Math Problem Statement

Find a basis for the solution space of the given homogeneous system of equations: \[ \begin{bmatrix} -2 & 0 & -10 & 4 \\ 1 & -2 & 3 & -1 \\ 1 & -4 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}. \]

Solution

To find a basis for the solution space of the given homogeneous system, we need to solve the system of linear equations:

-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}.$$ ### Step 1: Row Reduction To solve this system, we’ll row reduce the matrix to its reduced row echelon form (RREF) to identify the free variables. 1. Start with the matrix: $$\begin{bmatrix} -2 & 0 & -10 & 4 \\ 1 & -2 & 3 & -1 \\ 1 & -4 & 1 & 0 \end{bmatrix}.$$ 2. Perform row operations to simplify this matrix. ### Step 2: Identify Free Variables Once in RREF, we can determine which variables are leading (basic) and which are free. The free variables will correspond to the parameters in our general solution, forming the basis vectors. ### Step 3: Form Basis Vectors The basis vectors for the solution space will be the vectors corresponding to each free variable in terms of the leading variables. Let's complete the row reduction and solve this to give the final basis vectors. I'll proceed with the row reduction now.After row reducing, we obtain the matrix in reduced row echelon form (RREF): $$\begin{bmatrix} 1 & 0 & 5 & -2 \\ 0 & 1 & 1 & -\frac{1}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix}.$$ From this, we can identify the pivot columns (leading variables) as $$x_1$$ and $$x_2$$, while $$x_3$$ and $$x_4$$ are free variables. ### Expressing Variables in Terms of Free Variables 1. From the first row: $$x_1 = -5x_3 + 2x_4$$. 2. From the second row: $$x_2 = -x_3 + \frac{1}{2}x_4$$. Letting $$x_3 = t$$ and $$x_4 = s$$, where $$t$$ and $$s$$ are parameters, we can express the solution vector as: $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = t \begin{bmatrix} -5 \\ -1 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 2 \\ \frac{1}{2} \\ 0 \\ 1 \end{bmatrix}.$$ ### Basis for the Solution Space The basis for the solution space is thus: $$\left\{ \begin{bmatrix} -5 \\ -1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ \frac{1}{2} \\ 0 \\ 1 \end{bmatrix} \right\}.$$ Would you like further details on each step or have any questions? ### Additional Questions to Consider 1. How does row reduction help in determining the basis of a solution space? 2. Why are some variables chosen as free, while others are pivot variables? 3. What is the significance of a basis in a vector space? 4. How can this solution be interpreted geometrically? 5. How would the basis change if we altered the homogeneous system? ### Tip In linear algebra, the number of free variables in a homogeneous system is equal to the dimension of the solution space, which can help predict the number of basis vectors.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Homogeneous Systems
Basis of Solution Space
Row Reduction

Formulas

Matrix Row Reduction
Basis Vectors Identification

Theorems

Theorem of Homogeneous System Solutions

Suitable Grade Level

Undergraduate Level