Note: Native PDF, vector graphics, searchable, bookmarked per chapter. Information refers to the textbook that accompanies this solution manual.
about the y-axis. (See Figure 2.45.) −1 0 0 1 0 1 represents a reflection about the line x = y, while GC = 1 0 about the line x = −y. (See Figure 2.46.) 0 −1 2.3.49 AF = 2.3.50 CG = Figure 2.47.) . We have . 2.3.44 A rotation by π/2 given by the matrix A = 2.3.51 F J = JF = 1 1 −1 −1 1 −1 −1 0 represents the reflection represents a reflection both represent a rotation through 3π/4 combined with a scaling by 83 √ 2. (See Chapter 2 A F AF A F FA Figure 2.45: for Problem
2.4.20 Solving for x1 , x2 , and x3 in terms of y1 , y2 , and y3 we find that x1 x2 x3 = −8y1 − 15y2 + 12y3 = 4y1 + 6y2 − 5y3 = −y1 − y2 + y3 2.4.21 f (x) = x2 fails to be invertible, since the equation f (x) = x2 = 1 has two solutions, x = ±1. 91 Chapter 2 2.4.22 f (x) = 2x fails to be invertible, since the equation f (x) = 2x = 0 has no solution x. 2.4.23 Note that f ′ (x) = 3x2 + 1 is always positive; this implies that the function f (x) = x3 + x is increasing throughout. Therefore, the
arbitrary constants. t 2 0 2 1 Since there are invertible solutions S (for example, let y = t = 1), the matrices and are indeed 0 3 0 3 similar. 2x 2y 3z 3t 2x 2z = −y x + 3y . The solutions are of the form S = 0 z + 3t 3.4.60 First we find the matrices S = y t x z y t x . The solutions are of the form S = z such that x z y t = x z y t 0 1 , or, 1 0 x −z y −t = y , where y and t are arbitrary constants. Since there are t 0 1 1 0 are indeed similar. and invertible solutions S
k 0 2 1 0 0 0 1 .. . 2 1 .. → . 0 0 .. −k(I) 0 . 1 6 − 4k 2k 0 2 − 6k + 4k 2 .. . .. . .. . 2 0 1 − 2k 2 6 1 2k −2k 2 − 6k 1 0 → 0 1 0 0 .. . .. . .. . .. . 0 .. swap : . 2 I ↔ II → .. . 1 1 2k 2 6 0 2 2 0 1 − 2k −2(II) → +2k(II) 6 − 4k 2k 2(2k − 1)(k − 1) .. . 2 .. . . 0 .. . −(2k − 1) We see that there will be a unique solution when the 2(2k − 1)(k − 1) term is not equal to zero, when 2k − 1 = 0 and k −
is the conic 25−10x−10y +x2 +y 2 = 0, a circle of radius 5 centered at (5, 5). y 12 8 (5, 5) 4 x -4 4 -4 26 8 12 Section 1.2 1.2.58 The system of linear equations is c1 + c2 + c4 = 0 c1 + 2c2 + 4c4 c1 + 2c2 + 2c3 + 4c4 + 4c5 + 4c6 = = 0 0 c1 + 5c2 + 2c3 + 25c4 + 10c5 + 4c6 c1 + 5c2 + 6c3 + 25c4 + 30c5 + 36c6 = = 0 0 It has the solution (c1 , c2 , c3 , c4 , c5 , c6 ) = (2c6 , −3c6 , 2c6 , c6 , −2c6 , c6 ). This is the conic 2−3x+2y+x2 −2xy+y 2 = 0. y 6 4 2 x -2 2 4 6 -2
