2. Prove that a 2 x 2 matrix A with entries in Z26 is invertible if and only if gcd(det(A), 26) = 1.
2. Prove that a 2 x 2 matrix A with entries in Z26 is invertible if and only if gcd(det(A), 26) = 1.
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.5: Iterative Methods For Computing Eigenvalues
Problem 51EQ
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![### Problem 2:
**Objective:**
Prove that a 2 × 2 matrix \( A \) with entries in \( \mathbb{Z}_{26} \) is invertible if and only if \( \gcd(\text{det}(A), 26) = 1 \).
**Detailed Steps:**
1. **Define the Matrix \( A \):**
Let \( A \) be a \( 2 \times 2 \) matrix with entries in \( \mathbb{Z}_{26} \). This means that each element of \( A \) is an integer between 0 and 25 (inclusive).
2. **Determinant Condition:**
The determinant of \( A \), denoted as \( \text{det}(A) \), is calculated as follows for a general 2 × 2 matrix:
\[
A = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\]
Then,
\[
\text{det}(A) = ad - bc
\]
3. **Invertibility in \( \mathbb{Z}_{26} \):**
A matrix \( A \) is considered invertible if there exists another matrix \( B \) such that \( AB = BA = I \) where \( I \) is the identity matrix. This condition must hold in the modular arithmetic of \( \mathbb{Z}_{26} \).
4. **Greatest Common Divisor (GCD):**
The matrix \( A \) is invertible in \( \mathbb{Z}_{26} \) if and only if the determinant \( \text{det}(A) \) is coprime with 26, i.e., \( \gcd(\text{det}(A), 26) = 1 \).
**Proof Outline:**
- If \( \gcd(\text{det}(A), 26) = 1 \), it implies that the determinant \( \text{det}(A) \) has a multiplicative inverse in \( \mathbb{Z}_{26} \).
- This multiplicative inverse is crucial for constructing the inverse matrix \( A^{-1} \), as each element of \( A^{-1} \) involves the determinant's multiplicative inverse.
- Conversely, if \( \gcd(\text{det}(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbb66fa7-7c22-4982-a22f-aaed542f65b3%2F2c03bf72-7411-4354-89f0-ef137c22bd29%2F4uw1thh_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 2:
**Objective:**
Prove that a 2 × 2 matrix \( A \) with entries in \( \mathbb{Z}_{26} \) is invertible if and only if \( \gcd(\text{det}(A), 26) = 1 \).
**Detailed Steps:**
1. **Define the Matrix \( A \):**
Let \( A \) be a \( 2 \times 2 \) matrix with entries in \( \mathbb{Z}_{26} \). This means that each element of \( A \) is an integer between 0 and 25 (inclusive).
2. **Determinant Condition:**
The determinant of \( A \), denoted as \( \text{det}(A) \), is calculated as follows for a general 2 × 2 matrix:
\[
A = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\]
Then,
\[
\text{det}(A) = ad - bc
\]
3. **Invertibility in \( \mathbb{Z}_{26} \):**
A matrix \( A \) is considered invertible if there exists another matrix \( B \) such that \( AB = BA = I \) where \( I \) is the identity matrix. This condition must hold in the modular arithmetic of \( \mathbb{Z}_{26} \).
4. **Greatest Common Divisor (GCD):**
The matrix \( A \) is invertible in \( \mathbb{Z}_{26} \) if and only if the determinant \( \text{det}(A) \) is coprime with 26, i.e., \( \gcd(\text{det}(A), 26) = 1 \).
**Proof Outline:**
- If \( \gcd(\text{det}(A), 26) = 1 \), it implies that the determinant \( \text{det}(A) \) has a multiplicative inverse in \( \mathbb{Z}_{26} \).
- This multiplicative inverse is crucial for constructing the inverse matrix \( A^{-1} \), as each element of \( A^{-1} \) involves the determinant's multiplicative inverse.
- Conversely, if \( \gcd(\text{det}(
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