In Exercises 3-6, the vector X is in a subspace H with a basis B = {b₁,b₂}. Find the B-coordinate vector of x. 3. bi 4. bi b₁ -3 = [_] » - [ ] × [ ] , b2 X = -3 - [3] D = [ 3 ] - [ ] b2 X = -3 5 5
In Exercises 3-6, the vector X is in a subspace H with a basis B = {b₁,b₂}. Find the B-coordinate vector of x. 3. bi 4. bi b₁ -3 = [_] » - [ ] × [ ] , b2 X = -3 - [3] D = [ 3 ] - [ ] b2 X = -3 5 5
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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![**Exercises 3–6: Finding Coordinate Vectors in Subspace**
In Exercises 3–6, the vector \( \mathbf{x} \) is in a subspace \( H \) with a basis \( \mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2\} \). Find the \( \mathcal{B} \)-coordinate vector of \( \mathbf{x} \).
**Exercise 3:**
\[ \mathbf{b}_1 = \begin{bmatrix} 1 \\ -4 \end{bmatrix}, \mathbf{b}_2 = \begin{bmatrix} -2 \\ 7 \end{bmatrix}, \mathbf{x} = \begin{bmatrix} -3 \\ 7 \end{bmatrix} \]
**Exercise 4:**
\[ \mathbf{b}_1 = \begin{bmatrix} 1 \\ -3 \end{bmatrix}, \mathbf{b}_2 = \begin{bmatrix} -3 \\ 5 \end{bmatrix}, \mathbf{x} = \begin{bmatrix} -7 \\ 5 \end{bmatrix} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F324f16c4-c4e5-4b06-b074-f261b39e021a%2F5f1e49d9-3301-4d91-8948-00ae5d0cf368%2F232epk9_processed.png&w=3840&q=75)
Transcribed Image Text:**Exercises 3–6: Finding Coordinate Vectors in Subspace**
In Exercises 3–6, the vector \( \mathbf{x} \) is in a subspace \( H \) with a basis \( \mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2\} \). Find the \( \mathcal{B} \)-coordinate vector of \( \mathbf{x} \).
**Exercise 3:**
\[ \mathbf{b}_1 = \begin{bmatrix} 1 \\ -4 \end{bmatrix}, \mathbf{b}_2 = \begin{bmatrix} -2 \\ 7 \end{bmatrix}, \mathbf{x} = \begin{bmatrix} -3 \\ 7 \end{bmatrix} \]
**Exercise 4:**
\[ \mathbf{b}_1 = \begin{bmatrix} 1 \\ -3 \end{bmatrix}, \mathbf{b}_2 = \begin{bmatrix} -3 \\ 5 \end{bmatrix}, \mathbf{x} = \begin{bmatrix} -7 \\ 5 \end{bmatrix} \]
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