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Solving a system of equations

Solving a system of equations. 1 x 1 + 3 x 2 + 6 x 3 = 3 2 x 1 + 4 x 2 + 7 x 3 = 2 3 x 1 + 5 x 2 + 8 x 3 = 3. Solving a system of equations. 1 x 1 + 3 x 2 + 6 x 3 = 3 2 x 1 + 4 x 2 + 7 x 3 = 2 3 x 1 + 5 x 2 + 8 x 3 = 3. 1. -2. 1. 0 = 2. no solution.

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Solving a system of equations

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  1. Solving a system of equations 1 x1 + 3 x2 + 6 x3 = 3 2 x1 + 4 x2 + 7 x3 = 2 3 x1 + 5 x2 + 8 x3 = 3

  2. Solving a system of equations 1 x1 + 3 x2 + 6 x3 = 3 2 x1 + 4 x2 + 7 x3 = 2 3 x1 + 5 x2 + 8 x3 = 3 1 -2 1 0 = 2 no solution

  3. Solving a system of equations a11 x1 + a12 x2 + a13 x3 = b1 a21 x1 + a22 x2 + a23 x3 = b2 a31 x1 + a32 x2 + a33 x3 = b3 A x = b has a solution iff rank (A) = rank (A b)

  4. Solving a system of equations A x = b has a solution iff rank (A) = rank (A b) rank (A) < rank (A b) then  y such that yT A = 0 and yT b  0

  5. Solving a system of equations A x = b has a solution iff rank (A) = rank (A b) rank (A) < rank (A b) then  y such that yT A = 0 and yT b  0 y is a witness that A x = b does not have a solution

  6. Solving a system of equations yT A = 0 and yT b  0 y is a witness that A x = b does not have a solution A x = b  yT (A x) = yT b  (yT A) x = yT b  0 = yT b  a contradiction

  7. Solving a system of equations Theorem: If Ax=b doesn’t have solution then  y such that yT A =0 and yT b  0

  8. Solving a system of equations 1 x1 + 3 x2 + 6 x3 = 3 2 x1 + 4 x2 + 7 x3 = 2 3 x1 + 5 x2 + 8 x3 = 1

  9. Solving a system of equations 1 x1 + 3 x2 + 6 x3 = 3 2 x1 + 4 x2 + 7 x3 = 2 3 x1 + 5 x2 + 8 x3 = 1 x1 = -3 x2 = 2 x3 = 0

  10. Back to linear programming 1 x1 + 3 x2 + 6 x3 = 3 2 x1 + 4 x2 + 7 x3 = 2 3 x1 + 5 x2 + 8 x3 = 1 x1  0 x2  0 x3  0 I.e., we want a non-negative solution.

  11. Back to linear programming 1 x1 + 3 x2 + 6 x3 = 3 2 x1 + 4 x2 + 7 x3 = 2 3 x1 + 5 x2 + 8 x3 = 1 0 -1 1 x1 + x2 + x3 = -1 x1  0 x2  0 x3  0 I.e., we want a non-negative solution.

  12. Back to linear programming yTA  0 and yTb < 0 y is a witness that A x = b, x  0 does not have a solution A x = b  yT(A x) = yTb  (yTA) x = yTb non-negative = negative, a contradiction

  13. Back to linear programming Theorem (Farkas): If Ax=b, x  0 doesn’t have a solution then  y such that yTA  0 and yTb < 0 Theorem: If Ax=b doesn’t have solution then  y such that yTA =0 and yTb  0

  14. Idea of the proof Theorem (Farkas): If Ax=b, x  0 doesn’t have a solution then  y such that yTA  0 and yTb < 0 Ax, x  0 b

  15. S=convex, b not in S c such that ( x S)cT x > cT b Idea of the proof Theorem (Farkas): If Ax=b, x  0 doesn’t have a solution then  y such that yTA  0 and yTb < 0 cTx Ax, x  0 b separating hyperplane

  16. Theorem (Farkas): If Ax=b, x  0 doesn’t have a solution  y such that yTA  0 and yTb < 0 Duality max cT x Ax=b x  0 min yT b yT A  cT

  17. “” “=” and non-negativity a1 x1 + ... + an xn b a1 x1 + ... + an xn= b + y, y  0 a1 x1 + ... + an xn – y = b, y  0

  18. “=” “” a1 x1 + ... + an xn= b a1 x1 + ... + an xn b a1 x1 + ... + an xnb a1 x1 + ... + an xn b -a1 x1 - ... - an xn -b

  19. Duality max c1Tx1+c2Tx2 + c3Tx3 + c4Tx4 A1 x1=b1 A2 x2  b2 A3 x3 = b3 A4 x4  b4 x1  0,x2  0 min y1Tb1+y2Tb2+y3Tb3 + y4Tb4 y1T A1 c1T y2T A2 c2T y3T A3 = c3T y4T A4 = c4T y2 0, y4 0

  20. Solving linear programs Simplex (Danzig, 40’s) Ellipsoid (Khachiyan, 80’s) Interior point (Karmakar, 80’s)

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