Math Problem Statement
Determine whether the following vectors are linearly dependent. (If the set is linearly independent, enter INDEPENDENT. If the set is linearly dependent, enter your answer as a linear combination of v1, v2, and v3 that equals the zero vector.) v1 =
−8
17 7 , v2 =
−3
6 3 , v3 =
−1
1 2
Solution
To determine whether the vectors , , and are linearly dependent or independent, we solve the equation:
This expands to:
Equivalently, it gives a system of linear equations:
-8c_1 - 3c_2 - c_3 &= 0, \\ 17c_1 + 6c_2 + c_3 &= 0, \\ 7c_1 + 3c_2 + 2c_3 &= 0. \end{aligned}$$ ### Step 1: Represent the system in augmented matrix form $$\begin{bmatrix} -8 & -3 & -1 & 0 \\ 17 & 6 & 1 & 0 \\ 7 & 3 & 2 & 0 \end{bmatrix}.$$ ### Step 2: Solve the system using row reduction We will perform Gaussian elimination to determine whether a nontrivial solution exists for $$c_1, c_2, c_3$$. 1. **Normalize the first row** by dividing it by $$-8$$ (to make the pivot 1): $$\begin{bmatrix} 1 & \frac{3}{8} & \frac{1}{8} & 0 \\ 17 & 6 & 1 & 0 \\ 7 & 3 & 2 & 0 \end{bmatrix}.$$ 2. **Eliminate the first column of the second and third rows**: $$R_2 \to R_2 - 17R_1, \quad R_3 \to R_3 - 7R_1.$$ This gives: $$\begin{bmatrix} 1 & \frac{3}{8} & \frac{1}{8} & 0 \\ 0 & \frac{15}{8} & \frac{1}{8} & 0 \\ 0 & \frac{9}{8} & \frac{9}{8} & 0 \end{bmatrix}.$$ 3. **Normalize the second row** by dividing it by $$\frac{15}{8}$$ (to make the pivot 1): $$\begin{bmatrix} 1 & \frac{3}{8} & \frac{1}{8} & 0 \\ 0 & 1 & \frac{1}{15} & 0 \\ 0 & \frac{9}{8} & \frac{9}{8} & 0 \end{bmatrix}.$$ 4. **Eliminate the second column of the first and third rows**: $$R_1 \to R_1 - \frac{3}{8}R_2, \quad R_3 \to R_3 - \frac{9}{8}R_2.$$ This gives: $$\begin{bmatrix} 1 & 0 & \frac{1}{10} & 0 \\ 0 & 1 & \frac{1}{15} & 0 \\ 0 & 0 & \frac{1}{5} & 0 \end{bmatrix}.$$ 5. **Normalize the third row** by dividing it by $$\frac{1}{5}$$: $$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}.$$ ### Step 3: Interpretation The reduced row echelon form shows no free variables; hence, the only solution is $$c_1 = 0$$, $$c_2 = 0$$, and $$c_3 = 0$$. Thus, the vectors $$\mathbf{v}_1$$, $$\mathbf{v}_2$$, and $$\mathbf{v}_3$$ are **linearly independent**. --- Would you like further clarification on the steps or another example? 😊 ### Related Questions 1. What does it mean for vectors to be linearly dependent? 2. How do you find the rank of a matrix? 3. What is the geometric interpretation of linear independence? 4. Can you provide a similar example but with dependent vectors? 5. How is linear independence used in real-world applications? ### Tip: To check for linear independence quickly, calculate the determinant of the square matrix formed by the vectors (if applicable). A nonzero determinant implies independence!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Linear Independence
Systems of Linear Equations
Gaussian Elimination
Vector Spaces
Formulas
c1 * v1 + c2 * v2 + c3 * v3 = 0
Gaussian Elimination
Theorems
Theorem of Linear Independence
Rank of Matrix
Reduced Row Echelon Form (RREF)
Suitable Grade Level
Grades 11-12
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