Math Problem Statement

Consider the set of three vectors S={(1,2,−1),(3,1,0),(−1,2,c)} in R3 with usual addition and scalar multiplication. For which value of c will the above set S be linearly dependent

Solution

To determine the value of cc for which the set S={(1,2,1),(3,1,0),(1,2,c)}S = \{(1, 2, -1), (3, 1, 0), (-1, 2, c)\} is linearly dependent, we need to check the condition for linear dependence.

Linear Dependence Condition:

Three vectors v1=(1,2,1)\mathbf{v}_1 = (1, 2, -1), v2=(3,1,0)\mathbf{v}_2 = (3, 1, 0), and v3=(1,2,c)\mathbf{v}_3 = (-1, 2, c) are linearly dependent if there exist scalars α1,α2,α3\alpha_1, \alpha_2, \alpha_3, not all zero, such that:

α1v1+α2v2+α3v3=0\alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 + \alpha_3 \mathbf{v}_3 = 0

This can be rewritten as a system of linear equations:

α1(121)+α2(310)+α3(12c)=(000)\alpha_1 \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \alpha_2 \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + \alpha_3 \begin{pmatrix} -1 \\ 2 \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

This results in the following system of equations:

  1. α1+3α2α3=0\alpha_1 + 3\alpha_2 - \alpha_3 = 0
  2. 2α1+α2+2α3=02\alpha_1 + \alpha_2 + 2\alpha_3 = 0
  3. α1+cα3=0-\alpha_1 + c\alpha_3 = 0

For the vectors to be linearly dependent, the determinant of the matrix formed by the vectors must be zero. This matrix is:

1 & 3 & -1 \\ 2 & 1 & 2 \\ -1 & 0 & c \end{pmatrix}$$ We calculate the determinant of $$A$$: $$\text{det}(A) = 1 \cdot \begin{vmatrix} 1 & 2 \\ 0 & c \end{vmatrix} - 3 \cdot \begin{vmatrix} 2 & 2 \\ -1 & c \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & 1 \\ -1 & 0 \end{vmatrix}$$ Now, compute each of these 2x2 determinants: $$\begin{vmatrix} 1 & 2 \\ 0 & c \end{vmatrix} = 1 \cdot c - 2 \cdot 0 = c$$ $$\begin{vmatrix} 2 & 2 \\ -1 & c \end{vmatrix} = 2c - (-2) = 2c + 2$$ $$\begin{vmatrix} 2 & 1 \\ -1 & 0 \end{vmatrix} = 2 \cdot 0 - 1 \cdot (-1) = 1$$ Substituting these values back into the determinant formula: $$\text{det}(A) = 1 \cdot c - 3 \cdot (2c + 2) - 1 \cdot 1$$ Simplifying: $$\text{det}(A) = c - 3(2c + 2) - 1 = c - 6c - 6 - 1 = -5c - 7$$ For the vectors to be linearly dependent, we need: $$\text{det}(A) = 0$$ Solving for $$c$$: $$-5c - 7 = 0$$ $$-5c = 7$$ $$c = -\frac{7}{5}$$ ### Conclusion: The value of $$c$$ for which the set $$S = \{(1, 2, -1), (3, 1, 0), (-1, 2, c)\}$$ is linearly dependent is $$c = -\frac{7}{5}$$. --- Would you like further explanation or have any questions? Here are some follow-up questions: 1. How can we interpret linear dependence geometrically in $$\mathbb{R}^3$$? 2. How does changing the value of $$c$$ affect the vectors' span? 3. What happens if you add or subtract a scalar multiple of one vector from another in a set of vectors? 4. What does it mean for a set of vectors to span $$\mathbb{R}^3$$? 5. How can you determine if a set of vectors is linearly independent? **Tip:** To verify whether vectors are linearly independent or dependent, always check the determinant of the matrix formed by placing the vectors as columns in the matrix. If the determinant is zero, they are dependent.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Linear Dependence
Determinants

Formulas

Det(A) = 0 for linear dependence
Determinant of a 3x3 matrix: det(A) = a(ei − fh) − b(di − fg) + c(dh − eg)

Theorems

Condition for linear dependence: determinant of matrix formed by vectors must be zero

Suitable Grade Level

Undergraduate Level (Linear Algebra Course)