Section 1.5. Solution Sets of Linear Systems

Similar documents
Announcements Monday, September 18

Review for Chapter 1. Selected Topics

Linear Independence x

Section 1.2. Row Reduction and Echelon Forms

MA 242 LINEAR ALGEBRA C1, Solutions to First Midterm Exam

Linear Independence Reading: Lay 1.7

Linear Algebra MATH20F Midterm 1

Chapter 1. Vectors, Matrices, and Linear Spaces

Math 54 HW 4 solutions

Math 1553 Introduction to Linear Algebra. School of Mathematics Georgia Institute of Technology

Chapter 1: Systems of Linear Equations

Linear equations in linear algebra

Section Gaussian Elimination

Linear Equations in Linear Algebra

MAT 1302B Mathematical Methods II

1. Determine by inspection which of the following sets of vectors is linearly independent. 3 3.

Midterm 1 Review. Written by Victoria Kala SH 6432u Office Hours: R 12:30 1:30 pm Last updated 10/10/2015

It is the purpose of this chapter to establish procedures for efficiently determining the solution of a system of linear equations.

MATH 2331 Linear Algebra. Section 1.1 Systems of Linear Equations. Finding the solution to a set of two equations in two variables: Example 1: Solve:

MTH 464: Computational Linear Algebra

13. Systems of Linear Equations 1

MATH 152 Exam 1-Solutions 135 pts. Write your answers on separate paper. You do not need to copy the questions. Show your work!!!

Math 314H EXAM I. 1. (28 points) The row reduced echelon form of the augmented matrix for the system. is the matrix

Lecture 03. Math 22 Summer 2017 Section 2 June 26, 2017

1. TRUE or FALSE. 2. Find the complete solution set to the system:

MAT 1302B Mathematical Methods II

Chapter 5. Eigenvalues and Eigenvectors

Math 301 Test I. M. Randall Holmes. September 8, 2008

1 Last time: multiplying vectors matrices

(i) [7 points] Compute the determinant of the following matrix using cofactor expansion.

if b is a linear combination of u, v, w, i.e., if we can find scalars r, s, t so that ru + sv + tw = 0.

Lecture 6: Spanning Set & Linear Independency

Linear Equations in Linear Algebra

Announcements Wednesday, August 30

Section 3.3. Matrix Equations

Announcements Wednesday, August 30

Matrices and RRE Form

Math 2940: Prelim 1 Practice Solutions

Kernel and range. Definition: A homogeneous linear equation is an equation of the form A v = 0

Announcements Wednesday, November 01

Linear Algebra (wi1403lr) Lecture no.3

MAT 242 CHAPTER 4: SUBSPACES OF R n

1 Last time: inverses

Math 54. Selected Solutions for Week 5

Chapter 1: Linear Equations

2. Every linear system with the same number of equations as unknowns has a unique solution.

Linear Algebra Exam 1 Spring 2007

Announcements Wednesday, September 20

2018 Fall 2210Q Section 013 Midterm Exam I Solution

Week #4: Midterm 1 Review

Row Space, Column Space, and Nullspace

chapter 12 MORE MATRIX ALGEBRA 12.1 Systems of Linear Equations GOALS

Announcements Monday, September 17

MATH10212 Linear Algebra B Homework Week 4

Math 54 Homework 3 Solutions 9/

Solutions to Section 2.9 Homework Problems Problems 1 5, 7, 9, 10 15, (odd), and 38. S. F. Ellermeyer June 21, 2002

Math 3191 Applied Linear Algebra

Announcements Wednesday, November 01

Linear Algebra. Preliminary Lecture Notes

Section 4.5. Matrix Inverses

Chapter 1: Linear Equations

Announcements Wednesday, October 04

Span & Linear Independence (Pop Quiz)

Find the solution set of 2x 3y = 5. Answer: We solve for x = (5 + 3y)/2. Hence the solution space consists of all vectors of the form

1 Last time: determinants

Linear Algebra. Preliminary Lecture Notes

2018 Fall 2210Q Section 013 Midterm Exam II Solution

Review Notes for Linear Algebra True or False Last Updated: February 22, 2010

Multiple Choice Questions

Additional Problems for Midterm 1 Review

LECTURES 14/15: LINEAR INDEPENDENCE AND BASES

3 Fields, Elementary Matrices and Calculating Inverses

Math 2174: Practice Midterm 1

Announcements Monday, October 29

Section 1.8/1.9. Linear Transformations

Math 4A Notes. Written by Victoria Kala Last updated June 11, 2017

1 - Systems of Linear Equations

MATH 1553, SPRING 2018 SAMPLE MIDTERM 1: THROUGH SECTION 1.5

MATH 2331 Linear Algebra. Section 2.1 Matrix Operations. Definition: A : m n, B : n p. Example: Compute AB, if possible.

1 Last time: linear systems and row operations

MATH 2050 Assignment 6 Fall 2018 Due: Thursday, November 1. x + y + 2z = 2 x + y + z = c 4x + 2z = 2

Exam 2 Solutions. (a) Is W closed under addition? Why or why not? W is not closed under addition. For example,

Application of linear systems Linear Algebra with Computer Science Application

DEPARTMENT OF MATHEMATICS

Linear transformations

a 1n a 2n. a mn, a n = a 11 a 12 a 1j a 1n a 21 a 22 a 2j a m1 a m2 a mj a mn a 11 v 1 + a 12 v a 1n v n a 21 v 1 + a 22 v a 2n v n

Lecture 18: Section 4.3

Homework Set #8 Solutions

Review Solutions for Exam 1

Math 220: Summer Midterm 1 Questions

Chapter 2 Notes, Linear Algebra 5e Lay

(c)

Announcements Monday, September 25

Distances in R 3. Last time we figured out the (parametric) equation of a line and the (scalar) equation of a plane:

MTH 35, SPRING 2017 NIKOS APOSTOLAKIS

Work sheet / Things to know. Chapter 3

Contents. 1 Vectors, Lines and Planes 1. 2 Gaussian Elimination Matrices Vector Spaces and Subspaces 124

Lecture 4: Gaussian Elimination and Homogeneous Equations

MATH240: Linear Algebra Review for exam #1 6/10/2015 Page 1

web: HOMEWORK 1

Transcription:

Section 1.5 Solution Sets of Linear Systems

Plan For Today Today we will learn to describe and draw the solution set of an arbitrary system of linear equations Ax = b, using spans. Ax = b Recall: the solution set is the collection of all vectors x such that Ax = b is true.

Homogeneous Systems Everything is easier when b = 0, so we start with this case. Definition A system of linear equations of the form Ax = 0 is called homogeneous. These are linear equations where everything to the right of the = is zero. The opposite is: Definition A system of linear equations of the form Ax = b with b 0 is called nonhomogeneous or inhomogeneous. A homogeneous system always has the solution x = 0. This is called the trivial solution. The nonzero solutions are called nontrivial. Observation Ax = 0 has a nontrivial solution there is a free variable A has a column with no pivot.

Homogeneous Systems Example What is the solution set of Ax = 0, where 1 3 4 A = 2 1 2? 1 0 1 We know how to do this: first form an augmented matrix and row reduce. 1 3 4 0 2 1 2 0 row reduce 1 0 0 0 0 1 0 0. 1 0 1 0 0 0 1 0 The only solution is the trivial solution x = 0. Observation Since the last column (everything to the right of the =) was zero to begin, it will always stay zero! So it s not really necessary to write augmented matrices in the homogeneous case.

Homogeneous Systems Basic Example What is the solution set of Ax = 0, where ( ) 1 3 2 6 A = ( 1 3 2 6 row reduce equation )? ( ) 1 3 0 0 x 1 3x 2 = 0 parametric form { x 1 = 3x 2 x 2 = x 2 parametric vector form ( ) ( ) x1 3 x = = x 2. 1 This last equation is called the parametric vector form of the solution. It is obtained by listing equations for all the variables, in order, including the free ones, and making a vector equation. x 2

Homogeneous Systems Basic Example, continued What is the solution set of Ax = 0, where A = ( ) 3 Answer: x = x 2 for any x 1 2 in R. ( 1 3 2 6 )? {( )} 3 The solution set is Span. 1 Ax = 0 Note: one free variable means the solution set is a line in R 2 (2 = # variables = # columns).

Homogeneous Systems Example 1 What is the solution set of Ax = 0, where ( 1 1 ) 2 2 2 4 A = row reduce equation parametric form parametric vector form ( 1 1 2 2 2 4 )? ( 1 1 ) 2 0 0 0 x 1 x 2 + 2x 3 = 0 x 1 = x 2 2x 3 x 2 = x 2 x 3 = x 3 x 1 1 2 x = x 2 = x 2 1 + x 3 0. x 3 0 1

Homogeneous Systems Example 1, continued What is the solution set of Ax = 0, where A = ( 1 1 2 2 2 4 )? Answer: 1 2 Span 1, 0. 0 1 Ax = 0 [interactive] Note: two free variables means the solution set is a plane in R 3 (3 = # variables = # columns).

Homogeneous Systems Example 2 What is the solution set of Ax = 0, where A = 1 2 0 1 2 3 4 5 row reduce 1 0 8 7 0 1 4 3 2 4 0 2 0 0 0 0 equations { x1 8x 3 7x 4 = 0 x 2 + 4x 3 + 3x 4 = 0 parametric form parametric vector form x 1 = 8x 3 + 7x 4 x 2 = 4x 3 3x 4 x 3 = x 3 x 4 = x 4 x 1 8 7 x = x 2 x 3 = x3 4 1 + x4 3 0. x 4 0 1

Homogeneous Systems Example 2, continued What is the solution set of Ax = 0, where 1 2 0 1 A = 2 3 4 5? 2 4 0 2 Answer: 8 7 Span 4 1, 3 0. 0 1 [not pictured here] Note: two free variables means the solution set is a plane in R 4 (4 = # variables = # columns).

Parametric Vector Form Homogeneous systems Let A be an m n matrix. Suppose that the free variables in the homogeneous equation Ax = 0 are x i, x j, x k,... Then the solutions to Ax = 0 can be written in the form x = x i v i + x j v j + x k v k + for some vectors v i, v j, v k,... in R n, and any scalars x i, x j, x k,... The solution set is Span { v i, v j, v k,... }. The equation above is called the parametric vector form of the solution. It is obtained by listing equations for all the variables, in order, including the free ones, and making a vector equation.

Poll Poll How many solutions can there be to a homogeneous system with more equations than variables? A. 0 B. 1 C. The trivial solution is always a solution to a homogeneous system, so answer A is impossible. This matrix has only one solution to Ax = 0: 1 0 A = 0 1 0 0 This matrix has infinitely many solutions to Ax = 0: 1 1 A = 0 0 0 0

Nonhomogeneous Systems Basic Example (slightly modified) What is the solution set of Ax = b, where ( ) 1 3 A = and b = 2 6 ( 1 3 ) 3 2 6 6 row reduce equation ( ) 3? 6 ( 1 3 ) 3 0 0 0 x 1 3x 2 = 3 parametric form { x 1 = 3x 2 3 x 2 = x 2 + 0 parametric vector form ( ) ( ) x1 3 x = = x 2 + 1 The only difference from the homogeneous case is the constant vector p = ( ) 3 0. Note that p is itself a solution: take x 2 = 0. x 2 ( ) 3. 0

Nonhomogeneous Systems Basic Example (slightly modified) continued What is the solution set of Ax = b, where ( ) 1 3 A = and b = 2 6 ( ) 3? 6 ( ) ( ) 3 3 Answer: x = x 2 + for any x 1 0 2 in R. {( )} 3 This is a translate of Span : it is the parallel line through p = 1 ( ) 3. 0 p Ax = b Ax = 0 It can be written {( )} 3 Span + 1 ( ) 3. 0

Nonhomogeneous Systems Example 3 What is the solution set of Ax = b, where ( ) 1 1 2 A = 2 2 4 and b = ( 1 1 2 ) 1 2 2 4 2 row reduce ( ) 1? 2 ( 1 1 2 ) 1 0 0 0 0 equation x 1 x 2 + 2x 3 = 1 parametric form x 1 = x 2 2x 3 + 1 x 2 = x 2 x 3 = x 3 parametric vector form x 1 1 2 1 x = x 2 = x 2 1 + x 3 0 + 0. x 3 0 1 0

Homogeneous vs. Nonhomogeneous Systems Key Observation The set of solutions to Ax = b, if it is nonempty, is obtained by taking one specific or particular solution p to Ax = b, and adding all solutions to Ax = 0. Why? If Ap = b and Ax = 0, then so p + x is also a solution to Ax = b. A(p + x) = Ap + Ax = b + 0 = b, We know the solution set of Ax = 0 is a span. So the solution set of Ax = b is a translate of a span: it is parallel to a span. (Or it is empty.) Ax = b Ax = 0 This works for any specific solution p: it doesn t have to be the one produced by finding the parametric vector form and setting the free variables all to zero, as we did before. [interactive]

Homogeneous vs. Nonhomogeneous Systems Varying b If we understand the solution set of Ax = 0, then we understand the solution set of Ax = b for all b: they are all translates (or empty). ( ) 1 3 For instance, if A =, then the solution sets for varying b look like 2 6 this: Which b gives the solution set Ax = b in red in the picture? Ax = Ap p Choose p on the red line, and set b = Ap. Then p is a specific solution to Ax = b, so the solution set of Ax = b is the red line. Note the cool optical illusion! [interactive] For a matrix equation Ax = b, you now know how to find which b s are possible, and what the solution set looks like for all b, both using spans.

Solution Sets and Column Spans Very Important Let A be an m n matrix. There are now two completely different things you know how to describe using spans: The solution set: for fixed b, this is all x such that Ax = b. This is a span if b = 0, or a translate of a span in general (if it s consistent). Lives in R n. Computed by finding the parametric vector form. The column span: this is all b such that Ax = b is consistent. This is the span of the columns of A. Lives in R m. Don t confuse these two geometric objects! [interactive]