Problem 15
Question
For each matrix, find \(A^{-1}\) if it exists. Do not use a calculator. $$A=\left[\begin{array}{ll} 0.6 & 0.2 \\ 0.5 & 0.1 \end{array}\right]$$
Step-by-Step Solution
Verified Answer
The inverse of matrix \(A\) is \(\begin{bmatrix} -2.5 & 5 \\ 12.5 & -15 \end{bmatrix}\).
1Step 1: Find the Determinant
The first step to finding the inverse of a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is to compute its determinant \(\det(A)\). For matrix \(A = \begin{bmatrix} 0.6 & 0.2 \ 0.5 & 0.1 \end{bmatrix}\), the determinant is calculated as follows:\[\text{det}(A) = ad - bc = (0.6)(0.1) - (0.2)(0.5) = 0.06 - 0.1 = -0.04\]
2Step 2: Check if the Determinant is Zero
The inverse of a matrix exists if and only if its determinant is not zero. In this case, \(\det(A) = -0.04\), which is not zero. Therefore, the inverse of matrix \(A\) exists.
3Step 3: Compute the Inverse Matrix
The inverse of a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by:\[A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]Substitute the given values into the formula:\[A^{-1} = \frac{1}{-0.04} \begin{bmatrix} 0.1 & -0.2 \ -0.5 & 0.6 \end{bmatrix} = \begin{bmatrix} -2.5 & 5 \ 12.5 & -15 \end{bmatrix}\]So, the inverse of matrix \(A\) is \(\begin{bmatrix} -2.5 & 5 \ 12.5 & -15 \end{bmatrix}\).
Key Concepts
Determinant2x2 MatrixInverse CalculationMatrix Algebra
Determinant
The determinant is a special number that you can calculate from a square matrix. It plays a crucial role in matrix operations, especially when you need to find the inverse of a matrix.
For a 2x2 matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\]its determinant is computed by the formula:\[\text{det}(A) = ad - bc\]This determinant helps determine if a matrix is invertible.
If the determinant is zero, the matrix does not have an inverse. In the given exercise, the determinant is -0.04, which is non-zero. Therefore, an inverse exists for the matrix.
For a 2x2 matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\]its determinant is computed by the formula:\[\text{det}(A) = ad - bc\]This determinant helps determine if a matrix is invertible.
If the determinant is zero, the matrix does not have an inverse. In the given exercise, the determinant is -0.04, which is non-zero. Therefore, an inverse exists for the matrix.
2x2 Matrix
A 2x2 matrix is one of the simplest forms of matrices in linear algebra. It is composed of two rows and two columns.
Here’s an example format of a 2x2 matrix:\[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]Each element, like \(a\), \(b\), \(c\), and \(d\), represents a particular entry of the matrix.
Despite its simplicity, this type of matrix is fundamental in mathematics and engineering due to its straightforward operations and foundational learning potential.
Understanding the basic operations on a 2x2 matrix such as finding the determinant and inverses is essential for advancing to more complex topics in matrix algebra.
Here’s an example format of a 2x2 matrix:\[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]Each element, like \(a\), \(b\), \(c\), and \(d\), represents a particular entry of the matrix.
Despite its simplicity, this type of matrix is fundamental in mathematics and engineering due to its straightforward operations and foundational learning potential.
Understanding the basic operations on a 2x2 matrix such as finding the determinant and inverses is essential for advancing to more complex topics in matrix algebra.
Inverse Calculation
Finding the inverse of a matrix is like finding a reciprocal in numbers.
The inverse operation is important in solving matrix equations and finding linear transformations. For a 2x2 matrix:\[A = \begin{bmatrix} a & b \ c & d \end{bmatrix},\]the formula for its inverse, denoted as \(A^{-1}\), is:\[A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}.\]This formula involves swapping the positions of \(a\) and \(d\), changing the signs of \(b\) and \(c\), and dividing by the determinant.
In our exercise, this provides us an inverse of:\[\begin{bmatrix} -2.5 & 5 \ 12.5 & -15 \end{bmatrix}.\]This process is straightforward yet powerful in solving systems of linear equations.
The inverse operation is important in solving matrix equations and finding linear transformations. For a 2x2 matrix:\[A = \begin{bmatrix} a & b \ c & d \end{bmatrix},\]the formula for its inverse, denoted as \(A^{-1}\), is:\[A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}.\]This formula involves swapping the positions of \(a\) and \(d\), changing the signs of \(b\) and \(c\), and dividing by the determinant.
In our exercise, this provides us an inverse of:\[\begin{bmatrix} -2.5 & 5 \ 12.5 & -15 \end{bmatrix}.\]This process is straightforward yet powerful in solving systems of linear equations.
Matrix Algebra
Matrix algebra is a branch of mathematics that deals with matrix operations.
It encompasses various rules and operations applicable to matrices, such as addition, subtraction, multiplication, and inversion.
Understanding matrix algebra is crucial, as it forms the basis for more complicated calculations in linear transformations and systems of equations.
It encompasses various rules and operations applicable to matrices, such as addition, subtraction, multiplication, and inversion.
Understanding matrix algebra is crucial, as it forms the basis for more complicated calculations in linear transformations and systems of equations.
- Addition and Subtraction: Matrices can be added or subtracted if they have the same dimensions.
- Multiplication: The multiplication of matrices involves rows and columns, requiring careful alignment.
- Inversion: Finding the inverse is possible only for square matrices (e.g., 2x2).
Other exercises in this chapter
Problem 14
Solve each system by substitution. $$\begin{aligned}&7 x-y=-10\\\&3 y-x=10\end{aligned}$$
View solution Problem 15
Find the partial fraction decomposition for each rational expression. $$\frac{4}{x(1-x)}$$
View solution Problem 15
Write the system of equations associated with each augmented matrix. $$\left[\begin{array}{rr|r} 2 & 1 & 1 \\ 3 & -2 & -9 \end{array}\right]$$
View solution Problem 15
Your friend missed the lecture on adding matrices. In your own words, explain to her how to add two matrices.
View solution