Math Problem Statement
s the set { ( 3 6 9 ) ,
( 15 30 45 ) ,
( − 5 − 6 − 15 ) } ⎩ ⎨ ⎧
3 6 9
,
15 30 45
,
−5 −6 −15
⎭ ⎬ ⎫ linearly independent?
Solution
To determine if the set
\begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix}, \begin{pmatrix} 15 \\ 30 \\ 45 \end{pmatrix}, \begin{pmatrix} -5 \\ -6 \\ -15 \end{pmatrix} \right\}$$ is linearly independent, we need to check if there exist scalars $$c_1$$, $$c_2$$, and $$c_3$$, not all zero, such that: $$c_1 \begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix} + c_2 \begin{pmatrix} 15 \\ 30 \\ 45 \end{pmatrix} + c_3 \begin{pmatrix} -5 \\ -6 \\ -15 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.$$ This equation translates into a system of linear equations: $$\begin{cases} 3c_1 + 15c_2 - 5c_3 = 0, \\ 6c_1 + 30c_2 - 6c_3 = 0, \\ 9c_1 + 45c_2 - 15c_3 = 0. \end{cases}$$ ### Step 1: Set up the augmented matrix The augmented matrix for this system is: $$\begin{pmatrix} 3 & 15 & -5 & | & 0 \\ 6 & 30 & -6 & | & 0 \\ 9 & 45 & -15 & | & 0 \\ \end{pmatrix}$$ ### Step 2: Row reduce the matrix We can perform row operations to simplify this matrix and determine if there is a non-trivial solution (a solution where at least one of the $$c_i$$'s is not zero). 1. Divide the first row by 3 to simplify: $$\begin{pmatrix} 1 & 5 & -\frac{5}{3} & | & 0 \\ 6 & 30 & -6 & | & 0 \\ 9 & 45 & -15 & | & 0 \\ \end{pmatrix}$$ 2. Subtract 6 times the first row from the second row and 9 times the first row from the third row: $$\begin{pmatrix} 1 & 5 & -\frac{5}{3} & | & 0 \\ 0 & 0 & 4 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ \end{pmatrix}$$ From this row-reduced form, we see that the second row simplifies to $$0 = 0$$, which indicates a dependency among the vectors. Specifically, the matrix has only two pivot positions (one in the first column and one in the third column), suggesting that the rank of the matrix is less than 3. ### Conclusion Since the rank is less than the number of vectors, these vectors are **linearly dependent**. Therefore, the set is **not linearly independent**. Would you like further details or have questions about this solution? Here are 5 related questions for further exploration: 1. How does one determine linear independence in higher dimensions? 2. Can a set of two vectors in $$\mathbb{R}^3$$ be linearly independent? 3. How is linear dependence related to the concept of basis? 4. What is the significance of the rank of a matrix in terms of vector spaces? 5. How does one find a basis for the span of a set of dependent vectors? **Tip:** Linear dependence among vectors implies that one vector can be expressed as a combination of others.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Linear Independence
Vector Spaces
Formulas
c1 * v1 + c2 * v2 + c3 * v3 = 0 (for checking linear independence)
Matrix row reduction
Theorems
Linear Independence Theorem
Rank Theorem
Suitable Grade Level
Grades 11-12 and Undergraduate
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