Problem 63 - Precalculus with Limits

Question

In Exercises 63 and \(64,\) show that the matrix is invertible and find its inverse. $$A=\left[\begin{array}{rr}{\sin \theta} & {\cos \theta} \\ {-\cos \theta} & {\sin \theta}\end{array}\right]$$

Step-by-Step Solution

Answer

The matrix A is invertible and its inverse is \( A^{-1} = \left[ \begin{array}{r} \sin\ \theta & -\cos\ \theta \ \cos\ \theta & \sin\ \theta \end{array} \right] \)

Step 1: Calculate the determinant

The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by \( ad - bc \). For the matrix \( A = \left[ \begin{array}{r} \sin\ \theta & \cos\ \theta \ -\cos\ \theta & \sin\ \theta \end{array} \right] \), calculating the determinant gives \( (\sin^2 \theta) - (-\cos^2 \theta)= \sin^2\theta + \cos^2\theta = 1 \), which is not equal to zero, indicating that the matrix A is invertible.

Step 2: Apply the Inverse Matrix Formula

Using the formula for finding the inverse of a 2x2 matrix, the inverse of \( A \) is \( A^{-1} = \frac{1}{ad-bc}\left[ \begin{array}{r} d & -b \ -c & a \end{array} \right] \), where a=\(\sin \theta\), b=\(\cos \theta\), c=-\(\cos \theta\), d=\(\sin \theta\) and ad-bc= 1 in our case. Thus \( A^{-1} = \left[ \begin{array}{r} \sin\ \theta & -\cos\ \theta \ \cos\ \theta & \sin\ \theta \end{array} \right] \).

Step 3: Verification

To verify the validity of our solution, we perform a matrix multiplication between the matrix \( A \) and its inverse \( A^{-1} \). The result should be the identity matrix \( I \). Indeed, this is the case which confirms our solution.

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a special number that can tell us whether the matrix is invertible and give insights into many of its properties. For a 2x2 matrix, the determinant is calculated using the formula \( ad - bc \) for matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \). In the context of matrix \( A \), with elements given by trigonometric functions of \( \theta \), the determinant simplifies to \( \sin^2\theta + \cos^2\theta \) due to the identity \( \sin^2\theta + \cos^2\theta = 1 \) for all values of \( \theta \). This is crucial because if the determinant were zero, the matrix \( A \) wouldn't be invertible. Instead, we find it's always equal to 1, confirming that \( A \) is indeed invertible.

Inverse of a Matrix

The inverse of a matrix \( A \), denoted as \( A^{-1} \), is a matrix that, when multiplied with \( A \) yields the identity matrix \( I \). For a 2x2 matrix, \( A^{-1} \) is found using the formula \( \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix} \). Applying this to our matrix with \( a = \sin \theta \) and so on, the inverse becomes \( A^{-1} = \begin{bmatrix} \sin\theta & -\cos\theta \ \cos\theta & \sin\theta \end{bmatrix} \), which can be used in various applications such as solving systems of equations or transforming geometric objects.

Identity Matrix

The identity matrix, typically denoted by \( I \), serves as the equivalent of the number 1 in matrix algebra. For a 2x2 matrix, it's represented as \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \). It holds a unique property where any matrix \( A \) multiplied by \( I \) remains unchanged, such that \( AI = IA = A \). This property is used to verify the correctness of an inverse matrix, as with our matrix \( A \) and its inverse \( A^{-1} \); when multiplied together, they should yield the identity matrix, confirming that \( A^{-1} \) is indeed the true inverse of \( A \).

Matrix Multiplication

Matrix multiplication is not the same as multiplying numbers. It involves a row-by-column operation, yielding a new matrix whose elements come from the sum of products of elements in the respective row of the first matrix and the column of the second matrix. Importantly, matrix multiplication is not generally commutative, meaning that \( AB \) may not equal \( BA \). In our verification step for the invertible matrix \( A \), we multiply \( A \) by \( A^{-1} \) and, if all is correct, the product should be equal to the identity matrix \( I \), which it is in our case.

Sin and Cos Trigonometric Identities

The trigonometric identities involving \( \sin \) and \( \cos \) are fundamental to solving many mathematical problems. One such identity is \( \sin^2\theta + \cos^2\theta = 1 \), which is always true for any angle \( \theta \). These identities are instrumental in simplifying the expressions involving trigonometric functions. For instance, in finding the determinant of matrix \( A \), we rely on this identity to confirm that the value is 1, and thus, \( A \) is invertible. Understanding these identities is critical not only in matrix algebra but also in calculus, geometry, and even physics.