Instructor's Solutions Manual for Linear Algebra with by Otto and Bretscher and Kyle Burke
By Otto and Bretscher and Kyle Burke
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1.. 120 . 2 Thus, there are initially 120 liberal students, and 140 conservative students. Since the number of liberal students initially is the same as the number of conservative students in the end, the class ends with 120 conservative students and 140 liberal students. 64 Let x1 and x2 be the initial number of students in Sections A and B, respectively. Then, since there are 55 students total, x1 + x2 = 55. 3x2 , and in the process, lose 4 students. 0. Our matrix becomes . . 55 , which reduces to 1 0 ..
These points are on the line through the origin and the observer’s eye. 43 a T (x) = 3 · x2 = 2x1 + 3x2 + 4x3 = [2 3 4] x2 x3 4 x3 The transformation is indeed linear, with matrix [2 3 4]. v1 b If v = v2 , then T is linear with matrix [v1 v2 v3 ], as in part (a). v3 a a x1 x1 x1 c Let [a b c] be the matrix of T . Then T x2 = [a b c] x2 = ax1 + bx2 + cx3 = b · x2 , so that v = b c c x3 x3 x3 does the job. 44 T x2 = v2 × x2 = v3 x1 − v1 x3 = v3 −v2 v1 x2 − v2 x1 x3 v3 x3 0 −v3 v2 v3 0 −v1 .
5. 11 Undefined since the two vectors do not have the same number of components. 15 [1 2 3 4] = 5 · 1 + 6 · 2 + 7 · 3 + 4 · 8 = 70 either way. 17 Undefined, since the matrix has three columns, but the vector has only two components. 4, rref (A) = 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 . 1 0 0 0 . 25 In this case, rref(A) has a row of zeros, so that rank(A) < 4; there will be a free variable. 4. 26 From Example 3d we know that rank(A) = 3, so that rref(A) = 0 0 0 1 0 0 0 0 . 1 0 Since all variables are leading, the system Ax = c cannot have infinitely many solutions, but it could have a unique solution (for example, if c = b) or no solutions at all (compare with Example 3c).