# 1.3.10) Write a vector equation that is equivalent to the given system

1.3.10) Write a vector equation that equations: 4x1 + x2 + x1 − 7x2 − 8x1 + 6x2 − is equivalent to the given system of 3x3 = 9 2x3 = 2 5x3 = 15 Solution: 4 1 3 9 x1 1 + x2 −7 + x3 −2 = 2 8 6 −5 15 1.3.12) Determine if b is a linear combination of a1 , a2 , and a3 . a1 1 0 2 −5 = −2 a2 = 5 a3 = 0 b = 11 2 5 8 −7 Solution: To see if b is a linear combination of a1 , a2 , and a3 we must find a solution to the equation b = x 1 a1 + x 2 a2 + x 3 a3 Rewrite as a augmented matrix 1 0 2 −5 −2 5 0 11 2 5 8 −7 Now we will row reduce: R2 : 2R1 + R2 1 0 2 2 0 5 4 1 2 5 8 −7 R3 : R3 − 2R1 1 0 5 2 0 5 4 1 0 5 4 −11 1 R3 : −R2 + R3 1 0 5 2 0 5 4 1 0 0 0 −12 Since 0 6= −12 we have that b is not a linear combination of a1 , a2 , and a3 . 2 0 6 10 1.3.26) Let A = −1 8 5 , let b = 3 , and let W be the set 1 −2 1 3 of all linear combinations of the columns of A. a. Is b in W b. Show that the third column of A is in W Solution: a. First let a1 2 0 6 −1 8 5 a2 = a3 = = 1 −2 1 We want to find x1 , x2 , and x3 such that b = x 1 a1 + x 2 a2 + x 3 a3 As in 1.3.12 we will consider the Augmented matrix B = [a1 , a2 , a3 , b] and row reduce. Now 2 0 6 10 B = −1 8 5 3 1 −2 1 3 R1 : (1/2)R1 1 0 3 5 −1 8 5 3 1 −2 1 3 R2 : R1 + R2 1 0 3 5 0 8 8 8 1 −2 1 3 2 R2 : (1/8)R2 R3 : R3 − R1 1 0 3 5 0 1 1 1 0 −2 −2 −2 R3 : R3 + 2R2 1 0 3 5 0 1 1 1 0 0 0 0 From this we can see that there are an infinite number of solutions. Let x3 = t Then x3 = t x2 = 1 − t x2 = 5 − 3t is a solution. In particular if t = 0 we get b = 5a1 + 1a2 + 0a3 b. a3 is in W since a3 =0a1 + 0a2 +1a3 1.3.30) Let v be the center of mass of a system of point masses located at v1 , ..., vk as in Exercise 29. Is v in Span{v1 , ..., vk }? Solution: From exercise 29 we know v = (1/m)[m1 v1 + · · · + mk vk ] where m = m1 + · · · + mk hence v is a linear combination of , ..., vk whichimplies v1 v ∈ Span{v1 , ..., vk }. 7 3 6 1.4.26) Let u = 2 ,v = 1 , and w = 1 . It can be shown that 5 3 0 3u − 5v − w = 0. Use this fact to find x1 and x2 tat satisfy the equation 7 2 6 x1 2 1 = 1 x2 5 3 0 . Solution: 3 3u − 5v − w = 0 implies 3u − 5v = w this gives x1 = 3 x2 = −5 satisfy the equation. 1.5.6) Find the solution set to the homogeneous system x1 + 3x2 − 5x3 = 0 x1 + 4x2 − 8x3 = 0 −3x1 − 7x2 + 9x3 = 0 Solution: Let A be the matrix of coefficients of the system and row reduce the augmented matrix A 0 to echelon form 1 3 −5 0 1 4 −8 0 −3 −7 9 0 R2 : R2 − R1 R3 : R3 + 3R1 1 3 −5 0 0 1 −3 0 0 2 −6 0 R1 : R1 − 3R2 R3 : R3 − 2R1 1 0 4 0 0 1 −3 0 0 0 0 0 Since x3 is a free variable we have nontrival solutions. Our Solutions are: x3 = t x2 = 3t x1 = −4t 1.5.26) Suppose Ax = b has a solution. Explain why the solution is unique precisely when Ax = 0 has only the trivial solution. Solution: Suppose Ax = 0 has only the trivial solution, and assume a1 and a2 are solutions to Ax = b. We will show a1 = a2 . By assumption we have Aa1 = b and Aa2 = b. By Theorem 1.4.5 we have A(a1 − a2 ) = Aa1 − Aa2 = b − b = 0 4 Hence a1 − a2 is a solution to Ax = 0. So by assumption a1 − a2 = 0 or a1 = a2 . Hence Ax = b has a unique solution. Suppose Ax = b has a unique solution say a, and let c be a solution to Ax = 0. Then by Theorem 1.4.5 we have A(a + c) = Aa + Ac = b + 0 = b Hence a + c is a solution to Ax = b and since we assumed that a was a unique solution to Ax = b we must have a + c = a or c = 0. Hence Ax = 0 has only the trivial solution. 1.7.6) Determine if the columns of the matrix A form a linearly independent set, where −4 −3 0 0 −1 4 A= 1 0 3 5 4 6 Solution: We know from the chapter that the columns of a matrix A are linearly independent if and only if the equation Ax = 0 has only the trivial solution. So we will row reduce the augmented matrix −4 −3 0 0 0 −1 4 0 B= 1 0 3 0 5 4 6 0 R1 ↔ R2 R2 : −R2 1 0 3 0 0 1 −4 0 −4 −3 0 0 5 4 6 0 R3 : R3 + 4R1 R4 : R4 − 5R1 1 0 3 0 1 −4 0 −3 12 0 4 −9 R3 : R3 + 3R2 R4 : R4 − 4R1 5 0 0 0 0 1 0 0 0 0 3 0 1 −4 0 0 0 0 0 7 0 From here we can see Ax = 0 has only the trivial solution, hence the columns of A are linearly independent. 1.7.16) Determine whether the vectors are linearly independent. 4 6 −2 , −3 6 9 Solution: 4 6 −2 = (2/3) −3 6 9 So the vectors are linearly dependent. 1.7.36) If v1 , . . . , v4 are in R4 and v3 is not a linear combination of v1 , v2 , v4 , then {v1 , . . . , v4 } is linearly independent. Solution: False. 4 Let v1 = v2 = v4 = 0 and v3 = −2 6 The any linear combination of v1 , v2 , v4 is the 0 vector hence v3 6∈ Span{v1 , v2 , v4 } but v1 , . . . , v4 is not linearly independent by theorem 9 since it contains the zero vector. 6

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