MANE 3351
Lecture 23
Classroom Management
Agenda
- Gauss Jordan Elimination, Partial Pivoting, Matrix Inversion using Gauss Jordan
- Homework 7 (assigned 11/20/24, due 12/2/24 - no late submissions)
- Lab today for those who did not finish
Resources
Handouts
- Lecture 23 slides
- Lecture 23 slides marked
Calendar
| Date | Lecture Topic | Lab Topic |
|---|---|---|
| 11/18 | Lecture 22 - Row Echelon Form | Lab 10 - Row Echelon Form/ChatGPT |
| 11/20 | Lecture 23 - Gaussian Elimination | |
| 11/25 | ||
| 11/27 | ||
| 12/2 | ||
| 12/4 | Review | |
| 12/9 | Final exam 1:15 - 3:00 pm |
Assignments
- Homework 6 (assigned 11/13, due 11/20)
- Lab 10 (assigned 11/18, due 11/25 (before lab))
- Homework 7 (assigned 11/20, due 12/2 - no late submissions)
Gauss-Jordan Elimination
- A Step-by-Step Method for Solving Linear Systems
- An extension of row-echelon form
Introduction to Gauss-Jordan Elimination
- Gauss-Jordan elimination is a method to solve systems of linear equations.
- Goal: Transform the matrix into reduced row-echelon form (RREF).
- Key feature: The solution is read directly from the matrix.
- Distinguish from Row Echelon Form
Key Steps in Gauss-Jordan Elimination
- Row Operations:
- Swap two rows.
- Multiply a row by a nonzero scalar.
- Add or subtract multiples of one row to another.
- Transform to RREF:
- Each leading entry is 1 (pivot).
- Pivots are the only nonzero entries in their column.
Example Problem
Solve the system:
\[
\begin{aligned}
x + y + z &= 6 \\
2x + 3y + z &= 14 \\
y + 2z &= 8
\end{aligned}
\]
Step-by-Step Solution
- Convert the system to an augmented matrix.
- Use row operations to make the first pivot 1.
- Eliminate all other entries in the pivot column.
- Repeat for each subsequent pivot.
Final Result
- Matrix in reduced row-echelon form (RREF):
\[
\begin{bmatrix}
1 & 0 & 0 & a \\
0 & 1 & 0 & b \\
0 & 0 & 1 & c
\end{bmatrix}
\]
- Solution: \(x = a\), \(y = b\), \(z = c\).
Applications of Gauss-Jordan Elimination
- Solving systems of linear equations.
- Finding inverse matrices.
- Used in engineering, physics, computer science, etc.
Summary of Gauss-Jordan Elimination
- Gauss-Jordan elimination simplifies solving linear systems.
- Three row operations ensure clarity and consistency.
- Matrix RREF provides direct solutions.
Partial Pivoting
- A pivot is the leading non-zero element in a row used to simplify the matrix.
- Pivoting ensures numerical stability during elimination.
-
Types of pivoting:
-
Partial Pivoting: Swap rows to place the largest absolute value in the pivot position.
- Complete Pivoting: Reorder rows and columns to place the largest value in the pivot position.
Why Pivoting is Necessary
- Avoid division by small numbers (reduce round-off errors).
- Improve accuracy in solving systems.
- Ensure numerical stability for ill-conditioned matrices.
Example Problem
Solve the system of equations:
\[
\begin{aligned}
0.0001x + y + z &= 1 \\
x + y + z &= 6 \\
2x + y + 10z &= 20
\end{aligned}
\]
Form the Augmented Matrix
Represent the system as an augmented matrix:
\[
\begin{bmatrix}
0.0001 & 1 & 1 & | & 1 \\
1 & 1 & 1 & | & 6 \\
2 & 1 & 10 & | & 20
\end{bmatrix}
\]
- The first three columns are the coefficients of the variables \(x, y, z\).
- The fourth column is the constants on the right-hand side.
Perform Partial Pivoting
- Compare the absolute values in the first column:
\[
\text{Column 1: } |0.0001|, |1|, |2|
\]
Largest value: 2.
- Swap row 1 with row 3:
\[
\begin{bmatrix}
2 & 1 & 10 & | & 20 \\
1 & 1 & 1 & | & 6 \\
0.0001 & 1 & 1 & | & 1
\end{bmatrix}
\]
Elimination Step-by-Step
- Eliminate entries below the pivot in the first column:
\[
R_2 \to R_2 - \frac{1}{2} R_1, \quad R_3 \to R_3 - \frac{0.0001}{2} R_1
\]
- Resulting matrix:
\[
\begin{bmatrix}
2 & 1 & 10 & | & 20 \\
0 & 0.5 & -4 & | & -4 \\
0 & 0.99995 & 0.9995 & | & 0.9995
\end{bmatrix}
\]
- Repeat pivoting and elimination for columns 2 and 3.
Slide 8: Complete Solution
- After performing Gauss-Jordan elimination with pivoting:
\[
\begin{bmatrix}
1 & 0 & 0 & | & 2 \\
0 & 1 & 0 & | & 1 \\
0 & 0 & 1 & | & 1
\end{bmatrix}
\]
- Solution:
\[
x = 2, \quad y = 1, \quad z = 1
\]
Slide 9: Challenges and Limitations
- Pivoting increases computational complexity for larger matrices.
- Requires additional bookkeeping for row swaps.
- May not guarantee exact accuracy for ill-conditioned matrices.
Applications of Pivoting
- Engineering: Solving large systems of linear equations.
- Computer science: Numerical simulations and optimizations.
- Physics: Stability in solving differential equations.
Summary
- Pivoting improves the accuracy and stability of Gauss-Jordan elimination.
- Partial pivoting is a practical and efficient choice.
- Numerical stability is crucial for solving large or complex systems.
Using Gauss-Jordan Elimination to find Inverse Matrix
- To find the inverse of a matrix \(A\), we augment it with the identity matrix \(I\).
- Perform Gauss-Jordan elimination to transform \(A\) into \(I\), turning \(I\) into \(A^{-1}\).
- Works only if the matrix \(A\) is invertible (determinant \(\neq 0\)).
Key Steps
- Write the augmented matrix \([A | I]\).
- Use row operations to transform \(A\) into the identity matrix \(I\).
- The resulting augmented matrix will be \([I | A^{-1}]\).
Example Problem
Find the inverse of:
\[
A = \begin{bmatrix}
2 & 1 \\
5 & 3
\end{bmatrix}
\]
- Form augmented matrix:
\[
\begin{bmatrix}
2 & 1 & | & 1 & 0 \\
5 & 3 & | & 0 & 1
\end{bmatrix}
\]
Step-by-Step Solution
- Scale the first row to make the pivot 1:
\[
R_1 \to R_1 / 2
\]
Result:
\[
\begin{bmatrix}
1 & 0.5 & | & 0.5 & 0 \\
5 & 3 & | & 0 & 1
\end{bmatrix}
\]
- Eliminate the first column of \(R_2\):
\[
R_2 \to R_2 - 5 \cdot R_1
\]
Result:
\[
\begin{bmatrix}
1 & 0.5 & | & 0.5 & 0 \\
0 & 0.5 & | & -2.5 & 1
\end{bmatrix}
\]
- Scale the second row to make the pivot 1:
\[
R_2 \to R_2 / 0.5
\]
Result:
\[
\begin{bmatrix}
1 & 0.5 & | & 0.5 & 0 \\
0 & 1 & | & -5 & 2
\end{bmatrix}
\]
- Eliminate the second column of \(R_1\):
\[
R_1 \to R_1 - 0.5 \cdot R_2
\]
Result:
\[
\begin{bmatrix}
1 & 0 & | & 3 & -1 \\
0 & 1 & | & -5 & 2
\end{bmatrix}
\]
Final Result
- The final augmented matrix is:
\[
\begin{bmatrix}
1 & 0 & | & 3 & -1 \\
0 & 1 & | & -5 & 2
\end{bmatrix}
\]
- Inverse of \(A\):
\[
A^{-1} = \begin{bmatrix}
3 & -1 \\
-5 & 2
\end{bmatrix}
\]
Conditions for Invertibility
-
A matrix \(A\) is invertible if:
-
It is square (\(n \times n\)).
- Determinant of \(A \neq 0\).
- For \(2 \times 2\):
\[
\text{det}(A) = ad - bc \neq 0
\]
Summary
- Gauss-Jordan elimination transforms \(A\) into \(I\), revealing \(A^{-1}\).
- Method highlights the importance of row operations.
- Verifiable through matrix multiplication: \(A \cdot A^{-1} = I\).
Discussion of ChatGPT
- These slides were prepared using ChatGPT
- The partial pivoting slides contained an error that required a prompt to be modified
- There was one lab on Monday that created coded that gave a wrong answer