Math Problem Statement
Find the set of solutions of the homogeneous system Ax = 0, where A is a 4x7 matrix.
Solution
The given problem is asking to find the set of solutions for the homogeneous system , where the matrix is:
1 & 0 & 4 & 1 & 0 & 3 & 1 \\ 0 & 1 & 5 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 8 & 6 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$ ### Steps to solve: 1. **Set up the system**: You are asked to solve the equation $$A \mathbf{x} = 0$$. Let $$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \\ x_7 \end{pmatrix}$$. This implies the system of linear equations can be written as: $$\begin{aligned} 1. & \quad x_1 + 4x_3 + x_4 + 3x_6 + x_7 = 0 \\ 2. & \quad x_2 + 5x_3 + 2x_4 = 0 \\ 3. & \quad x_4 + 8x_5 + 6x_6 = 0 \\ 4. & \quad x_7 = 0 \end{aligned}$$ 2. **Simplify the equations**: - From equation 4, we immediately have $$x_7 = 0$$. - Substituting $$x_7 = 0$$ into equation 1 gives: $$x_1 + 4x_3 + x_4 + 3x_6 = 0$$ - Equation 2 remains the same: $$x_2 + 5x_3 + 2x_4 = 0$$. - Equation 3 remains: $$x_4 + 8x_5 + 6x_6 = 0$$. 3. **Express variables in terms of free variables**: - Let $$x_3$$, $$x_5$$, and $$x_6$$ be the free variables. - Solve for the dependent variables: - From equation 3: $$x_4 = -8x_5 - 6x_6$$. - Substitute $$x_4$$ into equation 2 to solve for $$x_2$$: $$x_2 + 5x_3 + 2(-8x_5 - 6x_6) = 0 \implies x_2 = -5x_3 + 16x_5 + 12x_6$$ - Substitute $$x_4$$ into equation 1 to solve for $$x_1$$: $$x_1 + 4x_3 + (-8x_5 - 6x_6) + 3x_6 = 0 \implies x_1 = -4x_3 + 8x_5 + 3x_6$$ 4. **Final solution**: The general solution to the system can be written in parametric form: $$\mathbf{x} = \begin{pmatrix} -4x_3 + 8x_5 + 3x_6 \\ -5x_3 + 16x_5 + 12x_6 \\ x_3 \\ -8x_5 - 6x_6 \\ x_5 \\ x_6 \\ 0 \end{pmatrix} = x_3 \begin{pmatrix} -4 \\ -5 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} + x_5 \begin{pmatrix} 8 \\ 16 \\ 0 \\ -8 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_6 \begin{pmatrix} 3 \\ 12 \\ 0 \\ -6 \\ 0 \\ 1 \\ 0 \end{pmatrix}$$ Thus, the solution set is: $$\mathbf{x} = x_3 \begin{pmatrix} -4 \\ -5 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} + x_5 \begin{pmatrix} 8 \\ 16 \\ 0 \\ -8 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_6 \begin{pmatrix} 3 \\ 12 \\ 0 \\ -6 \\ 0 \\ 1 \\ 0 \end{pmatrix}$$ The solution is a linear combination of three vectors, where $$x_3$$, $$x_5$$, and $$x_6$$ are free parameters. Would you like more details or have any questions? Here are 5 related questions to expand your understanding: 1. What is a homogeneous system of equations? 2. How do you determine free and dependent variables in a matrix system? 3. What are the implications of having free variables in a solution set? 4. Can this system be solved using Gaussian elimination? How would that work? 5. How does the rank of matrix $$A$$ relate to the number of free variables? **Tip:** In homogeneous systems, if the rank of the matrix is less than the number of variables, you will always have free variables, leading to an infinite number of solutions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Homogeneous Systems
Parametric Solutions
Linear Combinations
Formulas
Ax = 0 for homogeneous systems
Solution set as a linear combination of free variables
Theorems
Rank-Nullity Theorem
Suitable Grade Level
Undergraduate level (Linear Algebra)