Parts ii and iii please **Problem 2.** Vectors **v₁** = (4, 6, 7), **v₂** = (0, 1, 1) and **v₃** = (0, 1, 2) form a basis for the vector space ℝ³. Vectors **u₁** = (1, 1, 1), **u₂** = (1, 2, 2) and **u₃** = (2, 3, 4) form another basis for ℝ³.(i) Find the transition matrix from the standard basis **e₁, e₂, e₃** to the ordered basis **u₁, u₂, u₃**. (ii) Find the transition matrix from the ordered basis **v₁, v₂, v₃** to the ordered basis **u₁, u₂, u₃**. (iii) Find coordinates of the vector **w** = 2**v₁** + 3**v₂** - 4**v₃** relative to the basis **v₁, v₂, v₃**, coordinates of **w** relative to the basis **u₁, u₂, u₃**, and coordinates of **w** relative to the standard basis.

Answered: Problem 2. Vectors V₁ = (4,6,7), V₂ =…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Parts ii and iii please

**Problem 2.** Vectors **v₁** = (4, 6, 7), **v₂** = (0, 1, 1) and **v₃** = (0, 1, 2) form a basis for the vector space ℝ³. Vectors **u₁** = (1, 1, 1), **u₂** = (1, 2, 2) and **u₃** = (2, 3, 4) form another basis for ℝ³.

(i) Find the transition matrix from the standard basis **e₁, e₂, e₃** to the ordered basis **u₁, u₂, u₃**.
(ii) Find the transition matrix from the ordered basis **v₁, v₂, v₃** to the ordered basis **u₁, u₂, u₃**.
(iii) Find coordinates of the vector **w** = 2**v₁** + 3**v₂** - 4**v₃** relative to the basis **v₁, v₂, v₃**, coordinates of **w** relative to the basis **u₁, u₂, u₃**, and coordinates of **w** relative to the standard basis.

Expert Solution

Step 1: Finding the transition matrix

$v_{1} = (4, 6, 7), v_{2} = (0, 1, 1), v_{3} = (0, 1, 2) u_{1} = (1, 1, 1), u_{2} = (1, 2, 2), u_{3} = (2, 3, 4)$

Therefore

$u_{1} = a v_{1} + b v_{2} + c v_{3} u_{2} = d v_{1} + e v_{2} + f v_{3} u_{3} = g v_{1} + h v_{2} + i v_{3} ∴ (1, 1, 1) = a (4, 6, 7) + b (0, 1, 1) + c (0, 1, 2) = (4 a, 6 a + b + c, 7 a + b + 2 c) ∴ a = \frac{1}{4}, b = - \frac{1}{4}, c = - \frac{1}{4} (1, 2, 2) = d (4, 6, 7) + e (0, 1, 1) + f (0, 1, 2) = (4 d, 6 d + e + f, 7 d + e + 2 f) ∴ d = \frac{1}{4}, e = \frac{3}{4}, f = - \frac{1}{4} (2, 3, 4) = g (4, 6, 7) + h (0, 1, 1) + i (0, 1, 2) = (4 g, 6 g + h + i, 4 d + h + 2 i) ∴ g = \frac{1}{2}, h = - \frac{1}{2}, i = \frac{1}{2} ∴ M = [\begin{matrix} a & d & g \\ b & e & h \\ c & f & i \end{matrix}] M = [\begin{matrix} \frac{1}{4} & \frac{1}{4} & \frac{1}{2} \\ - \frac{1}{4} & \frac{3}{4} & - \frac{1}{2} \\ - \frac{1}{4} & - \frac{1}{4} & \frac{1}{2} \end{matrix}]$

Hence the matrix is $[\begin{matrix} \frac{1}{4} & \frac{1}{4} & \frac{1}{2} \\ - \frac{1}{4} & \frac{3}{4} & - \frac{1}{2} \\ - \frac{1}{4} & - \frac{1}{4} & \frac{1}{2} \end{matrix}]$