Chapter 8: Problem 6
In each case show that the matrix \(A\) is invertible over the given field, and find \(A^{-1}\). a. \(A=\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right]\) over \(\mathbb{Z}_{5} .\) b. \(A=\left[\begin{array}{ll}5 & 6 \\ 4 & 3\end{array}\right]\) over \(\mathbb{Z}_{7}\).
Short Answer
Expert verified
a. Invertible, \( A^{-1} = \begin{bmatrix} 3 & 3 \\ 4 & 3 \end{bmatrix} \mod 5 \).
b. Invertible, \( A^{-1} = \begin{bmatrix} 2 & 3 \\ 2 & 1 \end{bmatrix} \mod 7 \).
Step by step solution
01
Determine if Matrix A is Invertible Over \( \mathbb{Z}_5 \)
Matrix \( A \) is invertible if its determinant is non-zero in the given field. For matrix \( A = \begin{bmatrix} 1 & 4 \ 2 & 1 \end{bmatrix} \), compute the determinant:\[\text{det}(A) = 1 \times 1 - 4 \times 2 = 1 - 8 = -7. \]In \( \mathbb{Z}_5 \), \(-7 \equiv 3 \mod 5\), which is non-zero. Therefore, matrix \( A \) is invertible over \( \mathbb{Z}_5 \).
02
Find the Inverse of Matrix A over \( \mathbb{Z}_5 \)
To find \( A^{-1} \), we use the formula for the inverse of a 2x2 matrix:\[A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \quad \text{where} \quad A = \begin{bmatrix} a & b \ c & d \end{bmatrix}.\]Thus,\[A^{-1} = \frac{1}{3} \begin{bmatrix} 1 & -4 \ -2 & 1 \end{bmatrix} \equiv 3 \begin{bmatrix} 1 & 1 \ 3 & 1 \end{bmatrix} \mod 5.\]Calculate each element modulo 5:\[A^{-1} = \begin{bmatrix} 3 & 3 \ 4 & 3 \end{bmatrix} \mod 5.\]
03
Determine if Matrix A is Invertible Over \( \mathbb{Z}_7 \)
For matrix \( A = \begin{bmatrix} 5 & 6 \ 4 & 3 \end{bmatrix} \), compute the determinant:\[\text{det}(A) = 5 \times 3 - 6 \times 4 = 15 - 24 = -9.\]In \( \mathbb{Z}_7 \), \(-9 \equiv 5 \mod 7\), which is non-zero. Therefore, matrix \( A \) is invertible over \( \mathbb{Z}_7 \).
04
Find the Inverse of Matrix A over \( \mathbb{Z}_7 \)
To find \( A^{-1} \), use the formula for the inverse of the 2x2 matrix:\[A^{-1} = \frac{1}{5} \begin{bmatrix} 3 & -6 \ -4 & 5 \end{bmatrix} \equiv 3 \begin{bmatrix} 3 & 1 \ 3 & 5 \end{bmatrix} \mod 7.\]Compute each element modulo 7:\[A^{-1} = \begin{bmatrix} 2 & 3 \ 2 & 1 \end{bmatrix} \mod 7.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Modular Arithmetic
Modular arithmetic is a system of arithmetic for integers, where numbers wrap around after reaching a certain value, known as the modulus. Think of it like the face of a clock, where after reaching 12, you circle back to 1.
For example, when we say 11 is equivalent to 1 modulo 10, we mean when 11 is divided by 10, the remainder is 1. In our scenario, calculating determinants within a finite field, like \(\mathbb{Z}_5\) or \(\mathbb{Z}_7\), relies heavily on this concept.
For example, when we say 11 is equivalent to 1 modulo 10, we mean when 11 is divided by 10, the remainder is 1. In our scenario, calculating determinants within a finite field, like \(\mathbb{Z}_5\) or \(\mathbb{Z}_7\), relies heavily on this concept.
- When performing operations like addition or multiplication, you reduce your results "mod" the base.
- It simplifies calculations by constraining numbers within a fixed range.
- Understanding modular arithmetic is crucial for solving problems related to matrix inversion over finite fields.
Determinant
The determinant is a special number that can be computed from a square matrix. It provides important information about the matrix, including whether the matrix is invertible.
For a 2x2 matrix, the determinant is calculated as follows for matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \):
\[\text{det}(A) = ad - bc\]
For a 2x2 matrix, the determinant is calculated as follows for matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \):
\[\text{det}(A) = ad - bc\]
- If \(\text{det}(A) = 0\), the matrix is not invertible.
- If \(\text{det}(A) eq 0\), the matrix is invertible.
Invertible Matrix
An invertible matrix, also known as a non-singular or non-degenerate matrix, is one that possesses an inverse. This inverse matrix, denoted as \(A^{-1}\), can only exist if the determinant of the matrix is non-zero.
The inverse of a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is calculated using the formula:\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]
The inverse of a matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\) is calculated using the formula:\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]
- The presence of the determinant in the denominator indicates the critical condition for invertibility.
- If the determinant is reduced to zero within the field, an inverse cannot be calculated.
Field Theory
Field theory is a branch of mathematics dealing with fields, which are algebraic structures equipped with operations of addition, subtraction, multiplication, and division. Fields provide a framework in which concepts like modular arithmetic and matrix inversion function smoothly.
In the context of the exercise, the field \(\mathbb{Z}_5\) (integers modulo 5) or \(\mathbb{Z}_7\) (integers modulo 7) sets the rules for calculations. Here are some properties:
In the context of the exercise, the field \(\mathbb{Z}_5\) (integers modulo 5) or \(\mathbb{Z}_7\) (integers modulo 7) sets the rules for calculations. Here are some properties:
- Every non-zero element has a multiplicative inverse within the field, crucial for matrix inversion.
- The operations in a field comply with specific rules ensuring consistent results, such as commutativity and associativity.