HW2sol - (Partial) Solutions to Homework 2 Problem 4.1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (Partial) Solutions to Homework 2 Problem 4.1 Write down examples of augmented matrices corresponding to each of the five types of solution sets for systems of equations with three unknowns. (no solution, one solution, a line of solutions, a plane of solutions, all of R 3 .) Answer: The easiest way to write down these systems is using reduced row echelon form matrices (or even just row echelon form). Why is that? Remember, the number of free variables and pivot variables has something to do with the solution set. Let’s look at an example to figure out what they have to do with each other. Example: Let M = 1 0 1 0 1 0 0 0 0 , let V = 5 2 and let’s look for solutions to the system MX = V . Our matrix M is in reduced row echelon form, so it’s easy to solve. We have one free variable, corresponding to column 2. Let’s say our variables are x 1 ,x 2 , and x 3 , so that x 1 and x 3 are pivot variables, and x 2 is a free variable. Then to write the solution set, we give x 2 an arbitrary value, say x 2 = λ , and then we can write x 1 and x 3 in terms of λ (although in this case, x 3 does not need to be written in terms of λ , we know in fact that x 3 = 2). So any solution has x 3 = 2, x 2 = λ , and x 1 = 5- λ . In vector form, this can be written as: { 5- λ λ 2 } = { 5 2 + λ - 1 1 } What does this set look like, i.e., what does it represent? There is one degree of freedom, that is, we can make λ whatever we want it to be. Think about it – if we have a vector v , and we’re allowed to have any multiple of v , then that looks like a line. So this solution set looks like a line. You can do examples with different numbers of free variables to get a better idea of how they relate to geometric interpretation, but it turns out that the number of free variables is the dimension of the solution set. So in three variables, we have the following correspondence. If there is at least one solution , then • 3 pivot variables, no free variables ↔ one unique solution • 2 pivot variables, 1 free variable ↔ line of solutions = one dimension of solutions • 1 pivot variable, 2 free variables ↔ plane of solutions = two dimensions of solutions • 0 pivot variables, 3 free variables ↔ three dimensions of solutions But wait, what does 3 free variables mean? If we only have a 3x3 matrix, how do we have three free variables? Remember what a free variable is – it’show do we have three free variables?...
View Full Document

This note was uploaded on 03/23/2010 for the course MAT 022A taught by Professor Pon during the Spring '10 term at UC Davis.

Page1 / 4

HW2sol - (Partial) Solutions to Homework 2 Problem 4.1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online