being vector u (3, -3, 1), v (-1, 1, 1) and w (-a, -1, b). Determine all the real values of a and b so that: vector w belons to L{(u, v)} with gauss

We are tasked with determining the real values of \( a \) and \( b \) such that the vector \( \mathbf{w} = (-a, -1, b) \) belongs to the span of vectors \( \mathbf{u} = (3, -3, 1) \) and \( \mathbf{v} = (-1, 1, 1) \). In other words, we need to find if there exist scalars \( \lambda \) and \( \mu \) such that:

\[
\mathbf{w} = \lambda \mathbf{u} + \mu \mathbf{v}
\]

This equation expands into the following system:

\[
(-a, -1, b) = \lambda (3, -3, 1) + \mu (-1, 1, 1)
\]

Breaking it down component-wise:

1. For the \( x \)-component: 
   \[
   -a = 3\lambda - \mu
   \]

2. For the \( y \)-component:
   \[
   -1 = -3\lambda + \mu
   \]

3. For the \( z \)-component:
   \[
   b = \lambda + \mu
   \]

We now have a system of three equations:

\[
\begin{aligned}
-a &= 3\lambda - \mu \\
-1 &= -3\lambda + \mu \\
b &= \lambda + \mu
\end{aligned}
\]

### Step 1: Solve for \( \lambda \) and \( \mu \)
From the second equation, solve for \( \mu \):

\[
\mu = -1 + 3\lambda
\]

Substitute this expression for \( \mu \) into the first and third equations.

### Step 2: Substitute \( \mu \) into the first equation
Substituting into the first equation:

\[
-a = 3\lambda - (-1 + 3\lambda)
\]

Simplifying:

\[
-a = 3\lambda + 1 - 3\lambda \quad \Rightarrow \quad -a = 1
\]

Thus, \( a = -1 \).

### Step 3: Substitute \( \mu \) into the third equation
Substituting into the third equation:

\[
b = \lambda + (-1 + 3\lambda)
\]

Simplifying:

\[
b = \lambda - 1 + 3\lambda \quad \Rightarrow \quad b = 4\lambda - 1
\]

### Step 4: Solve for \( \lambda \) and \( \mu \)
We now have \( a = -1 \). To find \( b \), we need the value of \( \lambda \). Using the second equation:

\[
-1 = -3\lambda + (-1 + 3\lambda)
\]

Simplifying:

\[
-1 = -3\lambda - 1 + 3\lambda \quad \Rightarrow \quad -1 = -1
\]

This is always true, which means there are no further restrictions on \( \lambda \). Therefore, \( b = 4\lambda - 1 \) is a free parameter depending on \( \lambda \).

### Conclusion
The solution is:

- \( a = -1 \)
- \( b = 4\lambda - 1 \), where \( \lambda \) is any real number.

Thus, \( a = -1 \) and \( b \) can take any real value of the form \( b = 4\lambda - 1 \), where \( \lambda \) is free to vary.

Do you want further details or have any questions?

Here are 5 related questions:
1. How do you find the span of two vectors geometrically?
2. What are the conditions for linear dependence of vectors?
3. How can we generalize this method for higher dimensions?
4. How does Gaussian elimination help in solving systems of equations?
5. What is the geometric interpretation of the system of equations in this problem?

**Tip**: Always check for consistency in systems of equations to avoid contradictions or unnecessary assumptions.

Being vector u (3, -3, 1), v (-1, 1, 1), and w (-a, -1, b). Determine all the real values of a and b so that vector w belongs to L{(u, v)} with Gauss.

Find all real values of a and b such that vector w (-a, -1, b) belongs to the span of vectors u (3, -3, 1) and v (-1, 1, 1) using Gaussian elimination. Learn key concepts in vector spaces and systems of equations. Suitable for high school students in Grades 10-12.

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