In Exercises 7-10, determine the values of the parameters for which the system has a unique solution, and describe the solution. 7. 6sx₁ + 4x2 = 5 9x + 28x₂ -2 =
In Exercises 7-10, determine the values of the parameters for which the system has a unique solution, and describe the solution. 7. 6sx₁ + 4x2 = 5 9x + 28x₂ -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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![### Unique Solution Parameters for Linear Systems
In this section, we will solve for the values of the parameter \( s \) that make the given systems of linear equations have a unique solution. We will also describe these solutions.
#### Exercise 7
\[ 6sx_1 + 4x_2 = 5 \]
\[ 9x_1 + 2sx_2 = -2 \]
#### Exercise 8
\[ 3sx_1 + 5x_2 = 3 \]
\[ 12x_1 + 5sx_2 = 2 \]
### Detailed Explanation:
To determine the values of \( s \) for which the system has a unique solution, we first need to ensure that the determinant of the coefficient matrix is non-zero. For each system, calculate the determinant of the \( 2 \times 2 \) matrix formed by the coefficients of \( x_1 \) and \( x_2 \).
#### Steps:
1. **Write down the coefficient matrix for each system.**
2. **Calculate the determinant** of this matrix. For a matrix of the form:
\[
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\]
The determinant is \( ad - bc \).
3. **Set the determinant to be non-zero** and solve for \( s \).
### Exercise 7 Detailed Calculation:
- Coefficient matrix:
\[
\begin{bmatrix}
6s & 4 \\
9 & 2s
\end{bmatrix}
\]
- Determinant is:
\[
(6s \cdot 2s) - (4 \cdot 9) = 12s^2 - 36
\]
- Set the determinant to be non-zero:
\[
12s^2 - 36 \neq 0
\]
- Solve for \( s \):
\[
12s^2 \neq 36 \implies s^2 \neq 3 \implies s \neq \pm \sqrt{3}
\]
### Exercise 8 Detailed Calculation:
- Coefficient matrix:
\[
\begin{bmatrix}
3s & 5 \\
12 & 5s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F324f16c4-c4e5-4b06-b074-f261b39e021a%2Fc7942a69-c51f-4925-b62f-04ea152328a5%2F5ccy4u_processed.png&w=3840&q=75)
Transcribed Image Text:### Unique Solution Parameters for Linear Systems
In this section, we will solve for the values of the parameter \( s \) that make the given systems of linear equations have a unique solution. We will also describe these solutions.
#### Exercise 7
\[ 6sx_1 + 4x_2 = 5 \]
\[ 9x_1 + 2sx_2 = -2 \]
#### Exercise 8
\[ 3sx_1 + 5x_2 = 3 \]
\[ 12x_1 + 5sx_2 = 2 \]
### Detailed Explanation:
To determine the values of \( s \) for which the system has a unique solution, we first need to ensure that the determinant of the coefficient matrix is non-zero. For each system, calculate the determinant of the \( 2 \times 2 \) matrix formed by the coefficients of \( x_1 \) and \( x_2 \).
#### Steps:
1. **Write down the coefficient matrix for each system.**
2. **Calculate the determinant** of this matrix. For a matrix of the form:
\[
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\]
The determinant is \( ad - bc \).
3. **Set the determinant to be non-zero** and solve for \( s \).
### Exercise 7 Detailed Calculation:
- Coefficient matrix:
\[
\begin{bmatrix}
6s & 4 \\
9 & 2s
\end{bmatrix}
\]
- Determinant is:
\[
(6s \cdot 2s) - (4 \cdot 9) = 12s^2 - 36
\]
- Set the determinant to be non-zero:
\[
12s^2 - 36 \neq 0
\]
- Solve for \( s \):
\[
12s^2 \neq 36 \implies s^2 \neq 3 \implies s \neq \pm \sqrt{3}
\]
### Exercise 8 Detailed Calculation:
- Coefficient matrix:
\[
\begin{bmatrix}
3s & 5 \\
12 & 5s
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