Consider the set $ S = \{2x^2 + 1, 2x - 4, -2x^2 + x\} $. The set $ S $ spans $ P_2 $.If $ t(x) = 6x^2 + 6x - 15 \in P_2 $,then $ t $ can be written as a linear combination of elements of $ S $.That is,\[ t(x) = a \cdot (2x^2 + 1) + b \cdot (2x - 4) + c \cdot (-2x^2 + x). \]Determine $ a, b, $ and $ c $.\[ a = \underline{\phantom{0}} \quad b = \underline{\phantom{0}} \quad c = \underline{\phantom{0}} \]

Answered: Image /qna-images/answer/45c1cb47-a18e-4fa1-9d03-12a700719d9a.jpg

Consider the set S = {2x² + 1, 2x – 4, 2x + x}. The set S spans P2. If t(a) — 6а? + 6х — 15 € Ра, 6x? + 6x – 15 e P2 , then t can be written as a linear combination of elements of S. That is, t(x) = — а - (2л + 1) +b . (2ӕ — 4) + с : (- 2»? + а). Determine a, b, and c a = 6 = C =

Consider the set S = {2x² + 1, 2x – 4, 2x + x}. The set S spans P2. If t(a) — 6а? + 6х — 15 € Ра, 6x? + 6x – 15 e P2 , then t can be written as a linear combination of elements of S. That is, t(x) = — а - (2л + 1) +b . (2ӕ — 4) + с : (- 2»? + а). Determine a, b, and c a = 6 = C =

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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Linear Algebra

In mathematics, problems can be solved by the arithmetic operators like addition, subtraction, multiplication, and division. In algebra, we represent the numbers by any letter called a variable. If these variables are of degree 1, then it is studied under linear algebra. The first-degree equations involving algebraic expressions are called linear equations. That is, if you represent it by drawing a graph you will get a straight line.

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Consider the set $ S = \{2x^2 + 1, 2x - 4, -2x^2 + x\} $. The set $ S $ spans $ P_2 $.

If $ t(x) = 6x^2 + 6x - 15 \in P_2 $,

then $ t $ can be written as a linear combination of elements of $ S $.

That is,
\[ t(x) = a \cdot (2x^2 + 1) + b \cdot (2x - 4) + c \cdot (-2x^2 + x). \]

Determine $ a, b, $ and $ c $.

\[ a = \underline{\phantom{0}} \quad b = \underline{\phantom{0}} \quad c = \underline{\phantom{0}} \]

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