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Chapter 2 SSM: Linear Algebra Chapter 2 SSM: Linear Algebra Chapter 2 2.1 1. Not a linear transformation, since y2 = x2 + 2 is not linear in our sense. 3. Not linear, since y2 = x1x3 is nonlinear. 5. By Fact 2.1.2, the three columns of the 2 3 matrix A are T (~e1); T (~e2), and T (~e3), so that × 7 6 13 A = : 11 9 −17 x1 7. Note that x ~v + + xm~vm = [~v : : : ~vm] , so that T is indeed linear, with matrix 1 1 · · · 1 2 · · · 3 xm [~v1 ~v2 ~vm]. 4 5 · · · y 2 3 x 9. We have to attempt to solve the equation 1 = 1 for x and x . Reducing y 6 9 x 1 2 2 2 2x + 3x = y x + 1:5x = 0:5y the system 1 2 1 we obtain 1 2 1 . 6x + 9x = y 0 = 3y + y 1 2 2 − 1 2 No unique solution (x1; x2) can be found for a given (y1; y2); the matrix is nonin vertible. y1 1 2 x1 11. We have to attempt to solve the equation = for x1 and x2. Reducing y2 3 9 x2 2 x1 + 2x2 = y1 x1 = 3y1 3 y2 the system we find that − 1 . The 3x1 + 9x2 = y2 x = y + y 2 1 3 2 3 2 − inverse matrix is − 3 . 1 1 − 3 y a b x 13. a. First suppose that a = 0. We have to attempt to solve the equation 1 = 1 6 y2 c d x2 for x1 and x2. ax + bx = y a x + b x = 1 y 1 2 1 ÷ 1 a 2 a 1 cx1 + dx2 = y2 ! cx1 + dx2 = y2 c(I) ! − b 1 x1 + a x2 = a y1 (d bc )x = c y + y ! a 2 a 1 2 − − 28 SSM: Linear Algebra Section 2.1 b 1 x1 + a x2 = a y1 ad bc c ( −a )x2 = a y1 + y2 − W e can solve this system for x and x if (and only if) ad bc = 0, as claimed. 1 2 − 6 If a = 0, then we have to consider the system bx = y cx + dx = y 2 1 swap : I II 1 2 2 cx1 + dx2 = y2 $ bx2 = y1 We can solve for x1 and x2 provided that b oth b and c are nonzero, that is if bc = 0. Since a = 0, this means that ad bc = 0, as claimed. 6 − 6 b. First suppose that ad bc = 0 and a = 0. Let D = ad bc for simplicity. We continue our work in part (a): − 6 6 − b 1 x1 + a x2 = a y1 D x = c y + y a ! a 2 a 1 2 · D − b 1 b x1 + a x2 = a y1 a (II) c a − x2 = y1 + y2 ! − D D 1 bc b x1 = ( a + aD )y1 D y2 x = c y +− a y 2 D 1 D 2 − x = d y b y 1 D 1 − D 2 x = c y + a y 2 D 1 D 2 − 1 bc D+bc ad d Note that a + aD = aD = aD = D : 1 a b − 1 d b It follows that = ad bc − , as claimed. If ad bc = 0 and a = 0, c d − c a − 6 then we have tosolve the system − cx1+ dx2 = y2 c bx = y ÷b 2 1 ÷ d 1 d x1+ c x2 = c y2 c (II) x = 1 y − 2 b 1 x = d y + 1 y 1 − bc 1 c 2 x = 1 y 2 b 1 1 d 1 a b − bc c d b It follows that = − = 1 (recall that a = 0), as c d 1 ad bc c −a b 0 − claimed. − 29 Chapter 2 SSM: Linear Algebra a b 15. By Exercise 13a, the matrix − is invertible if (and only if) a2 + b2 = 0, which b a 6 a b 1 a b is the case unless a = b = 0. If is invertible, then its inverse is 2 2 , b −a a +b b a by Exercise 13b. − 1 0 17. If A = − , then A~x = ~x for all ~x in R2, so that A represents a reflection about 0 1 − the origin. − 1 This transformation is its own inverse: A− = A. 1 0 x x 19. If A = , then A 1 = 1 , so that A represents the orthogonal projection onto 0 0 x 0 2 the ~e1 axis. (See Figure 2.1.) This transformation is not invertible, since the equation 1 A~x = has infinitely many solutions ~x. 0 Figure 2.1: for Problem 2.1.19 . 21. Compare with Example 5. 0 1 x x If A = , then A 1 = 2 . Note that the vectors ~x and A~x are perpen- 1 0 x x − 2 − 1 dicular and have the same length. If ~x is in the first quadrant, then A~x is in the fourth. Therefore, A represents the rotation through an angle of 90◦ in the clockwise direction. 0 1 (See Figure 2.2.) The inverse A 1 = represents the rotation through 90 in − 1 −0 ◦ the counterclockwise direction. 23. Compare with Exercise 21. 0 1 Note that A = 2 , so that A represents a rotation through an angle of 90 in the 1 0 ◦ clockwise direction, −followed by a scaling by the factor of 2. 30 SSM: Linear Algebra Section 2.1 Figure 2.2: for Problem 2.1.21 . 0 1 The inverse A 1 = 2 represents a rotation through an angle of 90 in the − 1 −0 ◦ 2 1 counterclockwise direction, followed by a scaling by the factor of 2 . 25. The matrix represents a scaling by the factor of 2. (See Figure 2.3.) Figure 2.3: for Problem 2.1.25 . 27. This matrix represents a reflection about the ~e1 axis. (See Figure 2.4.) 29. This matrix represents a reflection about the origin. Compare with Exercise 17. (See Figure 2.5.) x 31. The image must be reflected about the ~e axis, that is 1 must be transformed 2 x 2 x into 1 : This can be accomplished by means of the linear transformation T (~x) = −x 2 31 Chapter 2 SSM: Linear Algebra Figure 2.4: for Problem 2.1.27 . Figure 2.5: for Problem 2.1.29 . 1 0 ~x. −0 1 1 0 33. By Fact 2.1.2, A = T T . (See Figure 2.6.) 0 1 Figure 2.6: for Problem 2.1.33 . 32 SSM: Linear Algebra Section 2.1 1 1 p p Therefore, A = 2 − 2 . 2 1 1 3 p2 p2 4 5 a b 5 89 6 88 35. We want to find a matrix A = such that A = and A = . c d 42 52 41 53 5a + 42b = 89 6a + 41b = 88 This amounts to solving the system . 5c + 42d = 52 6c + 41d = 53 (Here we really have two systems with two unknowns each.) 1 2 The unique solution is a = 1; b = 2; c = 2, and d = 1, so that A = . 2 1 37. Since ~x = ~v + k(w~ ~v), we have T (~x) = T (~v + k(w~ ~v)) = T (~v) + k(T (w~) T (~v)), by Fact 2.1.3 − − − Since k is between 0 and 1, the tip of this vector T (~x) is on the line segment connecting the tips of T (~v) and T (w~). (See Figure 2.7.) Figure 2.7: for Problem 2.1.37 . x x 1 T (~e ) : : : T (~e ) 1 39. By Fact 2.1.2, we have T : : : = 1 m : : : = x T (~e ) + + 2 3 2 3 2 3 1 1 · · · xm xm 4 5 6 7 4 5 xmT (~em). 4 5 33 Chapter 2 SSM: Linear Algebra x1 41. These linear transformations are of the form [y] = [a b] , or y = ax1 + bx2. The x2 graph of such a function is a plane through the origin. 2 x1 x1 43. a. T (~x) = 3 x = 2x + 3x + 4x = [2 3 4] x 2 3 · 2 2 3 1 2 3 2 2 3 4 x3 x3 4 5 4 5 4 5 The transformation is indeed linear, with matrix [2 3 4]. v1 b. If ~v = v , then T is linear with matrix [v v v ], as in part (a). 2 2 3 1 2 3 v3 4 5 x1 x1 c. Let [a b c] be the matrix of T . Then T x = [a b c] x = ax + bx + cx = 2 2 3 2 2 3 1 2 3 x3 x3 a x1 a 4 5 4 5 b x , so that ~v = b does the job. 2 3 · 2 2 3 2 3 c x3 c 4 5 4 5 4 5 45. Yes, ~z = L(T (~x)) is also linear, which we will verify using Fact 2.1.3. Part a holds, since L(T (~v + w~)) = L(T (~v) + T (w~)) = L(T (~v)) + L(T (w~)), and part b also works, because L(T (k~v)) = L(kT (~v)) = kL(T (~v)). 47. Write w~ as a linear combination of ~v1 and ~v2 : w~ = c1~v1 + c2~v2. (See Figure 2.8.) Figure 2.8: for Problem 2.1.47 . Measurements show that we have roughly w~ = 1:5~v1 + ~v2. 34 SSM: Linear Algebra Section 2.1 Figure 2.9: for Problem 2.1.47 . Therefore, by linearity, T (w~) = T (1:5~v1 + ~v2) = 1:5T (~v1) + T (~v2). (See Figure 2.9.) 49. a. Let x1 be the number of 2 Franc coins, and x2 be the number of 5 Franc coins.
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