Math Problem Statement

Validate this example: Example: The three vectors are linearly dependent because 1 2 3 a  [3 0 2 2], a  [6 42 24 54], a  [21  21 0 15] 1 2 3 6 1 0

Solution

Let's break down the example you provided.

It seems that you have three vectors v1,v2,v3\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 represented as:

\mathbf{v}_2 = \begin{bmatrix} -6 \\ 42 \\ 24 \\ 54 \end{bmatrix}, \quad \mathbf{v}_3 = \begin{bmatrix} 21 \\ -21 \\ 0 \\ -15 \end{bmatrix}$$ The claim is that these three vectors are **linearly dependent**. To verify this, we need to check whether there exist scalars $$c_1, c_2, c_3$$ (not all zero) such that the following equation holds: $$c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{0}$$ This results in the following system of linear equations: $$c_1 \begin{bmatrix} 3 \\ 0 \\ 2 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} -6 \\ 42 \\ 24 \\ 54 \end{bmatrix} + c_3 \begin{bmatrix} 21 \\ -21 \\ 0 \\ -15 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$ Breaking this into individual components: 1. $$3c_1 - 6c_2 + 21c_3 = 0$$ 2. $$42c_2 - 21c_3 = 0$$ 3. $$2c_1 + 24c_2 = 0$$ 4. $$2c_1 + 54c_2 - 15c_3 = 0$$ Now we will solve this system of equations. **Step 1: Solve for $$c_3$$ in terms of $$c_2$$ from the second equation:** $$42c_2 - 21c_3 = 0 \quad \Rightarrow \quad c_3 = 2c_2$$ **Step 2: Substitute $$c_3 = 2c_2$$ into the other equations.** - From the first equation: $$3c_1 - 6c_2 + 21(2c_2) = 0 \quad \Rightarrow \quad 3c_1 - 6c_2 + 42c_2 = 0 \quad \Rightarrow \quad 3c_1 + 36c_2 = 0 \quad \Rightarrow \quad c_1 = -12c_2$$ - From the third equation: $$2(-12c_2) + 24c_2 = 0 \quad \Rightarrow \quad -24c_2 + 24c_2 = 0$$ This equation is satisfied trivially. - Finally, substitute $$c_1 = -12c_2$$ and $$c_3 = 2c_2$$ into the fourth equation: $$2(-12c_2) + 54c_2 - 15(2c_2) = 0 \quad \Rightarrow \quad -24c_2 + 54c_2 - 30c_2 = 0 \quad \Rightarrow \quad 0 = 0$$ This is also satisfied trivially. **Conclusion:** We found that $$c_1 = -12c_2$$ and $$c_3 = 2c_2$$, where $$c_2$$ can be any scalar. Since we have found a non-trivial solution (not all $$c_1, c_2, c_3$$ are zero), the vectors are **linearly dependent**. --- Would you like more details on any part of the solution? Or would you like to explore similar concepts? Here are some related questions you might find interesting: 1. What is the geometric interpretation of linearly dependent vectors? 2. How can we check for linear dependence in higher dimensions (e.g., 5 or 6)? 3. What happens if a set of vectors is linearly independent? 4. How does the rank of a matrix relate to linear dependence? 5. Can a set of vectors be linearly dependent if some vectors are zero? **Tip:** Always check if there is a non-trivial solution to the linear system when testing for linear dependence.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Linear Dependence
Systems of Linear Equations

Formulas

c1 * v1 + c2 * v2 + c3 * v3 = 0
Solving a system of linear equations to check for linear dependence

Theorems

Linear Dependence Theorem

Suitable Grade Level

Undergraduate (College-Level Linear Algebra)