Chapter 3: Problem 63
Use the Gauss-Jordan method to find the inverse of the given matrix (if it exists). $$\left[\begin{array}{lll}1 & 5 & 0 \\\1 & 2 & 4 \\\3 & 6 & 1\end{array}\right] \text { over } \mathbb{Z}_{7}$$
Short Answer
Expert verified
The inverse matrix is \(\left[\begin{array}{ccc} 3 & 0 & 4 \\ 5 & 0 & 1 \\ 0 & 5 & 3 \end{array}\right]\).
Step by step solution
01
Augment the Matrix with Identity
First, write the given matrix along with the identity matrix on the right, creating an augmented matrix: \[\left[\begin{array}{ccc|ccc}1 & 5 & 0 & 1 & 0 & 0 \1 & 2 & 4 & 0 & 1 & 0 \3 & 6 & 1 & 0 & 0 & 1\end{array}\right]\] We will perform row operations to convert the left side into the identity matrix and transform the right side into the inverse.
02
Make Leading 1 in Row 1, Column 1
The already present 1 in position (1,1) serves as our pivot. No need for changes here. The row is \([1, 5, 0 | 1, 0, 0]\).
03
Eliminate Below Pivot using Row 1
Using \( R_2 = R_2 - R_1 \) and \( R_3 = R_3 - 3R_1 \), we get:\[\left[ \begin{array}{ccc|ccc} 1 & 5 & 0 & 1 & 0 & 0 \0 & -3 & 4 & -1 & 1 & 0 \0 & -9 & 1 & -3 & 0 & 1 \end{array} \right]\]Working in \(\mathbb{Z}_7\), we'll transform the negative numbers into positive numbers by adding 7 to each negative entry: \[\left[ \begin{array}{ccc|ccc} 1 & 5 & 0 & 1 & 0 & 0 \0 & 4 & 4 & 6 & 1 & 0 \0 & 5 & 1 & 4 & 0 & 1 \end{array} \right]\]
04
Make Leading 1 in Row 2, Column 2
Multiply \( R_2 \) by 2 (the modular inverse of 4 in \( \mathbb{Z}_7 \)) to make a leading 1:\[\left[ \begin{array}{ccc|ccc} 1 & 5 & 0 & 1 & 0 & 0 \0 & 1 & 2 & 5 & 2 & 0 \0 & 5 & 1 & 4 & 0 & 1 \end{array} \right]\]
05
Eliminate Above and Below New Pivot
Using \( R_1 = R_1 - 5R_2 \) and \( R_3 = R_3 - 5R_2 \), we have:\[\left[ \begin{array}{ccc|ccc} 1 & 0 & -10 & -24 & -10 & 0 \0 & 1 & 2 & 5 & 2 & 0 \0 & 0 & -9 & -21 & -10 & 1 \end{array} \right]\]Convert negative elements using modulo 7:\[\left[ \begin{array}{ccc|ccc} 1 & 0 & 4 & 3 & 4 & 0 \0 & 1 & 2 & 5 & 2 & 0 \0 & 0 & 5 & 0 & 4 & 1 \end{array} \right]\]
06
Make Leading 1 in Row 3, Column 3
Multiply \( R_3 \) by 3 (inverse of 5 in \( \mathbb{Z}_7 \)):\[\left[ \begin{array}{ccc|ccc} 1 & 0 & 4 & 3 & 4 & 0 \0 & 1 & 2 & 5 & 2 & 0 \0 & 0 & 1 & 0 & 5 & 3 \end{array} \right]\]
07
Eliminate Above Pivots in Column 3
Calculate \( R_1 = R_1 - 4R_3 \) and \( R_2 = R_2 - 2R_3 \) to achieve the identity matrix on the left side:\[\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 3 & 0 & 4 \0 & 1 & 0 & 5 & 0 & 1 \0 & 0 & 1 & 0 & 5 & 3 \end{array} \right]\]When done in \( \mathbb{Z}_7 \), the matrix remains the same because each operation considered the modulo 7 transformations.
08
Extract the Inverse Matrix
The inverse matrix is extracted from the right side of the augmented matrix:\[\begin{array}{ccc}3 & 0 & 4 \5 & 0 & 1 \0 & 5 & 3\end{array}\]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Matrix
An inverse matrix is like a magic key that unlocks the solution to a matrix equation. Imagine you have a matrix \( A \) and you want to find another matrix \( A^{-1} \) such that when they multiply together, the result is the identity matrix \( I \):\[A \times A^{-1} = I\]To find the inverse using the Gauss-Jordan method, you augment the given matrix with the identity matrix and perform row operations until the original matrix becomes the identity matrix itself.
- If you succeed, the right side of the augmented matrix will have transformed into the inverse matrix.
- If you cannot make the original matrix into the identity matrix, an inverse does not exist.
Modular Arithmetic
Modular arithmetic is like a circular clock for numbers. Instead of just adding and subtracting numbers normally, you wrap around once you reach a certain number, called the modulus. In this exercise, we're using \( \mathbb{Z}_7 \), which means everything is calculated modulo 7.In simpler terms, this means:
- Add or subtract numbers as usual, but if the result is 7 or more, you subtract 7 to "wrap it around".
- If the result is negative, add 7 until it becomes positive.
- All calculations during the row operations need to respect this modulo rule.
- This can change the signs and values of the numbers making sure they stay within the range 0 to 6.
Augmented Matrix
An augmented matrix is like a two-for-one deal in mathematics. It combines two matrices into one to help perform transformations easily.Here’s how it works:
- The given matrix (let's call it matrix \( A \)) is on the left side.
- The identity matrix \( I \) is placed on the right side.
- It simplifies the process of finding the inverse dramatically.
- Allows for systematic application of row operations that are easy to track and execute.