Chapter 7: Problem 69
Find the inverse of each matrix, if it exists. \(\left[\begin{array}{ll}{8} & {6} \\ {7} & {5}\end{array}\right]\)
Short Answer
Expert verified
The inverse of the matrix is \( \begin{pmatrix} -\frac{5}{2} & 3 \\ \frac{7}{2} & -4 \end{pmatrix} \).
Step by step solution
01
Determine the Determinant
To find the inverse of a 2x2 matrix \[ A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \] we first need to compute the determinant using the formula: \[ \text{det}(A) = ad - bc \] For the given matrix \[ \begin{pmatrix} 8 & 6 \ 7 & 5 \end{pmatrix} \] we calculate: \[ \text{det}(A) = (8)(5) - (6)(7) = 40 - 42 = -2 \] Since the determinant is not zero, the matrix has an inverse.
02
Apply the Inverse Formula for a 2x2 Matrix
For a 2x2 matrix \[ A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \] with a non-zero determinant, its inverse is given by: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \] For our matrix, this becomes: \[ A^{-1} = \frac{1}{-2} \begin{pmatrix} 5 & -6 \ -7 & 8 \end{pmatrix} \]
03
Simplify the Inverse Matrix
We simplify the expression for the inverse by multiplying each element by \( \frac{1}{-2} \): \[ A^{-1} = \begin{pmatrix} 5 \times \frac{-1}{2} & -6 \times \frac{-1}{2} \ -7 \times \frac{-1}{2} & 8 \times \frac{-1}{2} \end{pmatrix} \] This results in: \[ A^{-1} = \begin{pmatrix} -\frac{5}{2} & 3 \ \frac{7}{2} & -4 \end{pmatrix} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Determinant of a Matrix
The determinant of a matrix is an essential concept in linear algebra. It is a scalar value that can provide critical insights into the properties of a matrix. In particular, for a 2x2 matrix, the determinant can help determine if the matrix is invertible. The determinant of a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is calculated as \( ad - bc \).
For the matrix \( \begin{pmatrix} 8 & 6 \ 7 & 5 \end{pmatrix} \), the determinant is computed as \( 8 \times 5 - 6 \times 7 = 40 - 42 = -2 \).
A determinant of zero indicates a singular matrix, meaning it cannot be inverted. In this case, as the determinant is \(-2\), which is non-zero, the matrix is invertible, allowing us to proceed with finding its inverse.
For the matrix \( \begin{pmatrix} 8 & 6 \ 7 & 5 \end{pmatrix} \), the determinant is computed as \( 8 \times 5 - 6 \times 7 = 40 - 42 = -2 \).
A determinant of zero indicates a singular matrix, meaning it cannot be inverted. In this case, as the determinant is \(-2\), which is non-zero, the matrix is invertible, allowing us to proceed with finding its inverse.
2x2 Matrix
A 2x2 matrix is the simplest type of square matrix, consisting of two rows and two columns. Such matrices are commonly represented as \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \).
This layout is fundamental in matrix operations, providing a straightforward framework to explore concepts like determinants and inverse matrices without extensive computational effort. Some key characteristics of 2x2 matrices include:
This layout is fundamental in matrix operations, providing a straightforward framework to explore concepts like determinants and inverse matrices without extensive computational effort. Some key characteristics of 2x2 matrices include:
- The order of the matrix (number of rows and columns): 2x2 signifies two rows and two columns.
- Suitability for both basic algebraic operations and more complex applications like eigenvalues in physics.
- Serving as building blocks for larger matrices by extension.
Inverse Matrix Formula
The formula for finding the inverse of a 2x2 matrix is a key tool when working with systems of equations or transformations in spaces such as graphics.
For a matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) with determinant \( \text{det}(A) eq 0 \), the inverse is determined using:\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]Using our example matrix \( A = \begin{pmatrix} 8 & 6 \ 7 & 5 \end{pmatrix} \), with determinant \(-2\), the inverse is:\[ A^{-1} = \frac{1}{-2} \begin{pmatrix} 5 & -6 \ -7 & 8 \end{pmatrix} \]By multiplying each element in the cofactor matrix \( \begin{pmatrix} 5 & -6 \ -7 & 8 \end{pmatrix} \) by \( \frac{1}{-2} \), the inverse matrix becomes:\[ A^{-1} = \begin{pmatrix} -\frac{5}{2} & 3 \ \frac{7}{2} & -4 \end{pmatrix} \]This inverse matrix can then be used in further computations like solving linear systems.
For a matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) with determinant \( \text{det}(A) eq 0 \), the inverse is determined using:\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]Using our example matrix \( A = \begin{pmatrix} 8 & 6 \ 7 & 5 \end{pmatrix} \), with determinant \(-2\), the inverse is:\[ A^{-1} = \frac{1}{-2} \begin{pmatrix} 5 & -6 \ -7 & 8 \end{pmatrix} \]By multiplying each element in the cofactor matrix \( \begin{pmatrix} 5 & -6 \ -7 & 8 \end{pmatrix} \) by \( \frac{1}{-2} \), the inverse matrix becomes:\[ A^{-1} = \begin{pmatrix} -\frac{5}{2} & 3 \ \frac{7}{2} & -4 \end{pmatrix} \]This inverse matrix can then be used in further computations like solving linear systems.
Matrix Multiplication
Matrix multiplication is a fundamental operation that combines matrices to produce a new matrix. It involves multiplying rows by columns and summing up the products.
To understand how the inverse matrix can be utilized, consider that when you multiply a matrix by its inverse, it results in the identity matrix. Let's explore the matrix multiplication concept further:
To understand how the inverse matrix can be utilized, consider that when you multiply a matrix by its inverse, it results in the identity matrix. Let's explore the matrix multiplication concept further:
- Given two matrices \( A \) and \( B \), their product \( AB \) is computed by taking each row of \( A \) and each column of \( B \) and calculating the dot product.
- This operation is not commutative, meaning \( AB eq BA \) in general.
- For a 2x2 matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\) and its inverse \( A^{-1} = \begin{pmatrix} -\frac{5}{2} & 3 \ \frac{7}{2} & -4 \end{pmatrix} \), multiplying \( A \times A^{-1} \) yields the identity matrix \( \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \).