To determine whether the set of vectors is linearly dependent or independent in $\mathbb{R}^3$, we need to check if there exists a non-trivial linear combination of these vectors that equals the zero vector.

Let's consider the set of vectors:

$\mathbf{v}_1 = \begin{pmatrix} 1 \\ -3 \\ 5 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} 2 \\ 2 \\ 4 \end{pmatrix}, \quad \mathbf{v}_3 = \begin{pmatrix} 4 \\ -4 \\ 14 \end{pmatrix}.$

These vectors are linearly dependent if there exist scalars $c_1, c_2,$ and $c_3$, not all zero, such that:

$c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{0}.$

This equation expands to:

$c_1 \begin{pmatrix} 1 \\ -3 \\ 5 \end{pmatrix} + c_2 \begin{pmatrix} 2 \\ 2 \\ 4 \end{pmatrix} + c_3 \begin{pmatrix} 4 \\ -4 \\ 14 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.$

This can be rewritten as a system of linear equations:

c_1 + 2c_2 + 4c_3 = 0, \\ -3c_1 + 2c_2 - 4c_3 = 0, \\ 5c_1 + 4c_2 + 14c_3 = 0. \end{cases}$$ To determine if there is a non-trivial solution, we can form the augmented matrix and perform row reduction (Gaussian elimination): $$\left[ \begin{array}{ccc|c} 1 & 2 & 4 & 0 \\ -3 & 2 & -4 & 0 \\ 5 & 4 & 14 & 0 \end{array} \right].$$ Let's perform row operations to reduce this matrix to its row echelon form: 1. **Start with the matrix:** $$\left[ \begin{array}{ccc|c} 1 & 2 & 4 & 0 \\ -3 & 2 & -4 & 0 \\ 5 & 4 & 14 & 0 \end{array} \right].$$ 2. **Eliminate the first column below the pivot in the first row.** Add $$3$$ times the first row to the second row: $$R_2 = R_2 + 3R_1 \rightarrow \left[ \begin{array}{ccc|c} 1 & 2 & 4 & 0 \\ 0 & 8 & 8 & 0 \\ 5 & 4 & 14 & 0 \end{array} \right].$$ Subtract $$5$$ times the first row from the third row: $$R_3 = R_3 - 5R_1 \rightarrow \left[ \begin{array}{ccc|c} 1 & 2 & 4 & 0 \\ 0 & 8 & 8 & 0 \\ 0 & -6 & -6 & 0 \end{array} \right].$$ 3. **Simplify the second row by dividing it by 8:** $$R_2 = \frac{1}{8} R_2 \rightarrow \left[ \begin{array}{ccc|c} 1 & 2 & 4 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & -6 & -6 & 0 \end{array} \right].$$ 4. **Eliminate the second column below the pivot in the second row.** Add $$6$$ times the second row to the third row: $$R_3 = R_3 + 6R_2 \rightarrow \left[ \begin{array}{ccc|c} 1 & 2 & 4 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right].$$ The third row is now all zeros, indicating that the rank of the matrix is less than 3. Since there are fewer pivots than there are vectors, the vectors $$\mathbf{v}_1, \mathbf{v}_2,$$ and $$\mathbf{v}_3$$ are **linearly dependent**. ### Conclusion The set $$\left\{ \begin{pmatrix} 1 \\ -3 \\ 5 \end{pmatrix}, \begin{pmatrix} 2 \\ 2 \\ 4 \end{pmatrix}, \begin{pmatrix} 4 \\ -4 \\ 14 \end{pmatrix} \right\}$$ is **linearly dependent**. Would you like more details or have any questions? --- Here are 5 related questions you might find interesting: 1. How do you determine if a set of vectors in $$\mathbb{R}^n$$ is linearly dependent? 2. What is the geometric interpretation of linear dependence in $$\mathbb{R}^3$$? 3. How can the concept of linear dependence be extended to more than three vectors? 4. What does it mean if the determinant of a matrix formed by a set of vectors is zero? 5. How can the Gram-Schmidt process be used to obtain a set of linearly independent vectors from a dependent set? **Tip:** When solving for linear independence or dependence, checking the rank of the matrix or using the determinant (for square matrices) are efficient methods.