(d) If the vector b is the sum of the four columns of A, write down the complete solution to Ax = b. Answer: � ⎡ � ⎡ � ⎡ 1 −2 3 � ⎢ � ⎢ � ⎢ � 1 ⎢ � 1 ⎢ � −2 ⎢x = + x2 + x4 � 1 ⎣ � 0 ⎣ � 0 ⎣ 1 0 12. (11 points) This problem finds the curve y = C + D 2t which gives the best least squares fit to the points (t, y) = (0, 6), (1, 4), (2, 0). (a) Write down the 3 equations that would be satisfied if the curve went through all 3 points. Answer: C + 1D = 6 C + 2D = 4 C + 4D = 0 (b) Find the coefficients C and D of the best curve y = C + D2t . Answer: � ⎡ ⎦ � 1 1 ⎦ � 1 1 1 � ⎢ 3 7 ⎢AT A = � 1 2 = � ⎣1 2 4 7 21 1 4 � ⎡ ⎦ � 6 ⎦ � 1 1 1 � ⎢ 10 AT b = � 4 ⎢ = � ⎣1 2 4 14 0 Solve AT Axˆ = AT b : ⎤ ⎥⎤ ⎥ ⎤ ⎥ ⎤ ⎥ ⎤ ⎥⎤ ⎥ ⎤ ⎥ 3 7 C 10 C 121 10 8 = gives = −7= . 7 21 D 14 D 14 −7 3 14 −2 (c) What values should y have at times t = 0, 1, 2 so that the best curve is y = 0? Answer: The projection is p = (0, 0, 0) if AT b = 0. In this case, b = values of y = c(2, −3, 1).3. (11 points) Suppose Avi = bi for the vectors v1, . . . , vn and b1, . . . , bn in Rn . Put the v’s into the columns of V and put the b’s into the columns of B. (a) Write those equations Avi = bi in matrix form. What condition on which vectors allows A to be determined uniquely? Assuming this condition, find A from V and B. Answer: A [v1 vn] = [b1 bn] or AV = B. Then A = BV −1 if the v�s are independent. ··· ··· (b) Describe the column space of that matrix A in terms of the given vectors. Answer: The column space of A consists of all linear combinations of b1, , bn.··· (c) What additional condition on which vectors makes A an invertible matrix? Assuming this, find A−1 from V and B. Answer: If the b�s are independent, then B is invertible and A−1 = V B−1 .� � � � � � 4. (11 points) (a) Suppose xk is the fraction of MIT students who prefer calculus to linear algebra at year k. The remaining fraction yk = 1 − xk prefers linear algebra. At year k + 1, 1/5 of those who prefer calculus change their mind (possibly after taking 18.03). Also at year k + 1, 1/10 of those who prefer linear algebra change their mind (possibly because of this exam). xk+1 xk 1 Create the matrix A to give = A and find the limit of Ak as k � �. yk+1 yk 0 Answer: A = ⎤ .8 .2 .1 .9 ⎥ . ⎤ 1/3 ⎥ The eigenvector with � = 1 is . 2/3 ⎤ ⎥ 1 This is the steady state starting from . 0 2 of all students prefer linear algebra! I agree. 3 (b) Solve these differential equations, starting from x(0) = 1, y(0) = 0 : dx dt = 3x − 4y dy dt = 2x − 3y . Answer: ⎤ ⎥ A = 3 2 −4 −3 . has eigenvalues �1 = 1 and �2 = −1 with eigenvectors x1 = (2, 1) and x2 = (1, 1). The initial vector (x(0), y(0)) = (1, 0) is x1 − x2. So the solution is (x(t), y(t)) = et(2, 1) + e−t(1, 1).� � � � x(0) x(t)(c) For what initial conditions does the solution to this differential equation y(0) y(t) lie on a single straight line in R2 for all t? Answer: If the initial conditions are a multiple of either eigenvector (2, 1) or (1, 1), the solution is at all times a multiple of that eigenvector.5. (11 points) (a) Consider a 120� rotation around the axis x = y = z. Show that the vector i = (1, 0, 0) is rotated to the vector j = (0, 1, 0). (Similarly j is rotated to k = (0, 0, 1) and k is rotated to i.) How is j − i related to the vector (1, 1, 1) along the axis? Answer: � ⎡ −1 j − i = � 1 ⎣ 0 � ⎡ 1 is orthogonal to the axis vector � 1 ⎣ . 1 So are k − j and i − k. By symmetry the rotation takes i to j, j to k, k to i. (b) Find the matrix A that produces this rotation (so Av is the rotation of v). Explain why A3 = I. What are the eigenvalues of A? Answer: A3 = I because this is three 120� rotations (so 360�). The eigenvalues satisfy �3 = 1 so � = 1, e2�i/3, e−2�i/3 = e4�i/3 . (c) If a 3 by 3 matrix P projects every vector onto the plane x+2y+z = 0, find three eigenvalues and three independent eigenvectors of P. No need to compute P . Answer: The plane is perpendicular to the vector (1, 2, 1). This is an eigenvector of P with � = 0. The vectors (−2, 1, 0) and (1, −1, 1) are eigenvectors with � = 0.6. (11 points) This problem is about the matrix � ⎡ 1 2 A = � 2 4 ⎣ . 3 6 (a) Find the eigenvalues of AT A and also of AAT . For both matrices find a complete set of orthonormal eigenvectors. Answer: AT A = ⎤ 1 2 2 4 3 6 ⎥ � � 1 2 3 2 4 6 ⎡ ⎣ = ⎤ 14 28 28 56 ⎥ 1 ⎤ 1 ⎥ 1 ⎤ ⎥ −2 has �1 = 70 and �2 = 0 with eigenvectors x1 = �5 2 and x2 = �51 . � ⎡ � ⎡ 1 2 ⎤ ⎥ 5 10 15 1 2 3 AAT = � 2 4 ⎣ = � 10 20 30 ⎣ has �1 = 70, �2 = 0, �3 = 0 with 2 4 6 3 6 15 30 45 � ⎡ � ⎡ � ⎡ 1 −2 3 1 1 1 x1 = � 2 ⎣ and x2 = � 1 ⎣ and x3 = � 6 ⎣ .�14 �5 �703 0 −5 (b) If you apply the Gram-Schmidt process (orthonormalization) to the columns of this matrix A, what is the resulting output? Answer: Gram-Schmidt will find the unit vector � ⎡ 11 q1 = �14 � 23 ⎣ . But the construction of q2 fails because column 2 = 2 (column …
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