Math Problem Statement
Determine whether the vectors i) [5,0,0],[7,2,6],[9,4,-8] ii) [0,0,2],[0,5,-8],[-3,4,1] are linearly independent?
Solution
To determine if vectors are linearly independent, we check whether the only solution to the linear combination:
is .
This involves forming a matrix with the vectors as rows (or columns) and determining if the determinant is nonzero (for square matrices) or if the rank of the matrix equals the number of vectors.
Let’s examine both cases.
Case i: Vectors [5, 0, 0], [7, 2, 6], [9, 4, -8]
The matrix formed is:
5 & 0 & 0 \\ 7 & 2 & 6 \\ 9 & 4 & -8 \end{bmatrix}$$ We calculate the determinant of $$A$$: $$\text{det}(A) = 5 \begin{vmatrix} 2 & 6 \\ 4 & -8 \end{vmatrix} - 0 \cdot \begin{vmatrix} 7 & 6 \\ 9 & -8 \end{vmatrix} + 0 \cdot \begin{vmatrix} 7 & 2 \\ 9 & 4 \end{vmatrix}$$ $$\text{det}(A) = 5 \cdot ((2)(-8) - (6)(4)) = 5 \cdot (-16 - 24) = 5 \cdot (-40) = -200$$ Since the determinant is nonzero, the vectors in case i are **linearly independent**. --- ### **Case ii: Vectors [0, 0, 2], [0, 5, -8], [-3, 4, 1]** The matrix formed is: $$B = \begin{bmatrix} 0 & 0 & 2 \\ 0 & 5 & -8 \\ -3 & 4 & 1 \end{bmatrix}$$ We calculate the determinant of $$B$$: $$\text{det}(B) = 0 \cdot \begin{vmatrix} 5 & -8 \\ 4 & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 0 & -8 \\ -3 & 1 \end{vmatrix} + 2 \cdot \begin{vmatrix} 0 & 5 \\ -3 & 4 \end{vmatrix}$$ The first two terms vanish due to multiplication by 0. The remaining term is: $$\text{det}(B) = 2 \cdot ((0)(4) - (5)(-3)) = 2 \cdot (0 + 15) = 2 \cdot 15 = 30$$ Since the determinant is nonzero, the vectors in case ii are also **linearly independent**. --- ### Final Answer: - i) Vectors [5, 0, 0], [7, 2, 6], [9, 4, -8] are **linearly independent**. - ii) Vectors [0, 0, 2], [0, 5, -8], [-3, 4, 1] are **linearly independent**. Would you like a detailed explanation of the determinant calculation steps? Here are related questions to expand: --- 1. What does it mean geometrically for vectors to be linearly independent? 2. How does determinant relate to vector independence in 3D space? 3. What is the rank of a matrix, and how does it relate to independence? 4. Can linear independence be verified using row reduction instead of determinants? 5. How does the method change for more than 3 vectors? **Tip:** For 3D vectors, the determinant directly relates to the volume of the parallelepiped formed by the vectors. A nonzero determinant implies non-degenerate geometry and independence.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Linear Independence
Matrix Determinants
Formulas
Linear combination: c1*v1 + c2*v2 + c3*v3 = 0
Determinant formula for 3x3 matrix: det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)
Theorems
Linear Independence Criterion (Determinant-based)
Suitable Grade Level
Undergraduate (Introductory Linear Algebra)