Math Problem Statement
Consider the following system of linear equations: x1 + 2x2 + 3x3 + 4x4 = 0, 4x1 + 5x2 + 6x3 + 7x4 = 0, and 6x1 + 7x2 + 8x3 + 9x4 = 0. (a) (12 pts) Find a basis for the column space of the coecient matrix of this system.
Solution
To find a basis for the column space of the coefficient matrix of the system of linear equations, we will start by setting up the coefficient matrix for the system. Then, we will reduce it to find the linearly independent columns.
The system of equations is:
x_1 + 2x_2 + 3x_3 + 4x_4 = 0 \\ 4x_1 + 5x_2 + 6x_3 + 7x_4 = 0 \\ 6x_1 + 7x_2 + 8x_3 + 9x_4 = 0 \\ \end{cases}$$ The coefficient matrix $$A$$ is: $$A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 4 & 5 & 6 & 7 \\ 6 & 7 & 8 & 9 \end{bmatrix}$$ ### Step 1: Row Reduce the Matrix To identify the linearly independent columns, we will reduce $$A$$ to its row echelon form (REF). Starting with: $$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 4 & 5 & 6 & 7 \\ 6 & 7 & 8 & 9 \end{bmatrix}$$ 1. Make the entry below the pivot in the first column zero by row operations: $$R_2 = R_2 - 4R_1 \Rightarrow \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & -3 & -6 & -9 \\ 6 & 7 & 8 & 9 \end{bmatrix}$$ 2. Make the entry in the third row, first column zero: $$R_3 = R_3 - 6R_1 \Rightarrow \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & -3 & -6 & -9 \\ 0 & -5 & -10 & -15 \end{bmatrix}$$ 3. Divide $$R_2$$ by $$-3$$ to make the second pivot a 1: $$R_2 = \frac{1}{-3} R_2 \Rightarrow \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & -5 & -10 & -15 \end{bmatrix}$$ 4. Eliminate the entry below the second pivot: $$R_3 = R_3 + 5R_2 \Rightarrow \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ ### Step 2: Identify Pivot Columns The resulting row echelon form of the matrix is: $$\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ From this form, we see that the pivot columns are the first and second columns. ### Step 3: Extract Basis Vectors The pivot columns correspond to the original first and second columns of $$A$$. Therefore, a basis for the column space of $$A$$ is given by these columns: $$\text{Basis} = \left\{ \begin{bmatrix} 1 \\ 4 \\ 6 \end{bmatrix}, \begin{bmatrix} 2 \\ 5 \\ 7 \end{bmatrix} \right\}$$ Thus, the basis for the column space of the coefficient matrix is: $$\left\{ \begin{bmatrix} 1 \\ 4 \\ 6 \end{bmatrix}, \begin{bmatrix} 2 \\ 5 \\ 7 \end{bmatrix} \right\}$$ Would you like further details on how to verify this basis or on other properties related to the column space? Here are some related questions you may explore: 1. How can we determine the rank of the coefficient matrix? 2. What is the null space of the coefficient matrix? 3. How would we solve the homogeneous system of equations? 4. How can we check if a vector is in the column space of the matrix? 5. What are the applications of the column space in real-world problems? **Tip:** When finding the basis for the column space, focus on identifying pivot columns in the row-reduced form, as they represent the linearly independent columns of the original matrix.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Operations
Row Reduction
Column Space
Formulas
Row reduction to Row Echelon Form
Pivot column identification for basis determination
Theorems
Rank-Nullity Theorem
Suitable Grade Level
Undergraduate Mathematics (Linear Algebra)