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pset1-s09-soln

pset1-s09-soln - 18.06 Problem Set 1 Solution Due Wednesday...

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18.06 Problem Set 1 Solution Due Wednesday, 11 February 2009 at 4 pm in 2-106. Total: 145 points Problem 1: If k ~v k = 7 and k ~w k = 3, what are the smallest and largest possible values of k ~v + ~w k and ~v · ~w ? Solution (10 points) k ~v + ~w k ≤ k ~v k + k ~w k = 10. k ~v + ~w k ≥ k ~v k - k ~w k = 4. (5pts) | ~v · ~w | ≤ k ~v k · k ~w k = 21. So - 21 ~v · ~w 21. (5pts) The maximum is achieved when the vectors ~v and ~w are parallel and pointing to the same direction, for example, ~v = (7 , 0 , 0 , . . . ) and ~w = (3 , 0 , 0 , . . . ); the minimum is achieved when they are parallel but pointing to opposite directions, for instance, ~v = (7 , 0 , 0 , . . . ) and ~w = ( - 3 , 0 , 0 , . . . ). REMARK: We should try not to restrict ourselves to the 3-dimensional case. The statement of this problem works for vectors in arbitrary dimensional space. Problem 2: Let A and B be 4 × 4 matrices, and divide each of them into 2 × 2 chunks via A = A 1 A 2 A 3 A 4 and B = B 1 B 2 B 3 B 4 , where A 1 is the upper-left 2 × 2 corner, A 2 is the upper-right 2 × 2 corner, and so on. Let C = AB be the 4 × 4 product of A and B , and similarly divide C into 2 × 2 chunks as C = C 1 C 2 C 3 C 4 . (a) Give formulas for these 2 × 2 chunks C 1 ... 4 in terms of matrix products and sums of the chunks A 1 ... 4 and B 1 ... 4 (your final formulas should not reference the individual numbers within those chunks). (b) Justify your formulas by an example (come up with 4 × 4 matrices A and B with nonzero entries, multiply them to get C , and compare to your formulas in terms of 2 × 2 chunks—it is acceptable to check just one of the output 2 × 2 chunks, say C 2 ). Solution (15 points = 10+5) (a) Similar to usual matrix multiplication, matrix multiplication by blocks for- mally has the same form. C 1 = A 1 B 1 + A 2 B 3 , C 2 = A 1 B 2 + A 2 B 4 , C 3 = A 3 B 1 + A 4 B 3 , C 4 = A 3 B 2 + A 4 B 4 . 1
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In other words, A 1 A 2 A 3 A 4 B 1 B 2 B 3 B 4 = A 1 B 1 + A 2 B 3 A 1 B 2 + A 2 B 4 A 3 B 1 + A 4 B 3 A 3 B 2 + A 4 B 4 NOTE: the order of A * and B * cannot be reversed. (b) For example, A = 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 0 , B = 0 1 0 1 1 1 0 1 1 0 1 1 1 0 1 0 . Then C = AB = 1 1 1 1 3 1 2 2 2 2 1 2 2 1 1 2 and C 1 = 1 0 0 1 0 1 1 1 + 0 1 1 1 1 0 1 0 = 0 1 1 1 + 1 0 2 0 = 1 1 3 1 , C 2 = 1 0 0 1 0 1 0 1 + 0 1 1 1 1 1 1 0 = 0 1 0 1 + 1 0 2 1 = 1 1 2 2 , C 3 = 1 1 0 1 0 1 1 1 + 0 1 1 0 1 0 1 0 = 1 2 1 1 + 1 0 1 0 = 2 2 2 1 , C 4 = 1 1 0 1 0 1 0 1 + 0 1 1 0 1 1 1 0 = 0 2 0 1 + 1 0 1 1 = 1 2 1 2 . Problem 3: Invent a 3 × 3 “magic” matrix M 3 with entries 0 , 1 , . . . , 8, such that all rows and columns and diagonals add to 12 (e.g. the first row could be 7,2,3). Compute the products: M 3 1 1 1 , ( 1 1 1 ) M 3 , M 3 1 2 3 1 2 3 1 2 3 , M 3 1 1 1 2 2 2 3 3 3 . Solution (10 points) 2
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For example, take M 3 = 7 0 5 2 4 6 3 8 1 , we have M 3 1 1 1 = 12 12 12 , ( 1 1 1 ) M 3 = ( 12 12 12 ) , M 3 1 2 3 1 2 3 1 2 3 = 12 24 36 12 24 36 12 24 36 , M 3 1 1 1 2 2 2 3 3 3 = 22 22 22 28 28 28 22 22 22 .
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