= [₁ n = (1, 2) and D = -4 4 3 What is D.n? Enter your answer as a vector by filling in the boxes.
= [₁ n = (1, 2) and D = -4 4 3 What is D.n? Enter your answer as a vector by filling in the boxes.
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
n⃗ =⟨−1, −2⟩ and D=[−4423].
What is D⋅n⃗ ?
Enter your answer as a
![To solve the problem, we need to calculate the product of the matrix \( D \) and the vector \( \vec{n} \).
Given:
\[ \vec{n} = \langle -1, -2 \rangle \]
and
\[ D = \begin{bmatrix} -4 & 2 \\ 4 & 3 \end{bmatrix} \]
To find \( D \cdot \vec{n} \), we perform matrix multiplication.
Matrix \( D \) is a 2x2 matrix:
\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \]
Where:
\[ a = -4, \quad b = 2 \]
\[ c = 4, \quad d = 3 \]
The vector \( \vec{n} \) is a 2x1 vector:
\[ \begin{bmatrix} x \\ y \end{bmatrix} \]
Where:
\[ x = -1, \quad y = -2 \]
The result of multiplying matrix \( D \) by vector \( \vec{n} \) is a new 2x1 vector \( \vec{r} = \begin{bmatrix} r_1 \\ r_2 \end{bmatrix} \), where:
\[ r_1 = a \cdot x + b \cdot y \]
\[ r_2 = c \cdot x + d \cdot y \]
Plugging in the values:
\[ r_1 = (-4) \cdot (-1) + 2 \cdot (-2) \]
\[ r_1 = 4 + (-4) \]
\[ r_1 = 0 \]
\[ r_2 = 4 \cdot (-1) + 3 \cdot (-2) \]
\[ r_2 = -4 + (-6) \]
\[ r_2 = -10 \]
Therefore, the resulting vector \( D \cdot \vec{n} \) is:
\[ \vec{r} = \begin{bmatrix} 0 \\ -10 \end{bmatrix} \]
Enter your answer as a vector by filling in the boxes:
\[ \langle \boxed{0}, \boxed{-10} \rangle \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d0ba8a5-1946-40ed-9646-dde57f3e1261%2Fe3e38857-d155-4698-b556-844a7f0ac889%2Ffvumv1h_processed.png&w=3840&q=75)
Transcribed Image Text:To solve the problem, we need to calculate the product of the matrix \( D \) and the vector \( \vec{n} \).
Given:
\[ \vec{n} = \langle -1, -2 \rangle \]
and
\[ D = \begin{bmatrix} -4 & 2 \\ 4 & 3 \end{bmatrix} \]
To find \( D \cdot \vec{n} \), we perform matrix multiplication.
Matrix \( D \) is a 2x2 matrix:
\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \]
Where:
\[ a = -4, \quad b = 2 \]
\[ c = 4, \quad d = 3 \]
The vector \( \vec{n} \) is a 2x1 vector:
\[ \begin{bmatrix} x \\ y \end{bmatrix} \]
Where:
\[ x = -1, \quad y = -2 \]
The result of multiplying matrix \( D \) by vector \( \vec{n} \) is a new 2x1 vector \( \vec{r} = \begin{bmatrix} r_1 \\ r_2 \end{bmatrix} \), where:
\[ r_1 = a \cdot x + b \cdot y \]
\[ r_2 = c \cdot x + d \cdot y \]
Plugging in the values:
\[ r_1 = (-4) \cdot (-1) + 2 \cdot (-2) \]
\[ r_1 = 4 + (-4) \]
\[ r_1 = 0 \]
\[ r_2 = 4 \cdot (-1) + 3 \cdot (-2) \]
\[ r_2 = -4 + (-6) \]
\[ r_2 = -10 \]
Therefore, the resulting vector \( D \cdot \vec{n} \) is:
\[ \vec{r} = \begin{bmatrix} 0 \\ -10 \end{bmatrix} \]
Enter your answer as a vector by filling in the boxes:
\[ \langle \boxed{0}, \boxed{-10} \rangle \]
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