Problem 16 - A Graphical Approach to Precalculus with Limits

Question

For each matrix, find \(A^{-1}\) if it exists. Do not use a calculator. $$A=\left[\begin{array}{ll} 0.8 & -0.3 \\ 0.5 & -0.2 \end{array}\right]$$

Step-by-Step Solution

Answer

The inverse of matrix A is \( \begin{bmatrix} 20 & -30 \\ 50 & -80 \end{bmatrix} \).

Step 1: Determine the formula for the inverse of a 2x2 matrix

For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the inverse \( A^{-1} \) (if it exists) is given by \[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \] provided that the determinant \( ad-bc eq 0 \).

Step 2: Calculate the determinant of matrix A

Compute the determinant of matrix \( A \), where \( A = \begin{bmatrix} 0.8 & -0.3 \ 0.5 & -0.2 \end{bmatrix} \), using the formula \( ad - bc \). Thus, the determinant is \[ (0.8)(-0.2) - (-0.3)(0.5) = -0.16 + 0.15 = -0.01 \].

Step 3: Check if the determinant is non-zero

Since the determinant \( -0.01 eq 0 \), the matrix \( A \) has an inverse.

Step 4: Compute the inverse of matrix A

Use the formula for the inverse of a 2x2 matrix: \[ A^{-1} = \frac{1}{-0.01} \begin{bmatrix} -0.2 & 0.3 \ -0.5 & 0.8 \end{bmatrix} \]. Simplifying, we get \[ A^{-1} = \begin{bmatrix} 20 & -30 \ 50 & -80 \end{bmatrix} \].

Key Concepts

Determinant Calculation

The determinant of a matrix is an essential value that helps determine if a matrix has an inverse. When dealing with a 2x2 matrix, the determinant can be calculated very simply. Consider a matrix in the form:

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

For this specific arrangement, the determinant is given by the formula:

• \( ad - bc \)

This calculation involves multiplying the diagonal elements (from top-left to bottom-right) and subtracting the product of the opposite diagonal (from top-right to bottom-left).
In the exercise, we calculate the determinant of:

• \( A = \begin{bmatrix} 0.8 & -0.3 \ 0.5 & -0.2 \end{bmatrix} \)

Using the formula, we find:

• \((0.8)(-0.2) - (-0.3)(0.5) = -0.16 + 0.15 = -0.01\)

This non-zero result indicates that matrix \( A \) indeed has an inverse.

2x2 Matrices

2x2 matrices are fundamental in linear algebra because of their simplicity and because they help build an understanding of more complex matrices. These matrices are written in a compact, two-by-two form:

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

Each element \( a, b, c, \) and \( d \) represents a number from a data set. In practical terms, each value corresponds to a relationship or transformation, such as rotations or scaling in coordinate systems.
One crucial aspect of these matrices is the determinant. Understanding how to compute it, as in the exercise, helps determine whether the matrix has unique solutions or can be inverted. Additionally, because they are small, 2x2 matrices are an excellent stepping stone toward understanding larger-sized matrices. They appear often in mathematical problems and models, making understanding their properties quite essential.

Algebraic simplification

Algebraic simplification involves reducing mathematical expressions into their simplest form. This practice not only makes equations easier to work with but also provides clearer insights into their meaning or implications. When computing the inverse of a matrix, simplification is crucial.Given the inverse computation:

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

The division by the determinant \( ad-bc \) results in fractional forms when the determinant is not a neat number. In our exercise, the determinant was \(-0.01\), resulting in the matrix multiplying by \(-100\).
By simplifying, we multiply directly by \(-100\) which effectively speeds up manual computation and helps in avoiding potential errors in fractional arithmetic. The final result

• \( A^{-1} = \begin{bmatrix} 20 & -30 \ 50 & -80 \end{bmatrix} \)

is much clearer, demonstrating the benefits of algebraic simplification. This step assures that working through matrix problems remains manageable and clear.

Inverse Matrix Formula

The formula to find the inverse of a 2x2 matrix is a pivotal tool in linear algebra. The inverse of a matrix, when it exists, is like a mirror image that, when multiplied by the original matrix, gives the identity matrix. Specifically, for a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by:

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

This formula rearranges the elements of the matrix into a new configuration while also including a division by the determinant \( ad-bc \).
It's important to check that this determinant is non-zero because a zero determinant indicates that the matrix is singular, meaning it does not have an inverse. This formula reveals that the inverse effectively undoes whatever transformation the original matrix represents, allowing one to solve systems of linear equations.
In the given exercise, using this formula confirms the inverse matrix once the determinant is computed and found to be non-zero. Practicing this understanding ensures that one can handle inverse calculations for any 2x2 matrix consistently and accurately.