Math Problem Statement
solve pls
Solution
To solve the system of linear equations for a non-trivial solution, we first need to write the system as a matrix and then find the value of the parameter that makes the determinant of the coefficient matrix equal to zero. This condition ensures a non-trivial solution. Here are the equations:
2x_1 + x_2 + 2x_3 &= 0, \\ 4x_1 - x_2 + 7x_3 &= 0, \\ x_1 + \lambda x_2 + 2x_3 &= 0. \end{aligned}$$ ### Step 1: Write the system as a matrix equation The system can be represented as $$A \mathbf{x} = \mathbf{0}$$, where $$A$$ is the coefficient matrix and $$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$$. The matrix $$A$$ is: $$A = \begin{pmatrix} 2 & 1 & 2 \\ 4 & -1 & 7 \\ 1 & \lambda & 2 \end{pmatrix}$$ ### Step 2: Find the determinant of matrix $$A$$ To ensure a non-trivial solution, the determinant of $$A$$ must be zero. Let's compute the determinant of $$A$$: $$\text{det}(A) = \begin{vmatrix} 2 & 1 & 2 \\ 4 & -1 & 7 \\ 1 & \lambda & 2 \end{vmatrix}$$ I will calculate the determinant of this matrix.The determinant of the matrix $$A$$ is $$-6\lambda - 3$$. ### Step 3: Set the determinant to zero For the system to have a non-trivial solution, the determinant must be zero: $$-6\lambda - 3 = 0$$ Solving for $$\lambda$$: $$\lambda = -\frac{1}{2}$$ ### Step 4: Substitute $$\lambda = -\frac{1}{2}$$ into the system Now that we know $$\lambda = -\frac{1}{2}$$, we substitute it back into the system of equations and solve for $$x_1$$, $$x_2$$, and $$x_3$$. The system becomes: $$\begin{aligned} 2x_1 + x_2 + 2x_3 &= 0, \\ 4x_1 - x_2 + 7x_3 &= 0, \\ x_1 - \frac{1}{2}x_2 + 2x_3 &= 0. \end{aligned}$$ I will now solve this system for a non-trivial solution.The general solution to the system is: $$x_1 = -1.5x_3, \quad x_2 = x_3, \quad x_3 = x_3$$ For a non-trivial solution, $$x_3 \neq 0$$. Therefore, the solution can be written in terms of $$x_3$$, where $$x_3$$ is any non-zero scalar: $$\mathbf{x} = x_3 \begin{pmatrix} -1.5 \\ 1 \\ 1 \end{pmatrix}$$ Thus, the solution is: $$\mathbf{x} = \begin{pmatrix} -1.5 \\ 1 \\ 1 \end{pmatrix} x_3$$ where $$x_3 \neq 0$$. This gives a family of non-trivial solutions. Would you like more details or have any questions? Here are five related questions for practice: 1. What is the significance of a matrix having a zero determinant? 2. How do we identify whether a system of linear equations has a unique or infinite number of solutions? 3. Can a system have a trivial solution even if the determinant is non-zero? 4. How does changing a parameter affect the solvability of a linear system? 5. What are eigenvalues, and how are they related to the determinant of a matrix? **Tip:** Always check the determinant when solving systems involving parameters to determine if the system has non-trivial solutions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Systems of Linear Equations
Determinants
Parameterization
Formulas
Determinant of a 3x3 matrix
Setting the determinant to zero for non-trivial solutions
Theorems
Non-trivial solution condition (determinant equals zero)
Suitable Grade Level
College level
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