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Recitation Notes Spring 16, 21-241: Matrices and Linear Transformations January 19, 2016 Abstract [TOPICS COVERED] Row echelon form, Row reduced echelon form, Gauss-Jordan elimination, Logic and qualifiers, Negating statements. 1 Administrative Matters • Course textbook: David Poole Linear Algebra: a modern introduction 4th edition. CAUTION: Some exercises differ from the 3rd to the 4th edition. Be very careful to check that you are doing the right problems if you want to use other editions. • The homework assignment has problems from the textbook as well as from "Introduction to mathematical arguments" by Michael Hutchings. Do not miss out anything. • The problems from the pdf are Exercises 1 and 3 on page 12, NOT just Exercises 1(a) and 1(c). • Use the pdf definitions for even and odd. Do NOT just use your intuitive understanding of the terms. • Introduction to mathematical arguments, right at the bottom of Exercise 1 it says: What is the negation of each of these statements? You need to do this too. Do NOT miss this out. Do NOT write an essay, use the symbols Z, ∀, ∃, ∧, ∨, =⇒ , ⇐⇒. 2 Definitions 1. Row operations 2. Row echelon form 3. Reduced row echelon form (also known as Reduced echelon form. Keyword: Reduced) 4. Gauss-Jordan Elimination 5. Statement 6. Logic operators: ∧, ∨, =⇒ , ⇔, ¬ 7. Qualifiers: ∀, ∃ 8. Even, odd 1 3 Problems 1. David Poole following are 1 0 (a) 0 0 0 1 7 0 (b) 0 1 0 0 0 1 (c) 0 0 0 0 0 0 (d) 0 0 1 0 0 0 (e) 0 1 0 0 0 1 (f) 1 0 1 2 1 0 (g) 0 1 0 0 2 1 0 0 (h) 0 0 0 0 Linear Algebra: a modern introduction (4th Ed.) Ex 2.2.1-8. Which of the in row echelon form? 1 3 0 1 0 −1 4 0 0 3 0 0 1 0 0 0 3 −4 0 0 0 0 5 0 1 1 0 0 3 0 1 1 3 5 1 −1 0 3 0 0 Solution. (a) No (b) Yes, No (c) Yes, Yes (d) Yes, Yes (e) No (f) No (g) No (h) Yes, No Page 2 2. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 2.1.28,30. Express the following systems of linear equations in augmented matrix form (a) x1 + x2 + x3 = 1 x1 + x3 = 0 −x1 + 2x2 − 2x3 = 0 (b) a − 2b + d = 2 −a + b − c − 3d = 1 Solution. (a) We have: 1x1 +1x2 +1x3 = 1 1x1 +0x2 +1x3 = 0 −1x1 +2x2 −2x3 = 0 This gives the augmented matrix: 1 1 1 1 1 0 1 0 −1 2 −2 0 (b) We have: 1a −2b +0c +1d = 2 −1a +1b −1c −3d = 1 This gives the augmented matrix: 1 −2 0 1 2 −1 1 −1 −3 1 3. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 2.1.36. Solve the system in 2(b). Page 3 Solution. 1 −2 0 1 2 −1 1 −1 −3 1 1 2 −−−−−−−−→ 1 −2 0 r2 7→ r2 + r1 0 −1 −1 −2 3 −−−−−−→ 1 −2 0 1 2 r2 7→ −r2 0 1 1 2 −3 −−−−−−−−−→ 1 0 2 5 −4 r1 7→ r1 + 2r2 0 1 1 2 −3 Therefore we have (a, b, c, d) = (−4 − 2s − 5t, −3 − s − 2t, s, t), s, t ∈ R. 4. Solve 2(a) from the augmented matrix above using Gauss-Jordan Elimination. Solution. 1 1 1 1 1 0 1 0 −1 2 −2 0 1 1 1 1 −−−−−−−−→ 0 −1 0 −1 r2 7→ r2 − r1 −1 2 −2 0 1 1 1 1 −−−−−−−−→ r3 7→ r3 + r1 0 −1 0 −1 0 3 −1 1 1 1 1 1 −−−−−−→ r2 7→ −r2 0 1 0 1 0 3 −1 1 1 1 1 1 −−−−−−−−−→ 1 r3 7→ r3 − 3r2 0 1 0 0 0 −1 −2 1 1 1 1 −−−−−−→ r3 7→ −r3 0 1 0 1 0 0 1 2 1 1 0 −1 −−−−−−−−→ r1 7→ r1 − r3 0 1 0 1 0 0 1 2 1 0 0 −2 −−−−−−−−→ r1 7→ r1 − r2 0 1 0 1 0 0 1 2 Note. Gauss-Jordan Elimination may not necessarily be the fastest way to obtain the rref of a matrix. Page 4 5. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 2.2.40. For what values of k, if any, will the system kx + 2y = 3 2x − 4y = −6 have (a) no solution, (b) a unique solution, and (c) infinitely many solutions? Solution. Step 1: Express in augmented matrix form. 3 k 2 2 −4 −6 Step 2: Express in reduced row echelon form. k 2 3 2 −4 −6 2 −4 −6 − − − − − → r1 ↔ r2 k 2 3 −−−−−−→ 1 1 −2 −3 r1 7→ r1 k 2 3 2 −−−−−−−−−→ 1 −3 −2 r2 7→ r2 − kr1 0 2 + 2k 3 + 3k 1 −2 −3 −→ 0 2(1 + k) 3(1 + k) Here we need to split into two cases. Case 1: 1 + k = 0. Then we have 1 −2 1 −2 −3 −3 = 0 2(1 + k) 3(1 + k) 0 0 0 This gives us the solution (x, y) = (2t − 3, t) for t ∈ R. Case 2: 1 + k 6= 0. 1 −2 −3 0 2(1 + k) 3(1 + k) −−−−−−−−−→ 1 1 −2 −3 r2 7→ r2 0 2 3 1+k −−−−−−−−→ 1 0 0 r1 7→ r1 + r2 0 2 3 −−−−−−→ 1 1 0 0 r2 7→ r2 0 1 32 2 Page 5 Thus there is unique solution (x, y) = (0, 23 ). Step 3: Conclude. (a) k ∈ ∅ (b) k 6= −1 (c) k = −1 6. Negate the following: ∃x P (x). ∀y Q(y). ∀z R(x, y, z) Solution. ! ¬ ∃x P (x). ∀y Q(y). ∀z R(x, y, z) −→ ∀x P (x). ¬ ∀y Q(y). ∀z R(x, y, z) −→ ∀x P (x). ∃y Q(y). ¬ ∀z R(x, y, z) −→ ∀x P (x). ∃y Q(y). ∃z ¬ R(x, y, z) −→ ∀x P (x). ∃y Q(y). ∃z ¬R(x, y, z) 7. Prove the following statement: For every integer x, if x is odd, then x2 is odd. What is the negation of the statement? Solution. x is odd =⇒ ∃k ∈ Z. x = 2k + 1 =⇒ x2 = (2k + 1)2 =⇒ x2 = 4k 2 + 4k + 1 =⇒ x2 = 2(2k 2 + 2k) + 1 Since k ∈ Z, ` := 2k 2 + 2k ∈ Z. Thus ∃` ∈ Z. x2 = 2` + 1. Therefore x2 is odd. Note: Notice how we conclude ’odd’ by definition. Notice also how we justified why ` is an integer. Page 6 4 Additional Notes 1. In definitions, we often just write "if", but what we mean is actually "if and only if". Do not confuse this with implication statements. 2. Gauss-Jordan Elimination Step 1: Swap zero-rows to the bottom Step 2: Swap any row with left-most non-zero column to the top (if necessary) Step 3: Perform scalar multiplication to make leading entry 1 Step 4: Subtract multiple of top row to other rows such that entries under the left-most non-zero column are 0 Step 5: Repeat for remaining rows Gauss-Jordan Elimination may not be the fastest way to obtain a solution when doing manual computation. However, it is an algorithm, and is useful for programming a linear equations solver. 3. Logic operators ¬ F T A T F A T T F F B T F T F ∧ T F F F ∨ T T T F =⇒ T F T T ⇐⇒ T F F T 4. ¬(A =⇒ B) is equivalent to A ∧ ¬B 5. Familiarize with De Morgan’s Law, and distributivity and associativity of ∧ and ∨. 6. You cannot always commute qualifiers. For example, ∃x ∈ R+ . ∀y ∈ R+ . y > x is saying that "there exists a positive real number x such that all positive real numbers are greater than x". This is clearly untrue (x is not greater than x). ∀y ∈ R+ . ∃x ∈ R+ . y > x is saying that "for all positive real numbers, there is a positive real number that is less than it". This is true, since for any y ∈ R+, y2 is also a positive real number and it is less than y. Page 7