Problem II Consider a linear operator L : R4 → Rª defined by the formula below, where V₁ = (1, 1, 1, 1), V2 = (1, 1, 0, 0), V3 = (1, 2, 0, −1) and v4 = (0, 0, 1, 1) (the formula involves the dot product and scalar multiplication). Find a matrix M such that L(u) = Mu for every u € Rª, where u and L(u) are regarded as column vectors. - 21. L(u) = (u · V₁)V2 + (U · V3)V4. 22. L(u) = (u · V₁)V2 − (U · V3)V4. 23. L(u) = (u · v₁)V4 + (U · V3)V2. 24. L(u) = (u · V₁)V4 — (U · V3)V2. 25. L(u) = (u · V₂)V₁ + (U · V3)V4. 26. L(u) = (u · V2)V₁ − (U · V3)V4. 27. L(u) = (u · V₁)V2 + (U · V4)V3. 28. L(u) = (u · v₁)V2 − (U · V4)V3. - 31. L(u) = (u · V₁)V₁ + (U · V2)V2 + (V3 · V4)U. 32. L(u) = (u · V₁)V2 + (U · V2)V₁ + (V3 · V4)U. 33. L(u) = (u · V2)V2 + (U · V3)V3 + (V₁ · V4)U. 34. L(u) = (u · V₂)V3 + (U · V3)V2 + (V₁ · V4)U. 35. L(u) = (u · V₁)V₁ + (U · V₁)V4 + (V2 · V3)U. 36. L(u) = (u · V₁)V₁ + (U · V₁)V1 + (V2 · V3)U. 37. L(u) = (u · V₂)V2 + (U · V4)V4 + (V₁ · V3)U. 38. L(u) = (u · V2)V₁ + (U · V4)V2 + (V₁ · V3)U.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.7: The Inverse Of A Matrix
Problem 32E
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Problem II Consider a linear operator L: R4 → R4 defined by the formula below, where
V₁ = (1, 1, 1, 1), V₂ = (1, 1, 0, 0), v3 = (1, 2, 0, −1) and v4 = (0, 0, 1, 1) (the formula involves the
dot product and scalar multiplication).
=
Find a matrix M such that L(u)
as column vectors.
=
= Mu for every u € R¹, where u and L(u) are regarded
21. L(u)
=
=
4
V4
(u · V₁)V₂ + (u · V3)V4.
22. L(u) (u · V₁)V2 — (U · V3)V4.
23. L(u) = (u · V₁)V₁ + (U · V3)V2.
24. L(u) (u. V₁)V₁ – (u · V3)V2.
25. L(u) = (u · V₂)V₁ + (U · V3)V4.
26. L(u) = (u · V2)V₁ − (U · V3)V4.
27. L(u) = (u V₁)V₂ + (U · V₁) V3.
· 2
28. L(u) = (u · V₁)V₂ — (U · V4)V3.
29. L(u) = (u · V2)V3 + (U · V₁)V4.
30. L(u) = (u · V2)V3 — (U · V₁)V4.
=
31. L(u) = (u · V₁)V₁ + (U · V2)V2 + (V3 · V4)U.
32. L(u) (u · V₁)V2 + (U · V2)V1 + (V3 · V₁)u.
33. L(u) = (u · V2)V2 + (U · V3)V3 + (V₁ · V4)U.
34. L(u) = (u · V2)V3 + (U · V3)V2 + (V₁ · V4)U.
35. L(u) = (u · V₁)V₁ + (U · VĀ)V4 + (V2 · V3)U.
36. L(u) = (u · V₁)V4 + (U · V₁)V₁ + (V₂ · V3)U.
37. L(u) = (u · V₂)V2 + (U · V4)V4 + (V₁ · V3)U.
38. L(u) = (u · V₂)V₁ + (U · V₁)V2 + (V₁ · V3)u.
39. L(u) = (U · V3)V3 + (U · V4)V4 + (V₁ · V2)u.
40. L(u) = (u V3)V4 + (U · V4)V3 + (V₁ · V2)U.
4
=
Transcribed Image Text:Problem II Consider a linear operator L: R4 → R4 defined by the formula below, where V₁ = (1, 1, 1, 1), V₂ = (1, 1, 0, 0), v3 = (1, 2, 0, −1) and v4 = (0, 0, 1, 1) (the formula involves the dot product and scalar multiplication). = Find a matrix M such that L(u) as column vectors. = = Mu for every u € R¹, where u and L(u) are regarded 21. L(u) = = 4 V4 (u · V₁)V₂ + (u · V3)V4. 22. L(u) (u · V₁)V2 — (U · V3)V4. 23. L(u) = (u · V₁)V₁ + (U · V3)V2. 24. L(u) (u. V₁)V₁ – (u · V3)V2. 25. L(u) = (u · V₂)V₁ + (U · V3)V4. 26. L(u) = (u · V2)V₁ − (U · V3)V4. 27. L(u) = (u V₁)V₂ + (U · V₁) V3. · 2 28. L(u) = (u · V₁)V₂ — (U · V4)V3. 29. L(u) = (u · V2)V3 + (U · V₁)V4. 30. L(u) = (u · V2)V3 — (U · V₁)V4. = 31. L(u) = (u · V₁)V₁ + (U · V2)V2 + (V3 · V4)U. 32. L(u) (u · V₁)V2 + (U · V2)V1 + (V3 · V₁)u. 33. L(u) = (u · V2)V2 + (U · V3)V3 + (V₁ · V4)U. 34. L(u) = (u · V2)V3 + (U · V3)V2 + (V₁ · V4)U. 35. L(u) = (u · V₁)V₁ + (U · VĀ)V4 + (V2 · V3)U. 36. L(u) = (u · V₁)V4 + (U · V₁)V₁ + (V₂ · V3)U. 37. L(u) = (u · V₂)V2 + (U · V4)V4 + (V₁ · V3)U. 38. L(u) = (u · V₂)V₁ + (U · V₁)V2 + (V₁ · V3)u. 39. L(u) = (U · V3)V3 + (U · V4)V4 + (V₁ · V2)u. 40. L(u) = (u V3)V4 + (U · V4)V3 + (V₁ · V2)U. 4 =
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