Problem 16 - Precalculus: Mathematics for Calculus

Question

Finding the Inverse of a Matrix Find the inverse of the matrix if it exists. $$\left[\begin{array}{ll}\frac{1}{2} & \frac{1}{3} \\\5 & 4\end{array}\right]$$

Step-by-Step Solution

Answer

The inverse of the matrix is \( \begin{bmatrix} 12 & -1 \\ -15 & \frac{3}{2} \end{bmatrix} \).

Step 1: Understand the matrix

The matrix given is \( A = \left[ \begin{array}{cc} \frac{1}{2} & \frac{1}{3} \ 5 & 4 \end{array} \right] \). It is a 2x2 matrix.

Step 2: Check if the inverse exists

For a 2x2 matrix \( A = \left[ \begin{array}{cc} a & b \ c & d \end{array} \right] \), the matrix has an inverse if its determinant \( \Delta = ad - bc eq 0 \).

Step 3: Calculate determinant

Compute \( \Delta = \frac{1}{2} \times 4 - \frac{1}{3} \times 5 = 2 - \frac{5}{3} = \frac{6}{3} - \frac{5}{3} = \frac{1}{3} \). Since \( \Delta eq 0 \), the inverse exists.

Step 4: Find the inverse matrix formula

The inverse of a 2x2 matrix is given by \( A^{-1} = \frac{1}{\Delta} \left[ \begin{array}{cc} d & -b \ -c & a \end{array} \right] \).

Step 5: Substitute values into the formula

Plug in the values from matrix \( A \) and its determinant: \( A^{-1} = \frac{1}{\frac{1}{3}} \left[ \begin{array}{cc} 4 & -\frac{1}{3} \ -5 & \frac{1}{2} \end{array} \right] = 3 \left[ \begin{array}{cc} 4 & -\frac{1}{3} \ -5 & \frac{1}{2} \end{array} \right] \).

Step 6: Calculate the inverse

Multiply each element by 3: \( A^{-1} = \left[ \begin{array}{cc} 12 & -1 \ -15 & \frac{3}{2} \end{array} \right] \). This is the inverse of matrix \( A \).

Key Concepts

Determinant of a Matrix

The determinant is a special number that can be calculated from a square matrix. It provides crucial information about the matrix, like whether the matrix can be inverted or not. For a 2x2 matrix, the determinant is computed using a specific formula. If you have a matrix:

• \( \begin{bmatrix}a & b \c & d\end{bmatrix}\)

The determinant, denoted as \( \Delta \), is given by the formula: \( \Delta = ad - bc \). This calculation involves multiplying the diagonals and then subtracting the results.
If the determinant is zero, the matrix does not have an inverse. Conversely, a nonzero determinant indicates that the inverse exists. In our exercise, with the matrix \( A = \begin{bmatrix} \frac{1}{2} & \frac{1}{3} \ 5 & 4 \end{bmatrix} \), we calculated its determinant to be \( \frac{1}{3} \), confirming that the inverse is possible.

2x2 Matrix

A 2x2 matrix is a simple form of a matrix structure that consists of two rows and two columns. It is often the starting point for learning matrix operations because its properties are straightforward and calculations are manageable. These matrices are written in a compact form:

• \( \begin{bmatrix}a & b \c & d\end{bmatrix}\)

Such matrices are pivotal in linear algebra and used widely in various applications like computer graphics and engineering. In our problem, the matrix \( A \) is represented by:
\( A = \begin{bmatrix} \frac{1}{2} & \frac{1}{3} \ 5 & 4 \end{bmatrix} \). This format helps in understanding and applying operations like determinants and inverses, enabling a foundation for more complex matrices. Studying the 2x2 matrix gives insights into manipulating larger matrices.

Inverse Matrix Formula

Once you have determined that a matrix has a non-zero determinant, you can proceed to find its inverse. The process to find the inverse of a 2x2 matrix is relatively simple with a given formula. For a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is calculated as:

• \( A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\)

This formula involves swapping the positions of \( a \) and \( d \), negating \( b \) and \( c \), and then scaling the entire matrix by the reciprocal of the determinant \((ad-bc)\).
In the given exercise, the inverse matrix was computed as:
\( A^{-1} = \begin{bmatrix} 12 & -1 \ -15 & \frac{3}{2} \end{bmatrix} \) using its determined value from our previous steps in the solution. Always ensure to validate your inverse by confirming the determinant isn't zero before applying this formula.