Chapter 4: Problem 14
Find the inverse of each matrix, if it exists. $$ \left[\begin{array}{ll}{1} & {2} \\ {2} & {1}\end{array}\right] $$
Short Answer
Expert verified
The inverse of the matrix is \( \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \).
Step by step solution
01
Determine if the inverse exists
For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), it has an inverse if and only if its determinant is non-zero. The determinant of this matrix is computed as \( ad - bc \). Thus, we have: \( 1 \times 1 - 2 \times 2 = 1 - 4 = -3 \). Since the determinant is -3 (not zero), an inverse exists.
02
Compute the inverse formula for a 2x2 matrix
The inverse of a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) with non-zero determinant \( ad - bc \) is given by \( A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).
03
Apply the formula to find the inverse
Using the components of the matrix \( A = \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix} \), plug them into the inverse formula: \( A^{-1} = \frac{1}{1 \cdot 1 - 2 \cdot 2} \begin{bmatrix} 1 & -2 \ -2 & 1 \end{bmatrix} \). Simplify: \( A^{-1} = \frac{1}{-3} \begin{bmatrix} 1 & -2 \ -2 & 1 \end{bmatrix} \).
04
Simplify the inverse matrix
Multiply each element of the matrix by the scalar \( \frac{1}{-3} \) to get the inverse: \( A^{-1} = \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Determinant
The determinant is a special number that can be calculated from a square matrix. It is a crucial value in matrix algebra because it helps to determine various properties of the matrix. For example, the determinant indicates whether a matrix has an inverse or not. In the case of a 2x2 matrix, the determinant is calculated using the formula:
\(det(A) = (1 \cdot 1) - (2 \cdot 2) = 1 - 4 = -3\).
Since the determinant is not zero, this matrix does indeed have an inverse.
- For a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is \( ad - bc \).
\(det(A) = (1 \cdot 1) - (2 \cdot 2) = 1 - 4 = -3\).
Since the determinant is not zero, this matrix does indeed have an inverse.
2x2 Matrix
A 2x2 matrix is one of the simplest forms of matrices and is structured as a square with two rows and two columns. The layout is represented as follows:
Understanding a 2x2 matrix is foundational as it is easier to compute determinants and perform matrix operations compared to larger matrices. In our example matrix \( A = \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix} \), familiarizing with this simple structure makes complex operations more approachable.
- A matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) consists of elements \( a, b, c, \) and \( d \).
Understanding a 2x2 matrix is foundational as it is easier to compute determinants and perform matrix operations compared to larger matrices. In our example matrix \( A = \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix} \), familiarizing with this simple structure makes complex operations more approachable.
Matrix Algebra
Matrix algebra is a collection of operations that can be performed on matrices. These operations include addition, subtraction, multiplication, and finding inverses. Here are some key points:
- Matrices of the same size can be added or subtracted by adding or subtracting corresponding elements.
- Matrix multiplication involves multiplying rows by columns and is not commutative (i.e., \( AB eq BA \) in general).
- The identity matrix \( I \) plays a special role as the multiplicative identity where \( AI = IA = A \).
Matrix Inversion
Matrix inversion is a procedure in matrix algebra used to find a matrix that, when multiplied with the original matrix, results in the identity matrix. For a 2x2 matrix, this can be achieved when the determinant is non-zero. The formula for finding the inverse is:
First, calculate the determinant \( -3 \). Since it's not zero, the inverse exists. Next, apply the formula:
\( A^{-1} = \frac{1}{-3} \begin{bmatrix} 1 & -2 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \).
Matrix inversion is essential for solving linear equations and transforming geometric data in various applications.
- For \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse \( A^{-1} \) is \( \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \).
First, calculate the determinant \( -3 \). Since it's not zero, the inverse exists. Next, apply the formula:
\( A^{-1} = \frac{1}{-3} \begin{bmatrix} 1 & -2 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \).
Matrix inversion is essential for solving linear equations and transforming geometric data in various applications.