This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 304, Fall 2011 Linear Algebra MATH 304 Linear Algebra Lecture 3: Some applications of systems of linear equations. Matrix algebra. Homework assignment Homework assignment 1 (due Thursday, September 8): Section 1.1: 1b, 1d, 3a, 3d, 6e, 6h Section 1.2: 1a, 1c, 1e, 1g, 3a, 3c, 5i, 9, 22b Additionally! Section 1.2 : 6b, 6c How to solve a system of linear equations • Order the variables • Write down the augmented matrix of the system • Convert the matrix to row echelon form • Check for consistency • Convert the matrix to reduced row echelon form • Write down the system corresponding to the reduced row echelon form • Determine leading and free variables • Rewrite the system so that the leading variables are on the left while everything else is on the right • Write down the general solution in parametric form System with a parameter y + 3 z = 0 x + y − 2 z = 0 x + 2 y + az = 0 ( a ∈ R ) The system is homogeneous (all righthand sides are zeros). Therefore it is consistent ( x = y = z = 0 is a solution). Augmented matrix: 0 1 3 1 1 − 2 1 2 a Since the 1st row cannot serve as a pivotal one, we interchange it with the 2nd row: 0 1 3 1 1 − 2 1 2 a → 1 1 − 2 0 1 3 1 2 a Now we can start the elimination. First subtract the 1st row from the 3rd row: 1 1 − 2 0 1 3 1 2 a → 1 1 − 2 0 1 3 0 1 a + 2 The 2nd row is our new pivotal row. Subtract the 2nd row from the 3rd row: 1 1 − 2 0 1 3 0 1 a + 2 → 1 1 − 2 0 1 3 0 0 a − 1 At this point row reduction splits into two cases. Case 1: a negationslash = 1. In this case, multiply the 3rd row by ( a − 1) − 1 : 1 1 − 2 0 1 3 0 0 a − 1 → 1 1 − 2 1 3 1 The matrix is converted into row echelon form. We proceed towards reduced row echelon form. Subtract 3 times the 3rd row from the 2nd row: 1 1 − 2 0 1 3 0 0 1 → 1 1 − 2 0 1 0 0 1 Add 2 times the 3rd row to the 1st row: 1 1 − 2 0 1 0 0 1 → 1 1 0 0 1 0 0 0 1 Finally, subtract the 2nd row from the 1st row:...
View
Full Document
 Spring '08
 HOBBS
 Linear Algebra, Algebra, Linear Equations, Equations, Systems Of Linear Equations, Row echelon form, Row

Click to edit the document details