Problem 15 - Algebra and Trigonometry

Question

Find the inverse of the matrix if it exists. $$ \left[\begin{array}{rr}{0.4} & {-1.2} \\ {0.3} & {0.6}\end{array}\right] $$

Step-by-Step Solution

Answer

The inverse of the matrix is approximately \[\begin{bmatrix} 1.0 & 2.0 \\ -0.5 & 0.67 \end{bmatrix}\].

Step 1: Check if the Determinant is Non-zero

To find the inverse of a matrix, we must first check if the matrix is invertible. A matrix is invertible if its determinant is not zero.For the matrix \[ A = \begin{bmatrix} 0.4 & -1.2 \ 0.3 & 0.6 \end{bmatrix} \]calculate the determinant using the formula:\[ \text{det}(A) = ad - bc \]where \( a = 0.4 \), \( b = -1.2 \), \( c = 0.3 \), and \( d = 0.6 \).\[ \text{det}(A) = (0.4 \times 0.6) - (-1.2 \times 0.3) = 0.24 + 0.36 = 0.6 \]

Step 2: Use the Formula for the Inverse of a 2x2 Matrix

Since the determinant is not zero, the matrix is invertible. We can now find the inverse using the formula for a 2x2 matrix:\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]Substitute the values:\[ A^{-1} = \frac{1}{0.6} \begin{bmatrix} 0.6 & 1.2 \ -0.3 & 0.4 \end{bmatrix} \]

Step 3: Simplify the Inverse Matrix

Finally, simplify the matrix by multiplying each element by \( \frac{1}{0.6} \) which is the same as multiplying by \( \frac{5}{3} \):\[ A^{-1} = \begin{bmatrix} 0.6 \times \frac{5}{3} & 1.2 \times \frac{5}{3} \ -0.3 \times \frac{5}{3} & 0.4 \times \frac{5}{3} \end{bmatrix} \]\[ A^{-1} = \begin{bmatrix} 1.0 & 2.0 \ -0.5 & \approx 0.67 \end{bmatrix} \]

Key Concepts

2x2 matrix

A 2x2 matrix is one of the simplest forms of a matrix. It consists of two rows and two columns. Each element in the matrix is specified by its position in these rows and columns. For instance, in the matrix \( \begin{bmatrix} 0.4 & -1.2 \ 0.3 & 0.6 \end{bmatrix} \), the first element 0.4 is in the first row and first column, while -1.2 is in the first row, second column.

This kind of matrix is often used in linear algebra for transformations and solving systems of equations. It provides a manageable introduction to the concept of matrices before moving on to larger matrices. In 2x2 matrices, calculations such as determinants and inverses are relatively straightforward compared to larger matrices. This makes them a popular topic for introductory courses in mathematics.

To understand the behavior of a 2x2 matrix, we can apply operations like addition, subtraction, and multiplication with other matrices or scalars. However, not every 2x2 matrix can be inverted, which is an essential concept to grasp.

determinant calculation

To determine whether a 2x2 matrix is invertible, we need to calculate its determinant. The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is calculated using the formula:

• \( \text{det}(A) = ad - bc \)

This formula might look simple, but it plays a crucial role in understanding the properties of a matrix.

In our example matrix \( \begin{bmatrix} 0.4 & -1.2 \ 0.3 & 0.6 \end{bmatrix} \), the determinant was calculated as \( 0.24 + 0.36 = 0.6 \). Because the determinant is not zero, the matrix is invertible. A zero determinant would imply a singular matrix, which does not have an inverse. Checking the determinant is the first step before calculating the inverse as it determines the feasibility.

invertible matrix

An invertible matrix, also known as a non-singular or non-degenerate matrix, is one that has an inverse. For a 2x2 matrix, having a non-zero determinant is enough to ensure that the inverse exists. Let's explore how to find the inverse of a 2x2 matrix once we know it's invertible.

The formula to find the inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is:

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

This formula flips the diagonal elements, changes the signs of the off-diagonal elements, and scales the entire matrix by the reciprocal of the determinant.

In our specific example, after verifying our matrix has a determinant of 0.6, the inverse is calculated as:

• \( A^{-1} = \frac{1}{0.6} \begin{bmatrix} 0.6 & 1.2 \ -0.3 & 0.4 \end{bmatrix} \)

By multiplying each of these elements by \( \frac{5}{3} \), the inverse matrix becomes:

• \( A^{-1} = \begin{bmatrix} 1.0 & 2.0 \ -0.5 & \approx 0.67 \end{bmatrix} \)

This inverse matrix can now be used for a variety of applications, including solving systems of linear equations or transforming geometric data.