To determine the value of $c$ that will make the set of vectors linearly dependent, we need to analyze the three vectors in the set $S = \{(1, 2, -1), (3, 1, 0), (-1, 2, c)\}$. The vectors will be linearly dependent if there is a non-trivial linear combination of these vectors that equals the zero vector.

In other words, we want to find whether there exist scalars $\lambda_1, \lambda_2, \lambda_3$, not all zero, such that:

$\lambda_1(1, 2, -1) + \lambda_2(3, 1, 0) + \lambda_3(-1, 2, c) = (0, 0, 0)$

This equation can be broken down into three separate component equations (one for each component of the vector):

$\lambda_1(1) + \lambda_2(3) + \lambda_3(-1) = 0 \quad \text{(for the first component)}$ $\lambda_1(2) + \lambda_2(1) + \lambda_3(2) = 0 \quad \text{(for the second component)}$ $\lambda_1(-1) + \lambda_2(0) + \lambda_3(c) = 0 \quad \text{(for the third component)}$

This gives us the following system of linear equations:

$\lambda_1 + 3\lambda_2 - \lambda_3 = 0$ $2\lambda_1 + \lambda_2 + 2\lambda_3 = 0$ $-\lambda_1 + c\lambda_3 = 0$

Now, let's solve this system to find the condition on $c$ for which a non-trivial solution exists.

Step 1: Solve the third equation

From the third equation, we get:

$-\lambda_1 + c\lambda_3 = 0 \quad \Rightarrow \quad \lambda_1 = c\lambda_3$

Step 2: Substitute into the first two equations

Substitute $\lambda_1 = c\lambda_3$ into the first two equations:

1. $c\lambda_3 + 3\lambda_2 - \lambda_3 = 0 \Rightarrow (c - 1)\lambda_3 + 3\lambda_2 = 0$
2. $2(c\lambda_3) + \lambda_2 + 2\lambda_3 = 0 \Rightarrow 2c\lambda_3 + \lambda_2 + 2\lambda_3 = 0 \Rightarrow (2c + 2)\lambda_3 + \lambda_2 = 0$

Now we have the system:

$(c - 1)\lambda_3 + 3\lambda_2 = 0 \quad \text{(Equation A)}$ $(2c + 2)\lambda_3 + \lambda_2 = 0 \quad \text{(Equation B)}$

Step 3: Solve the system for $\lambda_2$ and $\lambda_3$

From Equation B, solve for $\lambda_2$:

$\lambda_2 = -(2c + 2)\lambda_3$

Substitute this expression for $\lambda_2$ into Equation A:

$(c - 1)\lambda_3 + 3(-(2c + 2)\lambda_3) = 0$ $(c - 1)\lambda_3 - 3(2c + 2)\lambda_3 = 0$ $\left[(c - 1) - 6(c + 1)\right]\lambda_3 = 0$ $(c - 1) - 6c - 6 = 0$ $-5c - 7 = 0$ $c = -\frac{7}{5}$

Conclusion:

The set of vectors $S = \{(1, 2, -1), (3, 1, 0), (-1, 2, c)\}$ is linearly dependent when $c = -\frac{7}{5}$.

Would you like further details on this process or a different explanation?

Here are five related questions to deepen your understanding:

1. How can you determine if a set of vectors is linearly independent using the determinant?
2. What is the geometric interpretation of linear dependence in $\mathbb{R}^3$?
3. How does the rank of a matrix relate to linear independence of its rows or columns?
4. What happens if the value of $c$ is different from $-\frac{7}{5}$? Will the vectors be linearly independent?
5. How does one find the span of a set of vectors in $\mathbb{R}^3$?

Tip: A set of vectors is linearly dependent if one vector in the set can be expressed as a linear combination of the others.