**Solution:**To determine the vector $ \mathbf{x} $ given the provided coordinate vector $ [\mathbf{x}]_B $ and the basis $ B $, we use the following information:Given:\[ B = \left\{ \begin{pmatrix} 1 \\ -3 \end{pmatrix}, \begin{pmatrix} -5 \\ 3 \end{pmatrix} \right\} \]\[ [\mathbf{x}]_B = \begin{pmatrix} 5 \\ 2 \end{pmatrix} \]We can express $ \mathbf{x} $ as a linear combination of the basis vectors in $ B $. Specifically,\[ \mathbf{x} = 5 \begin{pmatrix} 1 \\ -3 \end{pmatrix} + 2 \begin{pmatrix} -5 \\ 3 \end{pmatrix} \]Let's perform the calculations step by step:1. Calculate $ 5 \begin{pmatrix} 1 \\ -3 \end{pmatrix} $: \[ 5 \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 5 \times 1 \\ 5 \times -3 \end{pmatrix} = \begin{pmatrix} 5 \\ -15 \end{pmatrix} \]2. Calculate $ 2 \begin{pmatrix} -5 \\ 3 \end{pmatrix} $: \[ 2 \begin{pmatrix} -5 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \times -5 \\ 2 \times 3 \end{pmatrix} = \begin{pmatrix} -10 \\ 6 \end{pmatrix} \]3. Add the resulting vectors: \[ \mathbf{x} = \begin{pmatrix} 5 \\ -15 \end{pmatrix} + \begin{pmatrix} -10 \\ 6 \end{pmatrix} = \begin{pmatrix} 5 + (-10) \\ -15 + 6 \end{pmatrix} = \begin{pmatrix} -5 \\ -9 \end{pmatrix} \]Thus, the vector $ \mathbf{x} $ is:\[ \mathbf{x} =

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Answered: Find the vector x determined by the…

Find the vector x determined by the given coordinate vector [x]p and the given basis B. 1. -5 -3 2. (Simplify your answers.)

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.

Question

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**Solution:**

To determine the vector $ \mathbf{x} $ given the provided coordinate vector $ [\mathbf{x}]_B $ and the basis $ B $, we use the following information:

Given:
\[ B = \left\{ \begin{pmatrix} 1 \\ -3 \end{pmatrix}, \begin{pmatrix} -5 \\ 3 \end{pmatrix} \right\} \]
\[ [\mathbf{x}]_B = \begin{pmatrix} 5 \\ 2 \end{pmatrix} \]

We can express $ \mathbf{x} $ as a linear combination of the basis vectors in $ B $. Specifically,

\[ \mathbf{x} = 5 \begin{pmatrix} 1 \\ -3 \end{pmatrix} + 2 \begin{pmatrix} -5 \\ 3 \end{pmatrix} \]

Let's perform the calculations step by step:

1. Calculate $ 5 \begin{pmatrix} 1 \\ -3 \end{pmatrix} $:

\[ 5 \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 5 \times 1 \\ 5 \times -3 \end{pmatrix} = \begin{pmatrix} 5 \\ -15 \end{pmatrix} \]

2. Calculate $ 2 \begin{pmatrix} -5 \\ 3 \end{pmatrix} $:

\[ 2 \begin{pmatrix} -5 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \times -5 \\ 2 \times 3 \end{pmatrix} = \begin{pmatrix} -10 \\ 6 \end{pmatrix} \]

3. Add the resulting vectors:

\[ \mathbf{x} = \begin{pmatrix} 5 \\ -15 \end{pmatrix} + \begin{pmatrix} -10 \\ 6 \end{pmatrix} = \begin{pmatrix} 5 + (-10) \\ -15 + 6 \end{pmatrix} = \begin{pmatrix} -5 \\ -9 \end{pmatrix} \]

Thus, the vector $ \mathbf{x} $ is:
\[ \mathbf{x} =

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