If then A-¹ = Given 6 = E 1 3 solve Ax = b. 0 x = A = 3 3 -1 -9 5 -8 5 3 -2
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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![The image presents a mathematical problem involving matrices. Here's the transcription and explanation fit for an educational website:
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**Problem Statement**
If
\[ A = \begin{bmatrix} 3 & -9 & 5 \\ 3 & -8 & 5 \\ -1 & 3 & -2 \end{bmatrix}, \]
then
\[ A^{-1} = \begin{bmatrix} \boxed{\phantom{x}} & \boxed{\phantom{x}} & \boxed{\phantom{x}} \\ \boxed{\phantom{x}} & \boxed{\phantom{x}} & \boxed{\phantom{x}} \\ \boxed{\phantom{x}} & \boxed{\phantom{x}} & \boxed{\phantom{x}} \end{bmatrix}. \]
**Task**
Given
\[ \vec{b} = \begin{bmatrix} 1 \\ 3 \\ -3 \end{bmatrix}, \]
solve the equation \( A\vec{x} = \vec{b} \).
**Solution**
\[ \vec{x} = \begin{bmatrix} \boxed{\phantom{x}} \\ \boxed{\phantom{x}} \\ \boxed{\phantom{x}} \end{bmatrix}. \]
**Explanation**
1. **Matrix \(A\)**: A 3x3 matrix is given.
2. **Inverse of Matrix \(A\) (\(A^{-1}\))**: The task is to find the inverse of matrix \(A\), represented by a 3x3 matrix filled with empty placeholders for each of its elements.
3. **Vector \(\vec{b}\)**: A 3x1 column vector with values 1, 3, and -3.
4. **Equation \(A\vec{x} = \vec{b}\)**: This equation needs to be solved for the vector \(\vec{x}\).
5. **Solution Vector \(\vec{x}\)**: A 3x1 column vector is shown with placeholders for the solutions.
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This problem requires understanding of linear algebra to find the inverse of a matrix and solve a system of linear equations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb898cad9-5347-4e0a-a74d-32f84bfad0f6%2F62c259aa-6f01-44da-a20e-b36dd4555fef%2F0barfj_processed.png&w=3840&q=75)
Transcribed Image Text:The image presents a mathematical problem involving matrices. Here's the transcription and explanation fit for an educational website:
---
**Problem Statement**
If
\[ A = \begin{bmatrix} 3 & -9 & 5 \\ 3 & -8 & 5 \\ -1 & 3 & -2 \end{bmatrix}, \]
then
\[ A^{-1} = \begin{bmatrix} \boxed{\phantom{x}} & \boxed{\phantom{x}} & \boxed{\phantom{x}} \\ \boxed{\phantom{x}} & \boxed{\phantom{x}} & \boxed{\phantom{x}} \\ \boxed{\phantom{x}} & \boxed{\phantom{x}} & \boxed{\phantom{x}} \end{bmatrix}. \]
**Task**
Given
\[ \vec{b} = \begin{bmatrix} 1 \\ 3 \\ -3 \end{bmatrix}, \]
solve the equation \( A\vec{x} = \vec{b} \).
**Solution**
\[ \vec{x} = \begin{bmatrix} \boxed{\phantom{x}} \\ \boxed{\phantom{x}} \\ \boxed{\phantom{x}} \end{bmatrix}. \]
**Explanation**
1. **Matrix \(A\)**: A 3x3 matrix is given.
2. **Inverse of Matrix \(A\) (\(A^{-1}\))**: The task is to find the inverse of matrix \(A\), represented by a 3x3 matrix filled with empty placeholders for each of its elements.
3. **Vector \(\vec{b}\)**: A 3x1 column vector with values 1, 3, and -3.
4. **Equation \(A\vec{x} = \vec{b}\)**: This equation needs to be solved for the vector \(\vec{x}\).
5. **Solution Vector \(\vec{x}\)**: A 3x1 column vector is shown with placeholders for the solutions.
---
This problem requires understanding of linear algebra to find the inverse of a matrix and solve a system of linear equations.
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