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2. Let V be the set of all real numbers with the two operations and defined by u v=u+v- 2 and ku = ku+ (1+ k) (a) Compute 201 Solution : 201 = (b) Compute 083 Solution : 083 = (c) Verify Ariom 4: Solution : We have u ) = role of %3! %3D = u for every u = u in V; thus the nu: the zero vector in V; so 0 = (d) Verify Axiom 5: Solution : For each u = u in V, we have %3D 0: thus the number

2. Let V be the set of all real numbers with the two operations and defined by u v=u+v- 2 and ku = ku+ (1+ k) (a) Compute 201 Solution : 201 = (b) Compute 083 Solution : 083 = (c) Verify Ariom 4: Solution : We have u ) = role of %3! %3D = u for every u = u in V; thus the nu: the zero vector in V; so 0 = (d) Verify Axiom 5: Solution : For each u = u in V, we have %3D 0: thus the number

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 32EQ
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2. Let V be the set of all real numbers with the two operations and defined by u Ov=u+v – 2 and kOu = ku + (1+ k) (a) Compute 201 Solution : 2Ð1 = (b) Compute 083 Solution : 083 = (c) Verify Axiom 4: Solution : We have u( ) = %3D = u for every u = uin V; thus the number 1 plays role of the zero vector in V; so 0 = (d) V erify Axiom 5: Solution : For each u = u in V, we have uÐ serves as the negative of u = u in V. ) = u + ( )– 2 = 0; thus the number %3D (e) Verify Axiom 7 : Solution : kO(uÐv) = k®(u+ v – 2) = but kOu O kOv so ko(uÐv) = k®uOkOv (f) Verify Axiom 8 : Solution : (k +1)Ou = %3D but so (k +1)Qu= k&u 1Ou (g) Verify Axiom 9 : Solution : kO(1Ou) = k® %3D %3D %3D but (kl)Ou = k®(1Ou) = (kl)®u. %3D so, *COPYRIGHT 64°F A rch IL || ||
Transcribed Image Text:2. Let V be the set of all real numbers with the two operations and defined by u Ov=u+v – 2 and kOu = ku + (1+ k) (a) Compute 201 Solution : 2Ð1 = (b) Compute 083 Solution : 083 = (c) Verify Axiom 4: Solution : We have u( ) = %3D = u for every u = uin V; thus the number 1 plays role of the zero vector in V; so 0 = (d) V erify Axiom 5: Solution : For each u = u in V, we have uÐ serves as the negative of u = u in V. ) = u + ( )– 2 = 0; thus the number %3D (e) Verify Axiom 7 : Solution : kO(uÐv) = k®(u+ v – 2) = but kOu O kOv so ko(uÐv) = k®uOkOv (f) Verify Axiom 8 : Solution : (k +1)Ou = %3D but so (k +1)Qu= k&u 1Ou (g) Verify Axiom 9 : Solution : kO(1Ou) = k® %3D %3D %3D but (kl)Ou = k®(1Ou) = (kl)®u. %3D so, *COPYRIGHT 64°F A rch IL || ||
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