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Homework # 2 Solutions Math 152, Fall 2014 Instructor: Dr. Doreen De Leon p. 48: 26, 38 26. Suppose A is the 3 × 3 zero matrix (with all zero entries). Describe the solution set of the equation Ax = 0. Solution: First, look at Ax: 0 0 0 x1 0 0 0 Ax = 0 0 0 x2 = x1 0 + x2 0 + x3 0 0 0 0 x3 0 0 0 0 = 0 = 0. 0 This is true for any x ∈ R3 , so the solution set of the equation is R3 . 38. Let A be an m × n matrix and let w be a vector in Rn that satisfies the equation Ax = 0. Show that for any scalar c, the vector cw also satisfies Ax = 0. Solution: We need to show that A(cw) = 0. A(cw) = c(Aw) = c(0) = 0. Therefore, cw solves the equation Ax = 0. p. 61: 12, 36 12. Find the value(s) of h for which the vectors are linearly dependent. Justify each answer. 3 −6 9 −6 , 4 , h 1 3 3 1 Solution: If we denote the vectors as v1 , v2 , and v3 , respectively, then the vectors are linearly dependent if there is a nontrivial solution of x1 v1 + x2 v2 + x3 v3 = 0. The solution of the vector equation is the same as the solution of the system augmented matrix is 3 −6 9 | 0 r → 1 r 1 −2 3 | 0 1 −2 3 | 1 r2 →r2 +6r1 3 1 −6 4 h | 0 −−−−→ −6 4 h | 0 −−−−−−→ 0 −8 h + 18 | r3 →r3 −r1 1 −3 3 | 0 1 −3 3 | 0 0 −1 0 | 1 −2 3 | 0 1 −2 3 | 0 r2 ↔r3 r3 →r3 −8r2 0 | 0 −−−−−−→ 0 −1 0 | 0 −−−→ 0 −1 0 −8 h + 18 | 0 0 0 h + 18 | 0 whose 0 0 0 This system has a nontrivial solution if h + 18 = 0. Therefore, the vectors are linearly dependent if h = −18 . 36. If v1 , v2 , v3 are in R3 and v3 is not a linear combination of v1 and v2 , then {v1 , v2 , v3 } is linearly independent. Solution: This is a false statement. Let 1 −1 1 v1 = 2 , v2 = −2 , v3 = 0 . 3 −3 0 Then v3 is not a linear combination of v1 and v2 . We can see this by solving c 1 v1 + c 2 v2 = v3 , which has the augmented matrix 1 −1 | 1 2 −2 | 0 . 3 −3 | 0 The echelon form of this matrix is 1 −1 | 1 0 0 | 2 . 0 0 | 0 Therefore, there is no solution. We also have 1v1 + 1v2 + 0v3 = 0, a linear dependence relation. Therefore, the vectors are linearly dependent. 2 p. 69-70: 24, 36 24. An affine transformation T : Rn → Rn has the form T (x) = Ax + b, with A an m × n matrix and b ∈ Rm . Show that T is not a linear transformation when b 6= 0. (Affine transformations are important in computer graphics.) Solution: For any vectors u, v ∈ Rn , T (u + v) = A(u + v) + b T (u) + T (v) = Au + b + Av + b = A(u + v) + 2b 6= T (u + v) unless b = 0. Therefore, T is not a linear transformation when b 6= 0. Alternate solution: T (0) = A(0) + b = b 6= 0, so T is not a linear transformation when b 6= 0. 36. Let T : Rn → Rm be a linear transformation. Suppose {u, v} is a linearly independent set, but {T (u), T (v)} is a linearly dependent set. Show that T (x) = 0 has a nontrivial solution. Solution: Since {T (u), T (v)} is linearly dependent, there exist constants c1 and c2 , not both zero, such that c1 T (u) + c2 T (u) = 0. Since T is linear, c1 T (u) + c2 T (u) = T (c1 u1 + c2 u2 ). So, x = c1 u + c2 v is such that T (x) = 0. Since {u, v} is linearly independent, we know that x 6= 0. (Otherwise, there would be a nontrivial linear combination of u and v that is zero, which is impossible.) Therefore, the equation T (x) = 0 has a nontrivial solution. 3