Suppose X 0

We are given the vector:

$X_0 = (a, b, 1 - a - b)$

and another vector $X_2$, and we know that the set $\{ X_0, X_2 \}$ is linearly dependent. This means that one of these vectors can be written as a scalar multiple of the other.

Step 1: Understanding Linear Dependence

Two vectors $X_0$ and $X_2$ are linearly dependent if there exists a scalar $\lambda$ such that:

$X_2 = \lambda X_0$

Step 2: Defining $X_2$

The problem does not directly specify $X_2$, but in many standard problems, $X_2$ is taken as:

$X_2 = (a^2, b^2, 1 - a^2 - b^2)$

Thus, for linear dependence:

$(a^2, b^2, 1 - a^2 - b^2) = \lambda (a, b, 1 - a - b)$

Step 3: Setting Up Equations

This gives us three component-wise equations:

1. $a^2 = \lambda a$
2. $b^2 = \lambda b$
3. $1 - a^2 - b^2 = \lambda (1 - a - b)$

From the first equation:

$\lambda = \frac{a^2}{a}, \quad \text{if } a \neq 0 \Rightarrow \lambda = a$

From the second equation:

$\lambda = \frac{b^2}{b}, \quad \text{if } b \neq 0 \Rightarrow \lambda = b$

Step 4: Solving for $b$

Since both expressions for $\lambda$ must be equal:

$a = b$

Using this in the constraint:

$a + b \leq 1$

$a + a \leq 1 \Rightarrow 2a \leq 1 \Rightarrow a \leq \frac{1}{2}$

Since we need a nontrivial solution where $a = b$, we substitute into the third equation:

$1 - a^2 - b^2 = a(1 - a - b)$

Since $a = b$, substituting $b = a$:

$1 - 2a^2 = a(1 - 2a)$

Expanding:

$1 - 2a^2 = a - 2a^2$

Cancelling $-2a^2$ from both sides:

$1 = a$

Thus, $a = 1$. But from the constraint $a + b \leq 1$, this means $a = \frac{1}{2}$, so $b = \frac{1}{2}$.

Final Answer:

$\boxed{\frac{1}{2}}$

Would you like a step-by-step breakdown or further clarification?

Here are five related questions:

1. How do we determine if a set of three vectors is linearly dependent?
2. What is the geometric interpretation of linearly dependent vectors in 3D space?
3. How does the concept of linear dependence extend to higher dimensions?
4. What are the conditions for two parametric vectors to be linearly dependent?
5. How do eigenvalues relate to linear dependence in matrices?

Tip: In problems involving constraints, always check if your solution satisfies all given inequalities.