Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
Given the matrices, solve for the matrix X in the equation A(X+3B) = C
![### Solving for Matrix \( X \) in the Matrix Equation
**Problem Statement:**
Given the matrices:
\[ A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}, \quad \text{and} \quad C = \begin{bmatrix} -3 & 0 \\ 3 & 2 \end{bmatrix}, \]
solve for the matrix \( X \) in the equation:
\[ A \left( X + 3B \right) = C. \]
**Detailed Solution:**
To find matrix \( X \), follow these steps:
1. **Step 1: Calculate \( 3B \)**
\[
3B = 3 \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 \cdot 1 & 3 \cdot 3 \\ 3 \cdot 3 & 3 \cdot 4 \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ 9 & 12 \end{bmatrix}
\]
2. **Step 2: Express the Equation in Terms of \( X \)**
\[ A \left( X + 3B \right) = C \]
\[ A \left( X + \begin{bmatrix} 3 & 9 \\ 9 & 12 \end{bmatrix} \right) = \begin{bmatrix} -3 & 0 \\ 3 & 2 \end{bmatrix} \]
3. **Step 3: Assume \( X \) is a 2x2 Matrix**
\[ X = \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} \]
4. **Step 4: Rewrite the Equation**
\[ A \left( \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} + \begin{bmatrix} 3 & 9 \\ 9 & 12 \end{bmatrix} \right) = \begin](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faca5ab5d-d1b4-4091-b3ec-a6eaf8cb8003%2F8cff5168-8ddd-410e-9be1-f9642205ba0b%2Fcjwik17_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving for Matrix \( X \) in the Matrix Equation
**Problem Statement:**
Given the matrices:
\[ A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}, \quad \text{and} \quad C = \begin{bmatrix} -3 & 0 \\ 3 & 2 \end{bmatrix}, \]
solve for the matrix \( X \) in the equation:
\[ A \left( X + 3B \right) = C. \]
**Detailed Solution:**
To find matrix \( X \), follow these steps:
1. **Step 1: Calculate \( 3B \)**
\[
3B = 3 \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 \cdot 1 & 3 \cdot 3 \\ 3 \cdot 3 & 3 \cdot 4 \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ 9 & 12 \end{bmatrix}
\]
2. **Step 2: Express the Equation in Terms of \( X \)**
\[ A \left( X + 3B \right) = C \]
\[ A \left( X + \begin{bmatrix} 3 & 9 \\ 9 & 12 \end{bmatrix} \right) = \begin{bmatrix} -3 & 0 \\ 3 & 2 \end{bmatrix} \]
3. **Step 3: Assume \( X \) is a 2x2 Matrix**
\[ X = \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} \]
4. **Step 4: Rewrite the Equation**
\[ A \left( \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} + \begin{bmatrix} 3 & 9 \\ 9 & 12 \end{bmatrix} \right) = \begin
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