Math Problem Statement

Problem 1.1: Find a combination x_1 w_1+x_2 w_2+x_3 w_3 that gives the zero vector: w_1=[■(2@4@6)], w_2=[■(3@6@9)], w_3=[■(1@2@3)] Those vectors are (independent) (dependent). The three vectors lie in a ______________. The matrix W with these columns is not invertible.

Solution

Problem 1.1 Solution

To solve the problem, we analyze the vectors w1,w2,w3\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3 given by:

w1=[246],w2=[369],w3=[123].\mathbf{w}_1 = \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}, \quad \mathbf{w}_2 = \begin{bmatrix} 3 \\ 6 \\ 9 \end{bmatrix}, \quad \mathbf{w}_3 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.

We need to find a linear combination x1w1+x2w2+x3w3=0x_1\mathbf{w}_1 + x_2\mathbf{w}_2 + x_3\mathbf{w}_3 = \mathbf{0}, where 0\mathbf{0} is the zero vector.

Step 1: Write the system as a matrix equation

We write the linear combination in matrix form: Wx=0,W \mathbf{x} = \mathbf{0}, where

2 & 3 & 1 \\ 4 & 6 & 2 \\ 6 & 9 & 3 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \quad \mathbf{0} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$ ### Step 2: Check linear dependence The columns of $$W$$ are linearly dependent because $$\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3$$ are scalar multiples of each other: $$\mathbf{w}_2 = \frac{3}{2}\mathbf{w}_1, \quad \mathbf{w}_3 = \frac{1}{2}\mathbf{w}_1.$$ Since the vectors are dependent, the determinant of $$W$$ is zero, and $$W$$ is not invertible. ### Step 3: Solve for $$\mathbf{x}$$ We parametrize the solution. Since the vectors are dependent, there is an infinite number of solutions. A simple solution can be found by inspection: $$x_1 = 3, \quad x_2 = -2, \quad x_3 = 0.$$ Verify: $$3\mathbf{w}_1 - 2\mathbf{w}_2 + 0\mathbf{w}_3 = \begin{bmatrix} 6 \\ 12 \\ 18 \end{bmatrix} - \begin{bmatrix} 6 \\ 12 \\ 18 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \mathbf{0}.$$ Thus, a solution is: $$\mathbf{x} = \begin{bmatrix} 3 \\ -2 \\ 0 \end{bmatrix}.$$ ### Interpretation 1. The vectors $$\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3$$ are **dependent**. 2. The three vectors lie in a **plane** in 3D space because they are spanned by two linearly independent vectors. ### Conclusion - The matrix $$W$$ is not invertible. Let me know if you want additional steps or clarifications! --- ### Related Questions: 1. How can you verify linear dependence of vectors? 2. What is the geometric meaning of linearly dependent vectors in $$\mathbb{R}^n$$? 3. How does the determinant of a matrix relate to its invertibility? 4. Why is the solution to a dependent vector system not unique? 5. Can you generalize the method to higher-dimensional matrices? ### Tip: To check for linear dependence, calculate the rank of the matrix formed by the vectors. If the rank is less than the

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Linear Dependence
Matrix Invertibility

Formulas

Linear combination: x_1 * w_1 + x_2 * w_2 + x_3 * w_3 = 0
Matrix equation: W * x = 0

Theorems

Linear Dependence Theorem
Invertibility Theorem (Determinant = 0 implies non-invertibility)

Suitable Grade Level

Grades 11-12 or College Level