Decide whether the following subset of R
3
is linearly dependent or linearly

independent.






1
−3
5

 ,


2
2
4

 ,


4
−4
14





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:

\[
\begin{cases}
c_1 + 2c_2 + 4c_3 = 0, \\
-3c_1 + 2c_2 - 4c_3 = 0, \\
5c_1 + 4c_2 + 14c_3 = 0.
\end{cases}
\]

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.

Learn how to determine if a set of vectors in R^3 is linearly dependent or independent with step-by-step solutions and Gaussian elimination. This problem explores the concepts of linear algebra, vector spaces, and matrix operations.

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