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# Math 340 - Week 7 Math 340

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This 3 page Class Notes was uploaded by Susan Ossareh on Sunday March 6, 2016. The Class Notes belongs to Math 340 at Colorado State University taught by in Spring 2016. Since its upload, it has received 22 views. For similar materials see Intro-Ordinary Differen Equatn in Math at Colorado State University.

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Date Created: 03/06/16

Math 340 Lecture – Introduction to Ordinary Differential Equations – February 29 , 2016 th What We Covered: 1. Course Content – Chapter 7: Matrix Algebra a. Section 7.5: Bases of a Subspace Continued i. Recap 1. Nullspace: ????????????????(????) ???????????? = ???? ∶ ???????? = 0 ???????????? ℝ ???? ???? 2. Definition: a subspace ???? ???? ℝ such that a. If ????,???? ???? ???? then x+y e V b. If xeV, ???? ???? ℝ then ???????? ???? ???? 3. Null(A) is a subspace of ℝ???? 4. Linear dependence and independence a. Definition: the vectors ???? ,…,???? ???? ℝ are linearly independent is 1 ???? ???? 1 1 ⋯+ ???? ???? =???? ???????????????????? ???????? ???? + ⋯+1???? = 0 ???? ii. Example: are ???? = (1,0) and ???? = (0,1) linearly independent? 1 2 ???????????????????????????? ????ℎ???????????? ???????????????????????? ???? 1???? 2ℝ ????????????ℎ ????ℎ???????? ???? ????1 1???? ???? 2 2 ???? 1,0 + ???? 0,1 = (0,0) 1 2 (????1,????2) = (0,0) ????1= 0 ???????????? ???? = 2 V1 and V2 are linearly independent because both a1 and a2 are equal ( ) iii. Example: are ???? 1 1,2 ???????????? ???? = (22,4) linearly independent? Suppose ????11,−2 + ???? −224 = 0.) (????1− 2???? 2−2???? + 1???? ) =2(0,0) ???? 1 2???? =20 − 2???? 1 4???? =20 1 −2 0 −2 4 0 1. Row Operation: 2R1+R2 1 −2 0 0 0 0 2. X2 is a free column so it’s an independent variable: ???? 2 ???? ????1− 2???? = 0 → ????1= 2???? ???? 1 = ????2 ????2 1 ???? = 2 ???? = 1 1 2 2 1,−2 + 1 −2,4 = (0,0) (2,−4 + −2,4 = (0,0) 3. They aren’t linearly independent, so they have to be linearly dependent iv. To find out if the vectors {x1, xk} are linearly independent, we find null(x), x=[x1, xk]. If null(x) is trivial then they are linearly independent, if not then they are dependent v. Bases of subspace 1. Definition: ???? = ???? ,???? ,…,???? ???????? ℝ ???? 1 2 ???? 2. B is a basis of V(subspace) if a. Span ???? ,…,???? )= ???? 1 ???? b. {????1,…,???? }????are linearly independent vi. Example: ???? = ℝ a basis for v is ???? = { 1,0 , 0,1 } 1. They are linearly independent 2. (????,????)????ℝ , then ????,???? = ???? 1,0 + ???? 0,1 ???? ???????????????? 1,0 ,(0,1) vii. Example: Find a basis for the null space of C 1 −1 0 2 0 ???? = 0 0 1 2 −1 1. The null space of C is all ????????ℝ 2. Cx=0, solve 1 1 0 2 0 0 0 0 1 2 −1 0 3. We can say x1 and x3 are pivot columns and the rest are free columns. Now we can back solve ????2= ????, ????4= 5, ???? =5???? ????3= ???? − 2???? ????1= ???? − 2???? X1 r-2s X2 R X3 = t-2s X4 S X5 t 4. V1, v2, v3 are linearly independent and null( c )= span (v1, v2, v3) so ???? = {????1,????2,????3} is a basis for null( c ) viii. In general, a basis for v is not unique but always has the same # of vectors. The number of vectors in B = dimensions of V b. Section 7.6: Square Matrices i. Definition: ????????????????− # ???????????????????????????? = # ???????????? ii. Matrix A is nonsingular if it’s a square matrix and Ax=b has a solution for any b. If not, its singular iii.Proposition: A is nonsingular if and only if A changes to a row echelon form without zeros in the diagonal iv. Proposition: If A is nonsingular then Ax=b has a unique solution v. Proposition: Ax=0 has only a trivial solution if and only if A is nonsingular. Hence the null(A)={0} Suggested Homework: Study for exam 1 Section 7.5: 5, 10, 20, 22, 30, 38 Section 7.6: 8, 14, 20, 24 st Math 340 Lab – Introduction to Ordinary Differential Equations – March 1 , 2016 What We Covered: 1. Went over the practice exam in class 2. Course Content – Chapter 7: Matrix Algebra a. Section 7.6: Inverse of a Matrix i. Definition: ???????????????? is invertible if there exis????????????such that AB= 1 = BA we call ???? = −1 ???? the inverse of A ii. Proposition: A is invertible if A is nonsingular 3 1 iii.Example: Compute ???? −1 for A= −1 2 3 1 | 0 −1 2 0 1 1. Row reduce to the identity a. R1 +R2 = R2 3 1 1 0 0 7⁄ |1/3 1 3 b. 1/3R1 and 3/7R2 1 1/3 1/3 0 | 0 1 1/7 3/7 c. –R2/3 + R1=R1 1 0 2/7 −1/7 | 0 1 1/7 3/7 2. So the inverse of A is… ???? −1= 1 2 −1 7 1 3 3. In general… ???? ????????????= ???? ???? ???? ???? −1 1 ???? −???? ???? = ???????? − ???????? −???? ???? 4. This is only valid for 2x2 matrix Suggested Homework: Study for exam 1 Section 7.6: 8, 14, 20, 24 Math 340 Lecture – Introduction to Ordinary Differential Equations – March 2nd, 2016 What We Covered: 1. More review for the exam a. Highlights i. Used the practice exam 2. Announcement a. No class on Friday!

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