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Chapter 1 Letting X And The Prize In coins Cows
Section 1.2 b We have to solve the system x1~v1+x2~v2+x3~v3+~ b=~x or 1.2.42 We want to find m1, m2, m3such that m1+m2+m3= 1 and 1 1m11 2+m22 3+m34 1=2 2, that is, we have to solve the system […]
Chapter 1 Note That The Last Two Equations Are
Section 1.1 Chapter 1 Section 1.1 1.1.1x+ 2y= 1 2x+ 3y= 1 −2×1st equation →x+ 2y= 1 −y=−1÷(−1) → x+ 2y= 1 y= 1 −2×2nd equation →x=−1 y= 1 , so that (x, y) = (−1,1). 1.1.32x+ 4y= 3 3x+ […]
Chapter 1 Undefined since the two vectors do not
Chapter 1 Figure 1.13: for Problem 1.3.5. 1.3.9 1 2 3 4 5 6 7 8 9 x y z = 1 4 9 1.3.13 1 2 3 4 7 11 […]
Chapter 2 every point on this line can be described as
Chapter 2 can let ~v =~e1and ~w =~e2instead.) 2.2.19 T(~e1) = ~e1,T(~e2) = ~e2, and T(~e3) = ~ 0, so that the matrix is 1 0 0 0 1 0 0 0 0 . 2.2.20 T(~e1) = […]
Chapter 2 The matrix A represents a reflection
Section 2.4 2.4.62 The determinant of Ais 1 and A−1=1 1 0 1 . Both Aand A−1represent horizontal shears. The determinant of Ais the area of the parallelogram spanned by ~v =1 0and ~w =−1 1. The angle from ~v […]
Chapter 2 we find that this equation has no solutions
Chapter 2 FJ JF and 2.3.60 Proceeding as in Exercise 59, we find that this equation has no solutions. 2.3.61 We need to solve the matrix equation 1 2 3 0 1 2 a b c d e […]
Chapter 2 We have to attempt to solve the equation
Chapter 2 Chapter 2 Section 2.1 2.1.1Not a linear transformation, since y2=x2+ 2 is not linear in our sense. 2.1.5By Theorem 2.1.2, the three columns of the 2 ×3 matrix Aare T(~e1), T (~e2), and T(~e3), so that A=7 6 […]
Chapter 3 Since these three vectors are parallel
Section 3.1 Chapter 3 Section 3.1 3.1.1Find all ~x such that A~x =~ 0: “1 2. . . 0 3 4. . . 0 #−→ “1 0. . . 0 0 1. . . 0 #, so that x1=x2= 0. […]
Chapter 3 We initially see that the first column is redunant
Section 3.3 3.3.17 For this problem, we again successively use Kyle Numbers to find our kernel, investigating the columns from left to right. We initially see that the first column is redunant: 1 0 0 0 0 0 0 1 […]
Chapter 3 We will use the commutative diagram method
Section 3.4 ~x =c11 2+c2−2 1−−→ T T (~x) = A~x =c1A1 2+c2A−2 1 =c15 10 +c210 −5= 5c11 2−5c2−2 1 3.4.21 aS=1−2 3 1 , and we find the inverse S−1to be equal to […]
Chapter 4 If A is the standard basis considered in Exercise
Section 4.3 4.3.44 a If Ais the standard basis considered in Exercise 13 and Bis the basis in Exercise 14, then S= 1 0 −1 0 A0 1 0−1A1 0 2 0 A0 1 0 2 A= 1 […]
Chapter 4 Since the spaces have the same dimension
Section 4.2 4.2.73 Yes. Tis an isomorphism; the inverse transformation is D(f(t)) = f′(t) = df dt , the derivative. We will check that the composite of Twith Dis the identity, in either order. Indeed D(T(f(t))) = d dt(T(f(t)) = […]
Chapter 4 Not a subspace since it does not contain
Section 4.1 Chapter 4 Section 4.1 4.1.1Not a subspace since it does not contain the neutral element, that is, the function f(t) = 0, for all t. 4.1.2This subset Vis a subspace of P2: •The neutral element f(t) = 0 […]
Chapter 5 Kepler’s third law of planetary motion
Section 5.4 5.4.36 We want a b c such that 5.4.37 a We want c0, c1such that c0+c1(35) = log 35 c0+c1(46) = log 46 c0+c1(59) = log 77 c0+c1(69) = log 133 or 1 […]
Chapter 5 representing a rotation or a reflection
Section 5.3 5.3.32 a No! As a counterexample, consider A= 1 0 0 1 0 0 (see Exercise 30). b Yes! More generally, if Aand Bare n×nmatrices such that BA =In, then AB =In, by Theorem 2.4.8c. 5.3.33 […]
Chapter 5 The idea is to perform the Gram-Schmidt process
Section 5.1 Chapter 5 Section 5.1 5.1.1k~vk=√72+ 112=√49 + 121 = √170 ≈13.04 5.1.2k~vk=√22+ 32+ 42=√4 + 9 + 16 = √29 ≈5.39 5.1.7Use the fact that ~u ·~v =k~ukk~vkcos θ, so that the angle is acute if ~u ·~v […]
Chapter 6 Take determinants on both sides of the equation
Chapter 6 We will evaluate the determinant of Bby expanding across the ith row (where iis neither pnor q). 6.2.61 We follow the hint: In0 −C A A B C D =A B −CA +AC −CB +AD =A B […]
Chapter 6 Will ensure that this matrix is invertible
Chapter 6 Chapter 6 Section 6.1 6.1.1Fails to be invertible; since det1 2 3 6 = 6 −6 = 0. 6.1.4Fails to be invertible; since det1 4 2 8 = 8 −8 = 0. 6.1.5Invertible; since det 257 0 […]
Chapter 7 Clearly Let’s Find The Kernel First Part
Chapter 7 Chapter 7 Section 7.1 7.1.1If ~v is an eigenvector of A, then A~v =λ~v. Hence A3~v =A2(A~v) = A2(λ~v) = A(Aλ~v) = A(λA~v) = A(λ2~v) = λ2A~v =λ3~v, so ~v is an eigenvector of A3with eigenvalue λ3. 7.1.4We […]
Chapter 7 Rotation-scaling matrices commute when multiplied
Section 7.6 7.5.45 If a6= 0, then there are two distinct eigenvalues, 1 ±√a, so that the matrix is diagonalizable. If a= 0, then 1 1 a1=1 1 0 1 fails to be diagonalizable. 7.5.47 If a6= 0, then there […]
Chapter 7 The Eigenvalues Are And Corresponding Eigenvectors
Chapter 7 7.2.46 aλ2 1+λ2 2= (λ1+λ2)2−2λ1λ2= (trA)2−2 det(A) = (a+d)2−2(ad −bc) = a2+d2+ 2bc. b Based on part (a), we need to show that a2+d2+ 2bc ≤a2+b2+c2+d2,or 2bc ≤b2+c2,or 0 ≤(b−c)2. But the last inequality is obvious. c By […]
Chapter 7 the other must be its complex conjugate
Chapter 7 7.4.49 The matrix of Twith respect to the standard basis 1, x, x2is B= 1−1 1 0 3 −6 . The eigenvalues of Bare 7.4.50 The matrix of Twith respect to the standard basis 1, x, […]
Chapter 8 Note that A is symmetric and orthogonal
Section 8.1 Chapter 8 Section 8.1 8.1.1~e1,~e2is an orthonormal eigenbasis. 8.1.21 √21 1,1 √21 −1is an orthonormal eigenbasis. 8.1.5Eigenvalues −1, −1, 2 Choose ~v1=1 √2 −1 1 0 in E−1and ~v2=1 √3 1 1 1 in E2and let ~v3=~v1×~v2=1 […]
Chapter 8 The columns of LT give us a basis with the desired
Section 8.2 It is required that xand zbe positive. This system has the unique solution 8.2.33 Use the formulas for x,y,zderived in Exercise 32. x=√a=√8 = 2√2 y=b √a=−2 2√2=−1 √2 z=qac−b2 a=q36 8=3 √2,so that L=2√2 0 −1 √2 […]
Chapter 9 It appears that the trajectories will be circles
Section 9.1 Chapter 9 Section 9.1 9.1.1x(t) = 7e5t, by Theorem 9.1.1. 9.1.2x(t) = −e·e−0.71t=−e1−0.71t, by Theorem 9.1.1. 9.1.5y(t) = −0.8e0.8t, by Theorem 9.1.1. 9.1.6x dx =dt x2 2=t+C, and 1 2= 0 + C, so that x2 2=t+1 2 […]
Chapter 9 that the presence of xn contributes to the decrease
Section 9.2 (–1, 1) (0, 1) g(λ) = λ3 + 2λ2 + λ + 1 9.2.14 a For i > 1, dxi dt =−kixi+xi−1. This means that in the absence of quantity xi−1(t), the quantity xi(t) will decay exponentially, but […]