Download Alex au pays des chiffres by Alex Bellos PDF

By Alex Bellos

ISBN-10: 2221122933

ISBN-13: 9782221122938

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Example text

It gives the unique solution for each of the two original linear systems. In particular, the solution for the system AX = B1 is (6, −51, 21), and the solution for 79 the system AX = B2 is ( 35 3 , −98, 2 ). (11) (b) Any nonhomogeneous system with two equations and two unknowns that has a unique solution will serve as a counterexample. For instance, consider x + y x − y = = 1 . 1 This system has a unique solution: (1, 0). Let (s1 , s2 ) and (t1 , t2 ) both equal (1, 0). Then the sum of solutions is not a solution in this case.

10 19 4 0 1 4 − 27 7 7 7 ⎢ ⎥ ⎢ 0 0 1 0 ⎥ 3 −15 −32 ⎢ ⎥ 2 ← (7) 2 ⎢ 0 0 6 12 20 2 0 ⎥ ⎣ ⎦ 7 7 7 7 10 26 0 0 0 − 37 − 97 7 7 Next, we target the (1,3), (3,3), and (4,3) ⎡ 1 4 1 ← (− 47 ) × 2 + 1 ⎢ ⎢ 0 0 3 ← (− 27 ) × 2 + 3 ⎣ 0 0 4 ← ( 37 ) × 2 + 4 0 0 entries. 0 −2 1 3 0 0 0 0 10 21 −15 −32 6 12 −5 −10 ⎤ 0 0 ⎥ ⎥ 0 ⎦ 0 We try to move the pivot to the third column but cannot due to the zeroes in the (3,4) and (4,4) entries. Hence, the pivot moves to the (3,5) entry. We convert that to 1. ⎤ ⎡ 0 1 4 0 −2 10 21 ⎢ 0 0 1 0 ⎥ 3 −15 −32 ⎥ ⎢ 3 ← ( 16 ) 3 ⎣ 0 0 0 1 0 2 0 ⎦ 0 0 0 0 0 −5 −10 Next, we target the (1,5), (2,5), and (4,5) ⎡ 1 4 1 ← (−10) × 3 + 1 ⎢ 0 0 ⎢ 2 ← (15) × 3 + 2 ⎣ 0 0 4 ← (5) × 3 + 4 0 0 entries.

0 0 1 3 ⎦ 0 0 0 0 We target the entries above the pivots to obtain the reduced row echelon form matrix. First, we use the (2,2) entry as the pivot and target the (1,2) entry. 2 Next, we use the (3,3) entry as a pivot and target the (1,3) and (2,3) entries. ⎡ ⎤ 1 0 0 6 ⎢ 0 1 0 1 ← ( 15 −1 ⎥ 8 )× 3 + 1 ⎢ ⎥ ⎣ 0 0 1 3 ⎦ 2 ← ( 14 ) × 3 + 2 0 0 0 0 Ignoring the last row, which gives the equation 0 = 0, this matrix yields the linear system ⎧ = 6 ⎨ x1 = −1 , x2 ⎩ 3 x3 = producing the unique solution (6, −1, 3).

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Alex au pays des chiffres by Alex Bellos


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