Problem 32 - Advanced Engineering Mathematics

Question

Find the inverse of \(\mathbf{A}=\left(\begin{array}{cc}\sin \theta & \cos \theta \\ -\cos \theta & \sin \theta\end{array}\right)\).

Step-by-Step Solution

Answer

The inverse of matrix \(\mathbf{A}\) is \(\begin{pmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{pmatrix}\).

Step 1: Understand the Structure of Matrix A

Matrix \(\mathbf{A}\) is a \(2 \times 2\) matrix given by: \[\mathbf{A} = \begin{pmatrix} \sin \theta & \cos \theta \ -\cos \theta & \sin \theta \end{pmatrix}\] We need to find the inverse of this matrix.

Step 2: Determine the Formula for the Inverse

The inverse of a \(2 \times 2\) matrix \[\begin{pmatrix} a & b \ c & d \end{pmatrix}\] is given by: \[\begin{pmatrix} d & -b \ -c & a \end{pmatrix} \times \frac{1}{ad - bc}\]Here, \(a = \sin \theta\), \(b = \cos \theta\), \(c = -\cos \theta\), and \(d = \sin \theta\).

Step 3: Compute the Determinant of A

The determinant of matrix \(\mathbf{A}\), denoted as \(\det(\mathbf{A})\) is calculated as follows:\[\det(\mathbf{A}) = (\sin \theta)(\sin \theta) - (\cos \theta)(-\cos \theta) = \sin^2 \theta + \cos^2 \theta\]Using the trigonometric identity, we know that \(\sin^2 \theta + \cos^2 \theta = 1\). Thus, \(\det(\mathbf{A}) = 1\).

Step 4: Apply the Inverse Formula

Using the inverse formula, we substitute the values of \(a, b, c,\) and \(d\) along with the determinant:\[\mathbf{A}^{-1} = \begin{pmatrix} \sin \theta & -\cos \theta \ \cos \theta & \sin \theta \end{pmatrix} \times \frac{1}{1}\]Since \(\det(\mathbf{A}) = 1\), we don't need to divide by anything other than 1.

Step 5: Verify the Inverse

To confirm that the matrix \(\mathbf{A}^{-1}\) is correct, multiply \(\mathbf{A}\) by \(\mathbf{A}^{-1}\) to ensure it results in the identity matrix, \(\mathbf{I}\):\[\mathbf{A} \times \mathbf{A}^{-1} = \begin{pmatrix} \sin \theta & \cos \theta \ -\cos \theta & \sin \theta \end{pmatrix} \times \begin{pmatrix} \sin \theta & -\cos \theta \ \cos \theta & \sin \theta \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\]This confirms that the computed inverse is correct as it results in an identity matrix.

Key Concepts

Determinant Calculation

One of the key steps in finding the inverse of a matrix is calculating its determinant. For a 2\( \times \)2 matrix, such as the one given by matrix \( \mathbf{A} \): \[ \mathbf{A} = \begin{pmatrix} \sin \theta & \cos \theta \ -\cos \theta & \sin \theta \end{pmatrix} \]The determinant \( \det(\mathbf{A}) \) is crucial. The determinant formula for a 2\( \times \)2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is \( ad - bc \). For this specific matrix: - \( a = \sin \theta \) - \( b = \cos \theta \) - \( c = -\cos \theta \) - \( d = \sin \theta \) Plugging these into the determinant formula:\[\det(\mathbf{A}) = (\sin \theta)(\sin \theta) - (\cos \theta)(-\cos \theta) \= \sin^2 \theta + \cos^2 \theta\]A remarkable property of sine and cosine comes into play here: \( \sin^2 \theta + \cos^2 \theta = 1 \), a well-known trigonometric identity from the Pythagorean theorem. Thus, \( \det(\mathbf{A}) = 1 \), ensuring the matrix has an inverse.

Trigonometric Identities

Trigonometric identities are expressions based on specific trigonometric properties and relationships. Among them, the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) is foundational. It states that for any angle \( \theta \), the square of the sine of \( \theta \) added to the square of the cosine of \( \theta \) equals 1. This identity is derived from the Pythagorean theorem, considering the unit circle's context. In our problem, this identity directly simplifies determinant computation, as explained in the matrix context:- It verifies that the determinant is \( 1 \), ensuring matrix invertibility.- Simplifies calculations by reducing complex trigonometric forms.Other important trigonometric identities include:

• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
• \( 1 + \tan^2 \theta = \sec^2 \theta \)

Utilizing trigonometric identities in calculations not only simplifies determining properties like determinants but also aids in simplifying expressions in calculus and other fields of mathematics.

Identity Matrix

An identity matrix is a special kind of matrix that plays a crucial role in linear algebra. It is the multiplicative identity for matrices, much like the number 1 in arithmetic. When a matrix is multiplied by the identity matrix, it remains unchanged.The identity matrix \( \mathbf{I} \) for a 2\( \times \)2 matrix is:\[ \mathbf{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \]In the context of finding the inverse of a matrix \( \mathbf{A} \), we verify our result by multiplying \( \mathbf{A} \) by its inverse \( \mathbf{A}^{-1} \). If the product \( \mathbf{A} \times \mathbf{A}^{-1} \) results in \( \mathbf{I} \), then \( \mathbf{A}^{-1} \) is correctly calculated. This property confirms that \( \mathbf{A} \) and \( \mathbf{A}^{-1} \) are true inverses. Thus, in our problem:\[\mathbf{A} \times \mathbf{A}^{-1} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \mathbf{I}\]Understanding the functionality of identity matrices is crucial, as they offer assurance of the matrix operations' validity and play a vital role in solving systems of linear equations, transformations, and more.