Transcription of 18.06 Linear Algebra, Final Exam Solution
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18.06 Linear Algebra, Final Exam Solution
(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 . + x2 1 + x4 2 .. x= . 1 0 0 . 1 0 1. 2. (11 points) This problem nds the curve y = C + D 2 t which gives the best least squares t to the points (t, y) = (0, 6), (1, 4), (2, 0). (a) Write down the 3 equations that would be satis ed if the curve went through all 3 points. Answer: C + 1D = 6. C + 2D = 4. C + 4D = 0. (b) Find the coe cients C and D of the best curve y = C + D2 t . Answer: . 1 1 . T. 1 1 1 3 7. A 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 Ax = AT b : . 3 7 C 10 C 1 21 7 10 8. = gives = = . 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.
Note: As stated there is no solution (my apologies!). All solutions to AT Av = 0 are multiples of (1, 1, 1, 1) which rules out v1 = 1 and v4 = 0. Intended problem: I meant to solve the reduced equations using KCL only at nodes 2 and 3. In fact symmetry gives v2 = v3 = 1 2. Then the currents are w1 = w2 = w5 = w6 = 2 1 around the sides and w 3 ...
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