Consider the set of three vectors S={(1,2,−1),(3,1,0),(−1,2,c)}*S*={(1,2,−1),(3,1,0),(−1,2,*c*)} in R3R3 with usual addition and scalar multiplication. For which value of c*c* will the above set S*S* be linearly dependent?

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.

Consider the set of three vectors S={(1,2,−1),(3,1,0),(−1,2,c)} in R^3 with usual addition and scalar multiplication. For which value of c will the above set S be linearly dependent?

Discover how to find the value of c that makes the set of vectors S = {(1, 2, -1), (3, 1, 0), (-1, 2, c)} in R^3 linearly dependent. This step-by-step solution explores concepts of linear dependence and system of equations. Suitable for students studying linear algebra.

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