Math Problem Statement

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

Solution

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

w=λu+μv\mathbf{w} = \lambda \mathbf{u} + \mu \mathbf{v}

This equation expands into the following system:

(a,1,b)=λ(3,3,1)+μ(1,1,1)(-a, -1, b) = \lambda (3, -3, 1) + \mu (-1, 1, 1)

Breaking it down component-wise:

  1. For the xx-component: a=3λμ-a = 3\lambda - \mu

  2. For the yy-component: 1=3λ+μ-1 = -3\lambda + \mu

  3. For the zz-component: b=λ+μb = \lambda + \mu

We now have a system of three equations:

-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.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Systems of Linear Equations

Formulas

w = λu + μv
Component-wise system of equations: -a = 3λ - μ, -1 = -3λ + μ, b = λ + μ

Theorems

Span of Vectors
Gaussian Elimination

Suitable Grade Level

Grades 10-12