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T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS Linear Mathematics 2010 Tutorial 1 — Solutions 1. 3 0 4 5 1 1 2 −2. a) Calculate the matrix product −1 3 5 4 b) Find the unique solution of the linear system of equations which corresponds to the augmented matrix 3 0 4 31 1 1 2 11 . 9 −1 3 5 Solution 3 0 4 5 31 a) 1 1 2 −2 = 11. −1 3 5 9 4 x1 3 0 4 31 x2 1 1 2 . Then we want to find x = 11 . Notice that b) Let A = −1 such that Ax = x3 3 5 9 the matrix A must be invertible because the question says that there is a unique solution. (It is easy to see that A is invertible because det A = −7 6= 0—check!) 5 The shortest way to solve this question is to observe that −2 is a solution by part (a). 4 5 −2 As det A = −7 the matrix A is invertible, so x = is the unique solution. 4 For completeness, we also find the solution using Gaussian elimination: 3 0 4 31 1 1 2 11 1 1 2 11 R1 ↔R2 R2 :=R2 −3R1 1 1 2 11 − −−−→ 3 0 4 31 −− −−−−−→ 0 −3 −2 −2 −1 3 5 9 5 9 −1 3 5 9 −1 3 1 1 2 11 1 1 2 11 R3 :=R3 +R1 R2 :=R2 +R1 −−−−−−−→ 0 −3 −2 −2 −−−−−−−→ 0 1 5 18 7 20 0 4 7 20 0 4 1 1 2 11 1 1 2 11 1 R3 :=− R3 R3 :=R3 −4R2 13 18 −−−−−−−→ 0 1 5 18 −−−−−−−→ 0 1 5 0 0 1 4 0 0 −13 −52 Therefore x3 = 4, x2 = 18−5x 3 = −2, x1 = 11−x2 −2x3 = 5. Using back substitution 5 the unique solution is −2 , as before. 4 2. Let A = 1 −3 2 2 −5 6 −1 0 −8 . x1 a) Use Gaussian elimination to find all solutions of the equation Ax = 0, where x = xx23 0 and 0 = 0 . 0 b) Hence show that some non–zero linear combination of the columns of A is equal to 0. Linear Mathematics c) Let b = b1 b2 b3 Tutorial 1 — Solutions Page 2 ∈ R3 and suppose that Ax = b has at least one solution. Using (b), show that the equation Ax = b has infinitely many solutions. Solution a) Using Gaussian elimination to reduce A to row echelon form we find 1 −3 2 1 −3 2 1 −3 2 R3 :=R3 +R1 R2 :=R2 −4R1 2 −5 6 0 1 2 −−−−−−−→ 0 1 2 −−−−−−−→ 0 −3 −6 −1 0 −8 −1 0 −8 1 −3 2 R3 :=R3 +3R2 −−−−−−−→ 0 1 2 0 0 0 Therefore, x3 is the free variable. Set x3 = t, where t is an arbitrary element of R. Then, by back substitution, x2 = −2t and x1 = 3x2 − 2x3 = −6t − 2t = −8t. That is, the general solution of the equation Ax = 0 is −8 −8t where t ∈ R. x = −2t = t −2 , t 1 b) Taking t = 1 in part (a), we see that x = −8 −2 1 is a solution of Ax = 0. Therefore, 1 −3 2 0 2 3 −8 −1 − 2 −5 + −8 =0= 0 , 0 0 is one linear combination of the columns of A which sums up to 0. In fact, 1 −3 2 2 3 −8t −1 − t −5 + t −8 = 0, 0 for any t ∈ R. x1 c) Suppose that x = xx23 is a solution to the equation Ax = b. Then, by part (b), A x1 −8t x2 −2t x3 +t = Ax + A −8t −2t t = Ax + 0 = b, for any t ∈ R. Hence, there are infinitely many solutions. 3. Let V be the set of all ordered pairs of real numbers, and consider the following addition and scalar multiplication operations on u = (u1 , u2 ) and v = (v1 , v2 ): u + v = (u1 + v1 , u2 + v2 ), ku = (ku1 , 0) a) Compute u + v and ku for u = (−1, 2), v = (3, 4) and k = 3. b) Explain in words why V is closed under addition and scalar multiplication. c) Since addition on V is the standard addition operation on R2 , certain vector space axioms hold for V because they are known to hold for R2 . Which axioms are they? d) Show that the axioms S2, S3 and S4 hold. e) Show that V is not a vector space under the given operations. Solution a) (−1, 2) + (3, 4) = (2, 6); 3(−1, 2) = (−3, 0). b) Adding two ordered pairs of real numbers produces another ordered pair of real numbers, so V is closed under addition. Multiplying an ordered pair of real numbers by a real number (using the operation defined above) produces an ordered pair of real numbers, so V is closed under scalar multiplication. Linear Mathematics Extra questions for outside the tutorial — Solutions Page 3 c) Axioms A1–5. d) S2: If u = (u1 , u2 ), v = (v1 , v2 ) ∈ A and k ∈ R then k(u + v) = k(u1 + v1 , u2 + v2 ) = (ku1 + kv1 , 0) = (ku1 , 0) + (kv1 , 0) = ku + kv. S3 If u ∈ A and k, l ∈ R then (k + l)u = ((k + l)u1 , 0) = (ku1 , 0) + (lu1 , 0) = ku + lu. S4 If u ∈ A and k, l ∈ R then (kl)u = ((kl)u1 , 0) = k(lu1 , 0) = k(lu). e) 1u = (u1 , 0) 6= u. So axiom S5 does not hold, and hence V is not a vector space. 4. Each of the following matrices is the reduced row echelon form of an augmented matrix belonging to a system of linear equations in the variables xi , (i = 1, 2, . . .). (Both the systems represented here have infinitely many solutions – why?) For each augmented matrix (i) determine the number of parameters needed to solve the system and (ii) express the solution of the system in parametric form. ! ! a) 1 0 4 0 0 1 −5 −1 0 0 0 0 b) 1 0 0 2 0 0 0 1 0 0 6 0 5 1 −1 Solution a) i) The variable x3 is a parameter. ii) Let x3 = t. Then x2 = −1 + 5t and x1 = −4t. x1 −4t 0 −4 x2 = −1 + 5t = −1 + t 5 , t ∈ R. x3 t 0 1 b) i) There is one parameter; namely, x2 . ii) Let x2 = t. Then x4 = −1, x3 = 5, x1 = 6 − 2t. x1 6 − 2t 6 −2 x 2 t 0 1 = x3 5 = 5 + t 0 , t ∈ R. x4 −1 −1 0 Extra questions for outside the tutorial — Solutions n x1 o x2 x1 , x2 , x3 ∈ R is a vector space. Describe each of the following x3 subsets of n R3geometrically and determineowhether or not each n xsubset is a vector space. o x1 1 x x 3 3 2 2 2 1 2 x2 x2 a) A = ∈ R = − = x . b) A = ∈ R x + x + x ≤ 1 . 3 1 2 3 2 3 x3 x3 5. Recall that R3 = Solution a) The set A is a line in R3 . We claim that A is a vector space. To check this we have to verify that each of the vector space axioms is satisfied. a d b A1 Suppose that c ∈ A and fe ∈ A. Then a2 = − 3b = c and d2 = − 3e = f , so a+d a d b e + 2 = − 3 − 3 = c + f . Hence, b+e ∈ A. Hence, A is closed under addition. 2 c+f A2 If u, v, x ∈ R3 then (u + v) + x = u + (v + x) so, in particular, this is true when u, v, x ∈ A. Hence, addition is associative in A. A3 If u, v ∈ R3 then u + v = v + u so, in particular, this is true when u, v ∈ A. Hence, addition is commutative in A. Linear Mathematics A4 The vector Tutorial 1 — Solutions 0 0 0 Page 4 ∈ A and a b c + 0 0 0 a = a b c = 0 0 0 + a b c a 0 ∈ R3 ; in particular, this holds if cb ∈ A, so 0 is a zero vector in A. 0 a −a a A5 If cb ∈ A then − cb = −b ∈ A since −a − (−b) − (−c) = −(a − b − c) = 0. −c Also a −a −a a 0 b + −b = 0 = −b + cb , c 0 −c −c a a so − cb is a negative of cb . a a ka = − kb = kc. S1 Suppose that cb ∈ A and k ∈ R. Then k cb = kb ∈ A since ka 2 3 kc S2 If u, v ∈ A and k ∈ R then k(u + v) = ku + kv since this is true for all points in R3 . S3 If u ∈ A and k, l ∈ R then (k + l)u = ku + lu since this is true in R3 . S4 If u ∈ A and k, l ∈ R then (kl)u = k(lu) since this is true in R3 . S5 If u ∈ A then (1)u = u since this is true in R3 . b) The set B is the unit ball of radius 1 and centre (0, 0, 0) in R3 , including its boundary. The set B is not a!vector space because it is not closed under scalar multiplication. For 1 2 1 1 example, a = 2 ∈ A but 2a = 1 ∈ / A. for all b c 1 2 1