Problem 26 - Algebra: Pure and Applied

Question

In the special linear group \(S L\left(2, \mathbb{Z}_{10}\right),\) let \(A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\) (a) Calculate \(A^{3}\) and \(A^{11}\). (b) Find the order of \(A\).

Step-by-Step Solution

Answer

\( A^3 = \begin{bmatrix} 1 & 6 \\ 0 & 1 \end{bmatrix} \), \( A^{11} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \); order of \( A \) is 5.

Step 1: Define the Matrix and its Properties

We have the matrix \( A \) defined as \( A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \) in the group \( SL(2, \mathbb{Z}_{10}) \). In this group, matrix multiplication and addition are performed modulo 10.

Step 2: Calculate \( A^2 \)

To find \( A^3 \), we first need \( A^2 \). Multiply \( A \) by itself and compute each element modulo 10: \[ A^2 = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} (1 \cdot 1 + 2 \cdot 0) & (1 \cdot 2 + 2 \cdot 1) \ (0 \cdot 1 + 1 \cdot 0) & (0 \cdot 2 + 1 \cdot 1) \end{bmatrix} = \begin{bmatrix} 1 & 4 \ 0 & 1 \end{bmatrix} \]

Step 3: Calculate \( A^3 \)

Use the result of \( A^2 \) to find \( A^3 \) by multiplying \( A^2 \) by \( A \): \[ A^3 = \begin{bmatrix} 1 & 4 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} (1 \cdot 1 + 4 \cdot 0) & (1 \cdot 2 + 4 \cdot 1) \ (0 \cdot 1 + 1 \cdot 0) & (0 \cdot 2 + 1 \cdot 1) \end{bmatrix} = \begin{bmatrix} 1 & 6 \ 0 & 1 \end{bmatrix} \]

Step 4: Calculate \( A^4 \) and Further Powers

Since we will also need \( A^{11} \), let's check if a pattern emerges by calculating successive powers, starting with \( A^4 \): \[ A^4 = A^3 \cdot A = \begin{bmatrix} 1 & 6 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 8 \ 0 & 1 \end{bmatrix} \] Continuing this pattern, you'll find: \( A^5 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \), the identity matrix. Thus \( A^6 = A \) again.

Step 5: Calculate \( A^{11} \)

Knowing \( A^5 = I \), the identity, we see that \( A^{11} = A^5 \cdot A^5 \cdot A = I \cdot A = A \). So, \( A^{11} = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \).

Step 6: Determine the Order of \( A \)

The order of a matrix in a group is the smallest positive integer \( m \) such that \( A^m = I \). From previous calculations, \( A^5 = I \), so the order of \( A \) is 5.

Key Concepts

Special Linear Group

The Special Linear Group, denoted as \( SL(n, R) \), is a fascinating area of linear algebra. It's primarily concerned with matrices that have a determinant equal to 1. This ensures that the transformations they represent are volume-preserving. In this particular exercise, we are looking at \( SL(2, \mathbb{Z}_{10}) \). Here, the matrices are 2x2 and behave under modulo 10 arithmetic.

In simple terms, all the matrices in this group are invertible. More importantly, they have the special property of being of determinant 1, crucial for maintaining the structure of the group. This specific exercise uses matrices that belong to a "modular" version of \( SL \), hence we're working modulo 10, a common practice when dealing with finite fields. Understanding how these groups operate can be key in many fields, like coding theory or cryptography.

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra. It is how we combine matrices to find new results. The process involves taking rows from the first matrix and columns from the second. You multiply corresponding elements and then sum them up.

It's important to remember that matrix multiplication is not commutative, meaning \( A \times B eq B \times A \) generally. In this exercise, multiplying the matrix \( A \) by itself is necessary to get powers like \( A^2 \) and \( A^3 \). Each entry of the product matrix is calculated by performing these multiplications and additions, all under modulo 10, which slightly alters the arithmetic involved.

• For example, to find \( (A^2)_{11} \), you calculate \( 1 \times 1 + 2 \times 0 = 1 \).
• Modulo 10 means any result above 10 will loop back, like \( 12 \) becoming \( 2 \).

Matrix Powers

The concept of raising a matrix to a power is similar to how we raise numbers to a power. For matrices, we multiply the matrix by itself when raising it to a higher power. To calculate \( A^{11} \) and \( A^3 \) in this exercise, you must sequentially multiply the matrix by itself.

Using previous powers can simplify this process. For instance, calculating \( A^3 = A \times A^2 \) uses a previously calculated \( A^2 \). By systematically building up to the desired power, you avoid repeated multiplication steps. It is especially efficient in this exercise since past steps like calculating \( A^5 \) show the resulting identity matrix. Recognizing these patterns means less computation and comprehension of fundamental matrix structures.

Modular Arithmetic

Modular arithmetic is a cornerstone in number theory and plays a key role in this exercise. When we operate with matrices in \( SL \(2, \mathbb{Z}_{10} \)\), every multiplication and addition is done modulo 10. This effectively means we only consider the remainder when dividing by 10.

This system creates a 'wrap-around' effect on numbers, making it particularly useful in titles like cryptography and computer science. For instance, in our example, when calculating matrix powers, \( 12 \) would convert to \( 2 \) because \( 12 \mod 10 = 2 \).

• In essence, modular arithmetic helps simplify calculations and can reveal cyclic patterns like periodic behaviors in matrix powers.
• Recognizing these dynamics is vital for understanding the behavior of such matrices and their applications across science and engineering.