Chapter 2.Solution of Linear Systems AX=B 华南师范大学数学科学学院谢珊玲
Chapter 2. Solution of Linear Systems AX=B 华南师范大学数学科学学院 谢骊玲
Example-Material Selection Manufacturing a product containing Manganese,silicon,and copper Required amounts [15;22;39]Ibs/ton manufactured 3 supplies exist with different concentrations of each ingredient Problem:How much to buy from each supplier? 华南师范大学数学科学学院谢删玲
Example – Material Selection ◼ Manufacturing a product containing Manganese, silicon, and copper Required amounts [15; 22; 39] lbs/ton manufactured 3 supplies exist with different concentrations of each ingredient ◼ Problem: How much to buy from each supplier? 华南师范大学数学科学学院 谢骊玲
Example-Continued X;=amount to buy from supplierj C=amount of ingredient i required per ton of product manufactured a=amount of ingredient i in each ton from supplier j Supplier 1 Supplier 2 Supplier 3 (lb/ton) (lb/ton) (Ib/ton) Manganese 1 3 2 Silicon 2 4 3 Copper 3 4 7 华南师范大学数学科学学院谢骊玲
Example - Continued ◼ Xj = amount to buy from supplier j ◼ Ci = amount of ingredient i required per ton of product manufactured ◼ aij = amount of ingredient i in each ton from supplier j Supplier 1 (lb/ton) Supplier 2 (lb/ton) Supplier 3 (lb/ton) Manganese 1 3 2 Silicon 2 4 3 Copper 3 4 7 华南师范大学数学科学学院 谢骊玲
Example -Cont. Relation among composition from suppliers,the amount needed from each supplier,and the composition of the final product 含4-ca12g X1+3X2+2X3=15 2X1+4X2+3X=22 3X,+4X2+7X3=39 华南师范大学数学科学学院谢骊玲
Example – Cont. ◼ Relation among composition from suppliers, the amount needed from each supplier, and the composition of the final product 1,2,3 3 1 = = = a X C for i i j ij j 3 4 7 39 2 4 3 22 3 2 15 1 2 3 1 2 3 1 2 3 + + = + + = + + = X X X X X X X X X 华南师范大学数学科学学院 谢骊玲
Example-Truss Truss-structure consisting of rigid members interconnected by Joints ■Example B 92 华南师范大学数学科学学院谢卿玲
Example - Truss ◼ Truss – structure consisting of rigid members interconnected by joints ◼ Example g1 g2 f f 8 4 f f5 5 f3 f2 f2 f6 f7 f1 f1 f3 f4 A B C D q f 华南师范大学数学科学学院 谢骊玲
Example -Cont. ■( Given values of the load forces,g and g2,find the values of the forces in the beams,fi,..../s ficos0+f2-fo=0 cos0 10 0 0-100f 0 fsin0+f方=0 sin 0 0 0 0010 2 0 -ficos0+f cosp+g=0 -cos0 0 0 coso 0 0 00 万 81 -fsin-∫-f4sinp=0 -sin 0 -1 -sin中 0 000 0 -3+∫5=0 0 -1 0 0 1 000 0 ∫-82=0 0 0 1 0 0 0 0 0 82 -fa coso-fs =0 0 0 0 -cosφ -1 0 0 0 f fasinp+fs=0 0 00 sino 0 00 16 0 华南师范大学数学科学学院谢骊玲
Example – Cont. ◼ Given values of the load forces, g1 and g2 , find the values of the forces in the beams, f1 , …, f8 − = − − − − − − − − 0 0 0 0 0 0 0 0 0 sin 0 0 0 1 0 0 0 cos 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 sin 0 1 sin 0 0 0 0 cos 0 0 cos 0 0 0 0 sin 0 0 0 0 0 1 0 cos 1 0 0 0 1 0 0 2 1 8 7 6 5 4 3 2 1 g g f f f f f f f f f f q f q f q q 华南师范大学数学科学学院 谢骊玲 sin 0 cos 0 0 0 sin sin 0 cos cos 0 sin 0 cos 0 4 8 4 5 3 2 2 5 1 3 4 1 4 1 1 7 1 2 6 + = − − = − = − + = − − − = − + + = + = + − = f f f f f g f f f f f f f g f f f f f q q q q
Systems of Linear Equations Consider the linear system Ax=b where A is an (nX n)matrix,x is the vector of (n)unknown solution values,and b is a column vector of constants a11 412 aln 为 21 a22 a2n 2 anl an2 an八xn a11+2x2+…+a1nxn=b1 a21+a22x2+…+a2nxn=b2 anlx1 an2x2 +.annxn bn 华南师范大学数学科学学院谢珊玲
Systems of Linear Equations ◼ Consider the linear system Ax = b ◼ where A is an (n × n) matrix, x is the vector of (n) unknown solution values, and b is a column vector of constants = n n nn n n n n b b b x x x a a a a a a a a a 2 1 2 1 1 2 21 22 2 11 12 1 n n nn n n n n n n a x a x a x b a x a x a x b a x a x a x b + + + = + + + = + + + = 1 1 2 2 21 1 22 2 2 2 11 1 12 2 1 1 华南师范大学数学科学学院 谢骊玲
Solution of a Linear System A formal way to obtain a solution using matrix algebra is to multiply each side of the equation by the inverse of 4 to yield A-Ax=A-Ib x=A-b 华南师范大学数学科学学院谢删玲
Solution of a Linear System ◼ A formal way to obtain a solution using matrix algebra is to multiply each side of the equation by the inverse of A to yield A Ax A b −1 −1 = x A b −1 = 华南师范大学数学科学学院 谢骊玲
Example 2 unknowns in 2 equations 14 3x灯+2x2=7 12 4刈+x2=1 + Solution 3 x2=-4灯1+1 华南师范大学数学科学学院谢骊玲
Example ◼ 2 unknowns in 2 equations -4 -2 0 2 4 6 8 10 12 14 -4 - 3 -2 - 1 0 1 2 x y Solution 4 1 3 2 7 1 2 1 2 x x x x + = + = 2 7 2 3 2 1 x x + − = 4 1 2 1 x = − x + 华南师范大学数学科学学院 谢骊玲
Upper-Triangular Linear Systems Def.2.1.An NXN matrix A=[a;]is called upper triangular provided that the elements satisfy a;-0 whenever ij.A is called lower triangular provided that a,0 wheneverij. Thm.2.1(Back Substitution).Suppose that AX-B is an upper- triangular system with the form as below.If a for k=1,2,....N,then there exists a unique solution for the system aux+a12x2 +a13x3+..+auwxy =b a2x2 +a23x3+..+a2nxy =b2 a33x3+…+a3wxN=b3 dNNXN =bN 华南师范大学数学科学学院谢骊玲
Upper-Triangular Linear Systems ◼ Def. 2.1. An N×N matrix A=[aij] is called upper triangular provided that the elements satisfy aij=0 whenever i>j. A is called lower triangular provided that aij=0 whenever i<j. ◼ Thm. 2.1(Back Substitution). Suppose that AX=B is an uppertriangular system with the form as below. If akk≠0 for k=1,2,…,N, then there exists a unique solution for the system. 华南师范大学数学科学学院 谢骊玲NN N N N N N N N N a x b a x a x b a x a x a x b a x a x a x a x b = + + = + + + = + + + + = 33 3 3 3 22 2 23 3 2 2 11 1 12 2 13 3 1 1
