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APSC 174J, Solutions #4 Posted: March 23, 2016 Section 5, Problem 1(c). To see if {~v2 } is linearly independent, we need to solve the vector equation: x · ~v2 = ~0 That is, x(3, 0) = (3x, 0) = (0, 0). The only solution is x = 0, so {~v1 } is linearly independent. Note: If a set contains only one vector, then it is linearly independent if and only if the vector is a non-zero vector. Section 5, Problem 1(g). To see if {~v3 , ~v4 } is linearly independent, we need to solve the vector equation: x1 · ~v3 + x2 · ~v4 = ~0 That is, x1 (0, 0) + x2 (1, 1) = (x2 , x2 ) = (0, 0). The solution is x2 = 0, and x1 is free, for example, we can choose x1 = 1, then 1 · ~v3 + 0 · ~v4 = ~0 Which means we find non-zero weights so that the linear combination of ~v3 , ~v4 equals the zero vector, so {~v3 , ~v4 } is linearly dependent. Section 5, Problem 1(k). To see if {~v2 , ~v4 , ~v5 } is linearly independent, we need to solve the vector equation: x1 · ~v2 + x2 · ~v4 + x3 · ~v5 = ~0 That is, x1 (3, 0) + x2 (1, 1) + x3 (2, 1) = (3x1 + x2 + 2x3 , x2 + x3 ) = (0, 0). The equations are ( 3x1 + x2 + 2x3 = 0 x2 + x3 = 0 The solution is x2 = −x3 , x1 = −(1/3)x3 and x3 is free, for example, if we choose x3 = 3, then x1 = −1, x2 = −3 and we will have −1 · ~v2 + (−3) · ~v4 + 3 · ~v5 = ~0 2 Which means we find non-zero weights so that the linear combination of ~v2 , ~v4 , ~v5 equals the zero vector, so {~v2 , ~v4 , ~v5 } is linearly dependent. Section 5, Problem 2(b). To see if {~v1 , ~v2 } is linearly independent, we need to solve the vector equation: x1 · ~v1 + x2 · ~v2 = ~0 1 2 0 0 1 0 That is, x1 + x2 = . This gives equations: 0 0 0 0 0 0 x1 + 2x2 = 0 x =0 2 0=0 0=0 The only solution is x1 = x2 = 0, so {~v1 , ~v2 } is linearly independent. Section 5, Problem 2(f ). To see if {~v2 , ~v4 } is linearly independent, we need to solve the vector equation: x1 · ~v2 + x2 · ~v4 = ~0 0 0 2 0 0 1 That is, x1 + x2 = . This gives equations: 1 0 0 0 0 1 2x1 = 0 x =0 1 x2 = 0 x2 = 0 The only solution is x1 = x2 = 0, so {~v2 , ~v4 } is linearly independent. Section 5, Problem 2(o). To see if {~v3 , ~v4 , ~v5 } is linearly independent, we need to solve the vector equation: x1 · ~v3 + x2 · ~v4 + x3~v5 = ~0 3 1 0 0 0 1 0 1 0 That is, x1 + x2 + x3 = . This gives equations: 1 1 0 0 0 1 1 0 x1 = 0 x + x = 0 1 3 x1 + x2 = 0 x2 + x3 = 0 The only solution is x1 = x2 = x3 = 0, so {~v3 , ~v4 , ~v5 } is linearly independent. Section 6, Problem 1(a). Since our system of linear equations have two equations, we consider vectors in the space R̂2 : 2 ~v = , 4 4 w ~ = 5 The system has a solution if and only if w ~ is in the linear span of ~v . Easy to see that w ~ is not in the linear span of ~v , so the system has no solutions. Section 6, Problem 1(b). Since our system of linear equations have two equations, we consider vectors in the space R̂2 : 2 ~v = , 4 4 w ~ = 8 The system has a solution if and only if w ~ is in the linear span of ~v . Since w ~ = 2~v , w ~ is in the linear span of ~v , so the system has solutions. To determine if there is a unique solution or infinitely many solutions, we need to see if {~v } is linearly independent or dependent. Since there is only one vector, and ~v 6= ~0, the set {~v } is linearly independent. Hence the system has a unique solution. Section 6, Problem 1(g). Since our system of linear equations have two equations, we consider vectors in the space R̂2 : 4 1 ~v1 = , 1 1 ~v2 = , 1 −1 ~v3 = , 0 −1 ~v4 = , 0 0 w ~ = 2 The system has a solution if and only if w ~ is in the linear span of {~v1 , ~v2 , ~v3 , ~v4 }. Observe that w ~ = ~v1 + ~v2 + ~v3 + ~v4 , so w ~ is in the linear span of {~v1 , ~v2 , ~v3 , ~v4 }, hence the system has solutions. To determine if there is a unique solution or infinitely many solutions, we need to determine if {~v1 , ~v2 , ~v3 , ~v4 } is linearly independent or dependent. Easy to see that ~v1 , ~v2 , ~v3 , ~v4 are linearly dependent, for example, 1 · ~v1 + (−1) · ~v2 + 0 · ~v3 + 0 · ~v4 = ~0 Which means we find non-zero weights so that the linear combination of ~v1 , ~v2 , ~v3 , ~v4 equals the zero vector, so {~v1 , ~v2 , ~v3 , ~v4 } is linearly dependent. Hence the system has infinitely many solutions.