Math Problem Statement
Cho m ∈ R và hệ phương trình (Given m ∈ R and the system of equations):
\[
\begin{cases}
x_1 + x_2 - 7x_3 = 1 \\
2x_1 - x_2 + mx_3 = 2 + m \\
-x_1 + 3x_2 - (m+1)x_3 = m^2 + 1
\end{cases}
\]
Find the values of m for which the system has a unique solution.
Solution
The image shows a system of linear equations with a parameter , and the task is to determine the values of such that the system has a unique solution. The system of equations is:
x_1 + x_2 - 7x_3 = 1 \\ 2x_1 - x_2 + mx_3 = 2 + m \\ -x_1 + 3x_2 - (m+1)x_3 = m^2 + 1 \end{cases}$$ ### Approach To determine for which values of $$m$$ the system has a unique solution, we need to check the determinant of the coefficient matrix. If the determinant is non-zero, the system has a unique solution. The coefficient matrix $$A$$ is: $$A = \begin{pmatrix} 1 & 1 & -7 \\ 2 & -1 & m \\ -1 & 3 & -(m+1) \end{pmatrix}$$ We can compute the determinant of matrix $$A$$ and solve for the values of $$m$$ that make this determinant non-zero. ### Determinant Calculation $$\text{det}(A) = \begin{vmatrix} 1 & 1 & -7 \\ 2 & -1 & m \\ -1 & 3 & -(m+1) \end{vmatrix}$$ Expanding this determinant along the first row: $$\text{det}(A) = 1 \cdot \begin{vmatrix} -1 & m \\ 3 & -(m+1) \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & m \\ -1 & -(m+1) \end{vmatrix} + (-7) \cdot \begin{vmatrix} 2 & -1 \\ -1 & 3 \end{vmatrix}$$ Now, we can compute each of the 2x2 minors: 1. First minor: $$\begin{vmatrix} -1 & m \\ 3 & -(m+1) \end{vmatrix} = (-1)(-(m+1)) - (m)(3) = m+1 - 3m = -2m + 1$$ 2. Second minor: $$\begin{vmatrix} 2 & m \\ -1 & -(m+1) \end{vmatrix} = (2)(-(m+1)) - (m)(-1) = -2(m+1) + m = -2m - 2 + m = -m - 2$$ 3. Third minor: $$\begin{vmatrix} 2 & -1 \\ -1 & 3 \end{vmatrix} = (2)(3) - (-1)(-1) = 6 - 1 = 5$$ Substitute these back into the determinant expression: $$\text{det}(A) = 1 \cdot (-2m + 1) - 1 \cdot (-m - 2) + (-7) \cdot 5$$ Simplifying: $$\text{det}(A) = (-2m + 1) + (m + 2) - 35 = -2m + m + 1 + 2 - 35 = -m - 32$$ So, the determinant is: $$\text{det}(A) = -m - 32$$ ### Condition for a Unique Solution For the system to have a unique solution, the determinant must be non-zero: $$-m - 32 \neq 0$$ $$m \neq -32$$ Thus, the system has a unique solution for all $$m \in \mathbb{R}$$ except $$m = -32$$. Would you like further details or have any questions about this problem? --- Here are five related questions you could explore: 1. What happens if $$m = -32$$? Does the system have no solution or infinitely many solutions? 2. How do we determine whether a system has no solution or infinitely many solutions when the determinant is zero? 3. Can we apply Cramer's rule to solve this system for specific values of $$m$$? 4. How does the parameter $$m$$ affect the behavior of the solution set geometrically? 5. What is the geometric interpretation of a system of three equations with three unknowns in 3D space? **Tip:** When analyzing systems with parameters, checking the determinant is a quick and powerful method to determine the conditions for unique, infinite, or no solutions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Equations
Determinants
Formulas
Determinant of a 3x3 matrix
Condition for a unique solution: det(A) ≠ 0
Theorems
Unique Solution Theorem
Suitable Grade Level
Grades 10-12
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