To determine whether the given vectors are linearly dependent or independent, we need to check if there exists a non-trivial linear combination of the vectors that equals the zero vector. That is, we want to see if there exist scalars $c_1, c_2, c_3$, not all zero, such that:

[ c_1 [2, -1, 3, 2] + c_2 [1, 3, 4, 2] + c_3 [3, -5, 2, 2] = [0, 0, 0, 0] ]

This can be expressed as the following system of equations by writing it component-wise:

$c_1(2) + c_2(1) + c_3(3) = 0 \quad \text{(1st component)}$ $c_1(-1) + c_2(3) + c_3(-5) = 0 \quad \text{(2nd component)}$ $c_1(3) + c_2(4) + c_3(2) = 0 \quad \text{(3rd component)}$ $c_1(2) + c_2(2) + c_3(2) = 0 \quad \text{(4th component)}$

This gives the system of equations: $2c_1 + c_2 + 3c_3 = 0 \quad \text{(1)}$ $-c_1 + 3c_2 - 5c_3 = 0 \quad \text{(2)}$ $3c_1 + 4c_2 + 2c_3 = 0 \quad \text{(3)}$ $2c_1 + 2c_2 + 2c_3 = 0 \quad \text{(4)}$

Solving the system:

To solve this, we can write the coefficient matrix and augment it with the zero vector:

2 & 1 & 3 \\ -1 & 3 & -5 \\ 3 & 4 & 2 \\ 2 & 2 & 2 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ We now reduce this matrix to row echelon form (REF) using Gaussian elimination to determine if the system has only the trivial solution $$c_1 = c_2 = c_3 = 0$$. Performing the row reduction, we get: $$\begin{pmatrix} 2 & 1 & 3 \\ -1 & 3 & -5 \\ 3 & 4 & 2 \\ 2 & 2 & 2 \end{pmatrix} \xrightarrow[]{RREF} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$ ### Conclusion: Since we obtain a row of zeros and three leading ones (pivots), there is only the **trivial solution** $$c_1 = 0$$, $$c_2 = 0$$, and $$c_3 = 0$$. Therefore, the given vectors are **linearly independent**. Would you like more details or have any questions? --- Here are 5 related questions to extend your understanding: 1. How can the determinant of the matrix formed by vectors help determine linear dependence? 2. What does it mean if there are free variables when solving a system of equations? 3. How can the concept of a basis relate to linearly independent vectors? 4. What is the geometric interpretation of linear independence in 3D or 4D space? 5. Can you explain why having more vectors than dimensions guarantees linear dependence? **Tip:** Linear independence of vectors is crucial for constructing bases in vector spaces, which are fundamental for many areas in mathematics, including solving systems of equations and performing transformations.