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Assignment 4 Answers Math 130 Linear Algebra of homogeneous equations has a solution, namely 0. 2d. Solve the system of linear equations. Assignment 4 answers Math 130 Linear Algebra x1 + 2x2 + 2x3 = 2 D Joyce, Fall 2013 x1 + 8x3 + 5x4 = −6 x1 + x2 + 5x2 + 5x4 = 3 Exercises from section 1.4, page 32, exercises 1, 2ad, 3ad, 5de, 6, 9, 14. You can solve this by hand in many ways using row operations. Here's one. You'd probably start by subtracting the first from each of the second and 1. True/false third rows to get a. The zero vector is a linear combination of any nonempty set of vectors. x1 + 2x2 + 2x3 = 2 True. It's 0 = 0v1 + ··· + 0vn. Moreover, an − 2x2 + 6x3 + 5x4 = −8 empty sum, that is, the sum of no vectors, is usually − x2 + 3x3 + 5x4 = 1 defined to be 0, and with that definition 0 is a linear combination of any set of vectors, empty or not. At this point there's more than one natural way to b. The span of the empty set ; is ;. continue. Since I don't like to work with fractions, False. It's the trivial vector space that includes I'll switch the second and third rows, then negate only the zero vector 0. For convenience, our text the new second row to get made that a special case of the definition. If you x + 2x + 2x = 2 define an empty sum to be 0, then you don't have 1 2 3 x − 3x − 5x = −1 to make it a special case. 2 3 4 − 2x2 + 6x3 + 5x4 = −8 c. If S is a subset of a vector space V , then span(S) is the intersection of all subspaces of V Next, I'll add twice the second row to the third row containing S. to get True. See theorem 5. d. In solving a system of linear equations, it x1 + 2x2 + 2x3 = 2 is permissible to multiply an equation by any con- x2 − 3x3 − 5x4 = −1 stant. − 5x4 = −10 False. It's almost true, but you can't multiply And divide the third row by −5. it by 0. That would be the same as erasing the equation. x1 + 2x2 + 2x3 = 2 e. In solving a system of linear equations, it is x2 − 3x3 − 5x4 = −1 permissible to add any multiple of one equation to x4 = 2 another. True. (But you can't add −1 times the equation The system is now in echelon form. I'm going to to itself.) continue to put it into reduced echelon form. First I'll add 5 times the third row to the second. f. Every system of linear equations has a solu- tion. x1 + 2x2 + 2x3 = 2 False. There are inconsistent systems like x−y = x2 − 3x3 + = 9 3 and x − y = 4. Note, however, that every system x4 = 2 1 Then subtract twice the second from the first. d. (2; −1; 0); (1; 2; −3); (1; −3; 2) Solve the vector equation x1 + + 8x3 = −16 x2 + −3x3 + = 9 x(1; 2; −3) + y(1; −3; 2) = (1; −1; 0) x = 2 4 by solving the system of linear equations It's in reduced echelon form. The third equation x + y = 1 tells me that x4 = 2; the second that x2 = 3x3 + 9; 2x − 3y = −1 and the first that x1 = −8x3 − 16. None of them −3x + 2y = 0 say what x3 is, and that's because x3 can be any which has the solution (x; y) = ( 2 ; 3 ), so the first number. Thus, the general solution is 5 5 vector is a lin combo of the other two. (x1; x2; x3; x4) = (−8x3 − 16; 3x3 + 9; x3; 2) 5. Determine if the given vector v is in the span where x3 can be any number. We've found the gen- of a set S. eral solution parameterized by the variable x3. d. v = (2; −1; 1; −3), You may have done it differently and found the S = f(1; 0; 1; −1); (0; 1; 1; 1)g. general solution parameterized some other way. This is the same question as exercise 3, but All solutions will describe the same set in R4. It's worded differently, so you can answer it in the same a particular straight line that doesn't pass through way, namely, by solving the vector equation the origin. That's easier to see if you write the solutions as x(1; 0; 1; −1) + y(0; 1; 1; 1) = (2; −1; 1; −3): x = t(−8; 3; 1; 0) + (−16; 9; 0; 2) This one is particularly easy to solve. x has to be 2 for the first coordinate to work out, and y has to where t = x3 is a parameter. be −1 for the second coordinate. And with those values of x and y, the third and fourth coordinates 3. Is the first vector a linear combination of the match, too. So there is a solution, and that means other two? v is in the span of S. 3 2 a. (−2; 0; 3); (1; 3; 0); (2; 4; −1) e. v = −x + 2x + 3x + 3, S = fx3 + x2 + x + 1; x2 + x + 1; x + 1g. The general method to solve questions like this is to solve a vector equation, in this case, the vector Here, the vector space is the set of polynomials equation of degree 3 or less. There are various shortcuts, but you could use the same method of solving the x(1; 3; 0) + y(2; 4; −1) = (−2; 0; 3): vector equation s(x3 + x2 + x + 1) + t(x2 + x + 1) + u(x + 1) That corresponds to a system of three equations in = −x3 + 2x2 + 3x + 3: two unknowns That corresponds to the system of linear equations x + 2y = −2 (one equation for each power of x) 3x + 4y = 0 s = −1 − y = 3 s + t = 2 s + t + u = 3 Then use whatever method you want to solve the s + t + u = 3 system. This particular system has a solution, (x; y) = (4; −3). Therefore, the first vector is a Since there's a solution, namely, (s; t; u) = (1; 1; 1), linear combination of the other two. therefore v is in the span of S. 2 6. Show that the vectors (1; 1; 0), (1; 0; 1), and span(S1) + span(S2) is the sum of two vectors, one 3 (0; 1; 1) generate F . being a lin combo of vectors in S1, the other being Here, F is the scalar field. We need to show that a lin combo of vectors in S2. every vector (a; b; c) in F 3 is a linear combination But a lin combo of vectors in S1 [ S2 is the sum of the three vectors. There are a couple approaches of a lin combo of vectors in S1 and a lin combo of you could take. vectors in S2, and vice versa. One way is to show that the three standard vec- We can add details to the argument as follows. tors (1; 0; 0), (0; 1; 0), and (0; 0; 1) are linear com- Let the vector w be the linear combination of vec- binations. Since all vectors in F 3 can be generated tors in S1 [ S2, and arrange the terms so that the by them, if we can generate them, we'll be able to vectors in S1 come before the vectors in S2. generate all vectors. (This depends on the theo- w = c v + ··· + c v + c v + ··· + c v rem: a linear combination of linear combinations 1 1 k k k+1 k+1 n n is a linear combination.) where v ;:::; v 2 S and v ;:::; v 2 S : Then So, how can you get (1; 0; 0)? You can find it by 1 k 1 k+1 n 2 w is the sum of the two vectors, c1v1 + ··· + ckvk the techniques of the earlier exercises or just look in the span of S , and c v + ··· + c v in the for it. You'll find (1; 0; 0) = 1 ((1; 1; 0) + (1; 0; 1) − 1 k+1 k+1 n n 2 span of S . And, of course, the sum of two such (0; 1; 1)). The other two standard vectors are simi- 2 vectors is a vector in the span of S1 [ S2. larly two of the given vectors minus the third. The primary advantage of the more detailed An alternate way is to show that an arbitrary vec- 3 proof is that it emphasizes that the terms in the tor (a; b; c) in F is a linear combination of the three sum need to be rearranged to express the sum w as vectors. In other words, solve the vector equation the sum of two vectors, one from span(S1) and the other from span(S ). x(1; 1; 0) + y(1; 0; 1) + z(0; 0; 1) = (a; b; c) 2 Math 130 Home Page at for x, y, and z in terms of a, b, and c. http://math.clarku.edu/~djoyce/ma130/ 1 0 0 1 9. Show that the four matrices , , 0 0 0 0 0 0 0 0 , and generate M (F ). 1 0 0 1 2×2 This is pretty easy. All you have to do is show a b that a generic matrix is a linear combination c d of those four matrices. But it is since it equals 1 0 0 1 0 0 0 0 a + b + c + d 0 0 0 0 1 0 0 1 14.
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