Question

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

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

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 \( \mathbb{Z}_5 \).

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:\[A^{-1} = \begin{bmatrix} 3 & 3 \ 4 & 3 \end{bmatrix} \mod 5.\]

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 \( \mathbb{Z}_7 \).

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:\[A^{-1} = \begin{bmatrix} 2 & 3 \ 2 & 1 \end{bmatrix} \mod 7.\]

Key concepts

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.

• 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.

In most cases, if the determinant is non-zero modulo a field, the matrix can have an inverse under that modulus.

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 \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \):
\[\text{det}(A) = ad - bc\]

• If \(\text{det}(A) = 0\), the matrix is not invertible.
• If \(\text{det}(A) eq 0\), the matrix is invertible.

In our original exercise, computing the determinant within the finite field ensures we can determine if the matrix is invertible under modular arithmetic rules. Remember, calculating the determinant correctly within the field modulus is crucial for determining matrix invertibility.

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 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.

Understanding whether a matrix is invertible in a specific modulus context ensures proper computation of the inverse, essential in solving mathematical problems involving matrices in modular arithmetic.

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:

• 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.

In matrix operations over finite fields, field theory is essential to ensure that determinants and inverse calculations adhere to these rules, enabling correct solutions. Thus, understanding field theory provides the broader mathematical context needed for matrix-related operations in modular arithmetic.