Show that the vectors (2, 3, 1)T , (2, 1, 3)T and (1, 1, 1)T are linearly dependent in R3

To show that the vectors \(\mathbf{v}_1 = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}\), \(\mathbf{v}_2 = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\), and \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\) are linearly dependent, we need to prove that 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 is equivalent to solving the equation:

\[
c_1 \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + c_2 \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} + c_3 \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\]

Writing this out in component form gives the following system of equations:

\[
2c_1 + 2c_2 + c_3 = 0 \tag{1}
\]
\[
3c_1 + c_2 + c_3 = 0 \tag{2}
\]
\[
c_1 + 3c_2 + c_3 = 0 \tag{3}
\]

### Step 1: Solve the system of equations
We can solve this system using standard algebraic techniques, such as substitution or elimination. Let's use elimination here.

Subtract equation (1) from equation (2):
\[
(3c_1 + c_2 + c_3) - (2c_1 + 2c_2 + c_3) = 0
\]
\[
c_1 - c_2 = 0 \quad \Rightarrow \quad c_1 = c_2 \tag{4}
\]

Now subtract equation (1) from equation (3):
\[
(c_1 + 3c_2 + c_3) - (2c_1 + 2c_2 + c_3) = 0
\]
\[
-c_1 + c_2 = 0 \quad \Rightarrow \quad c_1 = c_2 \tag{5}
\]

### Step 2: Substitute \( c_1 = c_2 \) into one of the original equations
Substitute \( c_1 = c_2 \) into equation (1):
\[
2c_1 + 2c_1 + c_3 = 0
\]
\[
4c_1 + c_3 = 0 \quad \Rightarrow \quad c_3 = -4c_1
\]

### Step 3: Conclusion
We have \( c_1 = c_2 \) and \( c_3 = -4c_1 \). Therefore, the solution to the system is:

\[
c_1 = t, \quad c_2 = t, \quad c_3 = -4t
\]

where \( t \) is any non-zero scalar. Since not all the coefficients \( c_1, c_2, c_3 \) are zero, the vectors are **linearly dependent**.

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Would you like more details or have any further questions? Here are some related topics:

1. What does it mean for vectors to be linearly independent?
2. How can we use matrix methods to check for linear dependence?
3. What is the geometric interpretation of linearly dependent vectors?
4. How does the rank of a matrix relate to linear dependence?
5. Can we extend this concept to higher dimensions?

**Tip:** The easiest way to check for linear dependence in \( \mathbb{R}^3 \) is to see if the determinant of the matrix formed by the vectors is zero!

Show that the vectors (2, 3, 1)T , (2, 1, 3)T and (1, 1, 1)T are linearly dependent in R3.

Learn how to prove that the vectors (2, 3, 1)T, (2, 1, 3)T, and (1, 1, 1)T are linearly dependent in R3. This step-by-step solution involves solving a system of equations and applying the linear dependence theorem, making it perfect for students studying linear algebra at the undergraduate level.

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