2. Let V be the set of all real numbers with the two operations O and O defined by u Ov=u+v - 2 and kOu = ku + (1+ k) (a) Compute 201 Solution : 201 = (b) Compute 083 Solution : 083 = (c) Verify Ariom 4: Solution : We have u( ) = %3D = u for every u = u in V; thus the number 1 plays role of the zero vector in V; so 0 = (d) Verify Axiom 5: Solution : For each u = u in V, we have u serves as the negative of u = ) = u + ( )- 2 = 0; thus the number u in V. u = (e) Verify Axiom 7: Solution : kO(uÐv) = k®(u+ v- 2) = %3D %3D but kOu O kOv %3D %3D so ko(uOv) = kOuOkOv (f)Verify Ariom 8 : Solution : (k +1)Ou = but kOu 1Ou = so (k +1)Ou= kOuIOu (g) Verify Ariom 9: Solution : kO(1®u) = k® %3D %3D but (kl)Ou = so, kO(1Ou) = (kl)®u.

2. Let V be the set of all real numbers with the two operations O and O defined by u Ov=u+v - 2 and kOu = ku + (1+ k) (a) Compute 201 Solution : 201 = (b) Compute 083 Solution : 083 = (c) Verify Ariom 4: Solution : We have u( ) = %3D = u for every u = u in V; thus the number 1 plays role of the zero vector in V; so 0 = (d) Verify Axiom 5: Solution : For each u = u in V, we have u serves as the negative of u = ) = u + ( )- 2 = 0; thus the number u in V. u = (e) Verify Axiom 7: Solution : kO(uÐv) = k®(u+ v- 2) = %3D %3D but kOu O kOv %3D %3D so ko(uOv) = kOuOkOv (f)Verify Ariom 8 : Solution : (k +1)Ou = but kOu 1Ou = so (k +1)Ou= kOuIOu (g) Verify Ariom 9: Solution : kO(1®u) = k® %3D %3D but (kl)Ou = so, kO(1Ou) = (kl)®u.

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter4: Eigenvalues And Eigenvectors

Section4.2: Determinants

Problem 19AEXP

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