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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Given the matrices, solve the matrix X in the equation (X - 5B) A = C
![### Solving for Matrix X in the given Equation
Given the matrices:
\[
A = \begin{bmatrix}
2 & 5 \\
2 & 4
\end{bmatrix}, \quad
B = \begin{bmatrix}
1 & 5 \\
1 & -5
\end{bmatrix}, \quad
\text{and} \quad
C = \begin{bmatrix}
-2 & 3 \\
4 & -1
\end{bmatrix},
\]
we need to solve for the matrix \( X \) in the matrix equation:
\[ (X - 5B)A = C. \]
### Solution Steps
1. **Expand and Simplify the Equation:**
Substitute \( B \) and distribute:
\[ (X - 5B)A = X A - 5 B A. \]
2. **Calculate \( 5B \):**
Multiply matrix \( B \) by scalar 5:
\[
5B = 5 \begin{bmatrix}
1 & 5 \\
1 & -5
\end{bmatrix}
= \begin{bmatrix}
5 & 25 \\
5 & -25
\end{bmatrix}.
\]
3. **Multiply \( 5B \cdot A \):**
\[
5B \cdot A = \begin{bmatrix}
5 & 25 \\
5 & -25
\end{bmatrix}
\begin{bmatrix}
2 & 5 \\
2 & 4
\end{bmatrix}
= \begin{bmatrix}
(5 \cdot 2 + 25 \cdot 2) & (5 \cdot 5 + 25 \cdot 4) \\
(5 \cdot 2 + (-25) \cdot 2) & (5 \cdot 5 + (-25) \cdot 4)
\end{bmatrix}
= \begin{bmatrix}
60 & 130 \\
-40 & -75
\end{bmatrix}.
\]
4. **Rewrite the equation using matrix \( C \)**:
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faca5ab5d-d1b4-4091-b3ec-a6eaf8cb8003%2F0c1e26e3-f7a4-4c45-8c6b-273e19085f51%2Fgk00bec_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving for Matrix X in the given Equation
Given the matrices:
\[
A = \begin{bmatrix}
2 & 5 \\
2 & 4
\end{bmatrix}, \quad
B = \begin{bmatrix}
1 & 5 \\
1 & -5
\end{bmatrix}, \quad
\text{and} \quad
C = \begin{bmatrix}
-2 & 3 \\
4 & -1
\end{bmatrix},
\]
we need to solve for the matrix \( X \) in the matrix equation:
\[ (X - 5B)A = C. \]
### Solution Steps
1. **Expand and Simplify the Equation:**
Substitute \( B \) and distribute:
\[ (X - 5B)A = X A - 5 B A. \]
2. **Calculate \( 5B \):**
Multiply matrix \( B \) by scalar 5:
\[
5B = 5 \begin{bmatrix}
1 & 5 \\
1 & -5
\end{bmatrix}
= \begin{bmatrix}
5 & 25 \\
5 & -25
\end{bmatrix}.
\]
3. **Multiply \( 5B \cdot A \):**
\[
5B \cdot A = \begin{bmatrix}
5 & 25 \\
5 & -25
\end{bmatrix}
\begin{bmatrix}
2 & 5 \\
2 & 4
\end{bmatrix}
= \begin{bmatrix}
(5 \cdot 2 + 25 \cdot 2) & (5 \cdot 5 + 25 \cdot 4) \\
(5 \cdot 2 + (-25) \cdot 2) & (5 \cdot 5 + (-25) \cdot 4)
\end{bmatrix}
= \begin{bmatrix}
60 & 130 \\
-40 & -75
\end{bmatrix}.
\]
4. **Rewrite the equation using matrix \( C \)**:
\[
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