**Problem Statement:**Let $ B = (1, x, x^2, x^3) $ be an ordered basis for $ P_3 $. Find the coordinate vector of $ f(x) = -8x^3 - 6x^2 - 9x + 3 $ relative to $ B $.**Solution:**To find the coordinate vector of the polynomial $ f(x) = -8x^3 - 6x^2 - 9x + 3 $ relative to the basis $ B = (1, x, x^2, x^3) $, we express $ f(x) $ as a linear combination of the basis elements: $ c_0 \cdot 1 + c_1 \cdot x + c_2 \cdot x^2 + c_3 \cdot x^3 $.Here, the coefficients of $ f(x) $ correspond directly to the basis elements:- $ c_3 = -8 $ (coefficient of $ x^3 $)- $ c_2 = -6 $ (coefficient of $ x^2 $)- $ c_1 = -9 $ (coefficient of $ x $)- $ c_0 = 3 $ (constant term)Thus, the coordinate vector of $ f(x) $ relative to $ B $ is:\[f_B = \begin{bmatrix} -8 \\ -6 \\ -9 \\ 3 \end{bmatrix}\]**Note:** The red 'X' at the bottom left might indicate an error or incorrect response symbol.

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Answered: Let B = (1, x, x², x*) be an ordered…

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Let B = (1, x, x², x*) be an ordered basis for P3. Find the coordinate vector of 8x – 6x2 – 9x + 3 relative to B.

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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**Problem Statement:**

Let $ B = (1, x, x^2, x^3) $ be an ordered basis for $ P_3 $. Find the coordinate vector of $ f(x) = -8x^3 - 6x^2 - 9x + 3 $ relative to $ B $.

**Solution:**

To find the coordinate vector of the polynomial $ f(x) = -8x^3 - 6x^2 - 9x + 3 $ relative to the basis $ B = (1, x, x^2, x^3) $, we express $ f(x) $ as a linear combination of the basis elements: $ c_0 \cdot 1 + c_1 \cdot x + c_2 \cdot x^2 + c_3 \cdot x^3 $.

Here, the coefficients of $ f(x) $ correspond directly to the basis elements:

- $ c_3 = -8 $ (coefficient of $ x^3 $)
- $ c_2 = -6 $ (coefficient of $ x^2 $)
- $ c_1 = -9 $ (coefficient of $ x $)
- $ c_0 = 3 $ (constant term)

Thus, the coordinate vector of $ f(x) $ relative to $ B $ is:

\[
f_B = \begin{bmatrix} -8 \\ -6 \\ -9 \\ 3 \end{bmatrix}
\]

**Note:** The red 'X' at the bottom left might indicate an error or incorrect response symbol.

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