Math Problem Statement
Find a basis for the solution space of the following homogeneous system of equations:
[[-2, 0, -10, 4], [1, 2, 3, -1], [1, -4, 1, 0]] * [x1, x2, x3, x4] = [0, 0, 0].
Solution
To solve this problem, we need to find a basis for the solution space of the given homogeneous system of linear equations. The system is represented in matrix form as:
-2 & 0 & -10 & 4 \\ 1 & 2 & 3 & -1 \\ 1 & -4 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ ### Steps to Solve: 1. **Set up the augmented matrix** for the system and use row reduction to bring it to row echelon form or reduced row echelon form. $$\begin{bmatrix} -2 & 0 & -10 & 4 & | & 0 \\ 1 & 2 & 3 & -1 & | & 0 \\ 1 & -4 & 1 & 0 & | & 0 \end{bmatrix}$$ 2. **Perform row operations** to simplify this matrix. The goal is to express some variables in terms of free variables if possible, as this will help us identify the basis vectors for the solution space. 3. **Identify pivot and free variables** from the row-reduced form. Variables that do not have pivots are free variables, and these will allow us to write the solution in terms of parameters. 4. **Express the solution in parametric vector form** and identify the basis vectors for the solution space, which correspond to the free variables. Let me perform these row operations for you. --- Starting with the augmented matrix: $$\begin{bmatrix} -2 & 0 & -10 & 4 & | & 0 \\ 1 & 2 & 3 & -1 & | & 0 \\ 1 & -4 & 1 & 0 & | & 0 \end{bmatrix}$$ #### Row Operations 1. Multiply the first row by $$-\frac{1}{2}$$: $$\begin{bmatrix} 1 & 0 & 5 & -2 & | & 0 \\ 1 & 2 & 3 & -1 & | & 0 \\ 1 & -4 & 1 & 0 & | & 0 \end{bmatrix}$$ 2. Subtract the first row from the second and third rows to create zeros below the first pivot: $$\begin{bmatrix} 1 & 0 & 5 & -2 & | & 0 \\ 0 & 2 & -2 & 1 & | & 0 \\ 0 & -4 & -4 & 2 & | & 0 \end{bmatrix}$$ 3. Divide the second row by 2: $$\begin{bmatrix} 1 & 0 & 5 & -2 & | & 0 \\ 0 & 1 & -1 & 0.5 & | & 0 \\ 0 & -4 & -4 & 2 & | & 0 \end{bmatrix}$$ 4. Add 4 times the second row to the third row to eliminate the $$y$$-term in the third row: $$\begin{bmatrix} 1 & 0 & 5 & -2 & | & 0 \\ 0 & 1 & -1 & 0.5 & | & 0 \\ 0 & 0 & 0 & 4 & | & 0 \end{bmatrix}$$ 5. Divide the third row by 4 to get a leading 1 in the fourth column: $$\begin{bmatrix} 1 & 0 & 5 & -2 & | & 0 \\ 0 & 1 & -1 & 0.5 & | & 0 \\ 0 & 0 & 0 & 1 & | & 0 \end{bmatrix}$$ 6. Now, we back-substitute to clear the entries above each pivot.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Homogeneous System of Linear Equations
Solution Space
Basis
Formulas
Matrix row reduction
Expressing solutions in parametric vector form
Theorems
Row Echelon Form
Basis for Solution Space
Suitable Grade Level
Undergraduate Level (Linear Algebra Course)