
- \( c = -\cos \theta \)
- \( d = \sin \theta \)
If the determinant is not zero, the 2x2 matrix has an inverse, which means the matrix transformation is reversible.
Pythagorean Identity
The Pythagorean identity is a fundamental trigonometric identity. It states that for any angle \( \theta \), \( \sin^2 \theta + \cos^2 \theta = 1 \).
Since the Pythagorean Identity confirms this is 1, the determinant is indeed non-zero.This concept assures us the matrix \( \mathbf{A} \) is invertible, meaning it can effectively reverse its transformation effect when multiplied by its inverse.This identity is crucial in linear algebra and trigonometry, often simplifying complex expressions involving trigonometric functions.
- This identity arises from the Pythagorean Theorem in a right-angled triangle, relating the sides of the triangle to the circle's radius.
Since the Pythagorean Identity confirms this is 1, the determinant is indeed non-zero.This concept assures us the matrix \( \mathbf{A} \) is invertible, meaning it can effectively reverse its transformation effect when multiplied by its inverse.This identity is crucial in linear algebra and trigonometry, often simplifying complex expressions involving trigonometric functions.
- Remember: inverting a matrix is only possible when the determinant is non-zero. The Pythagorean Identity gives us confidence in certain trigonometric matrices always having non-zero determinants, ensuring they are invertible.