Problem 46 - Algebra and Trigonometry

Question

Find the inverse of the matrix. For what value(s) of \(x\), if any, does the matrix have no inverse? \(\left[\begin{array}{ll}{\sec x} & {\tan x} \\ {\tan x} & {\sec x}\end{array}\right]\)

Step-by-Step Solution

Answer

The matrix is always invertible; its inverse is \( \begin{bmatrix} \sec x & -\tan x \\ -\tan x & \sec x \end{bmatrix} \).

Step 1: Determine the Determinant

For a 2x2 matrix in the form \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as \( ad - bc \). Here, the matrix is \( \begin{bmatrix} \sec x & \tan x \ \tan x & \sec x \end{bmatrix} \), so we have \( \det(A) = (\sec x)(\sec x) - (\tan x)(\tan x) = \sec^2 x - \tan^2 x \).

Step 2: Apply Trigonometric Identity

Recall the trigonometric identity \( \sec^2 x = 1 + \tan^2 x \). Substitute \( \sec^2 x \) in the determinant expression: \( \sec^2 x - \tan^2 x = (1 + \tan^2 x) - \tan^2 x = 1 \). The determinant is 1.

Step 3: Determine When the Determinant is Zero

A matrix is invertible if its determinant is not zero. In Step 2, we found the determinant is always 1. Therefore, there are no values of \( x \) for which the determinant is zero.

Step 4: Conclusion on Matrix Inversibility

Since the determinant is not zero for any \( x \), the matrix is always invertible. Thus, there is no value of \( x \) that makes the matrix non-invertible.

Step 5: Find the Inverse of the Matrix

The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \). Since the determinant is 1, the inverse of our matrix is simply \( \begin{bmatrix} \sec x & -\tan x \ -\tan x & \sec x \end{bmatrix} \).

Key Concepts

Determinant of a Matrix

The determinant of a matrix is a special number that can be calculated from its elements and tells us several things, including whether the matrix is invertible. For a 2x2 matrix, given in the form \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as \( ad - bc \).
For the matrix in the exercise, \( \begin{bmatrix} \sec x & \tan x \ \tan x & \sec x \end{bmatrix} \), the determinant is \( \sec^2 x - \tan^2 x \). This is a fundamental concept since when a matrix has a zero determinant, it means the matrix is not invertible, or in simpler terms, it doesn't have an inverse.
In our case, the calculation showed that the determinant is always 1, meaning it never equals zero, ensuring that the matrix is always invertible.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions and are true for all values of the variable. In this exercise, we use the identity \( \sec^2 x = 1 + \tan^2 x \).
This identity is crucial for simplifying the determinant of our matrix. Substituting \( \sec^2 x \) in the determinant expression gives us: \( \sec^2 x - \tan^2 x = (1 + \tan^2 x) - \tan^2 x = 1 \).
This step simplifies our calculations and confirms that the determinant is constant. These identities are very useful in making complex trigonometric calculations easier by reducing them to simpler forms.

Invertible Matrix

An invertible matrix is one that has a matrix inverse. The inverse of a matrix \( A \) is another matrix, often denoted as \( A^{-1} \), such that when it is multiplied with \( A \), it yields the identity matrix. This property is essential in solving systems of equations and performing other computations.
For a matrix to be invertible, its determinant must be non-zero. From our exercise, we found that the determinant is always 1, thus confirming the matrix is always invertible. Furthermore, calculating the inverse for a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the formula is \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).
Since the determinant is 1, finding the inverse becomes straightforward, resulting in \( \begin{bmatrix} \sec x & -\tan x \ -\tan x & \sec x \end{bmatrix} \). This inverse can then be used for various mathematical purposes, ensuring that any operations involving the matrix can be completed efficiently.