Math Problem Statement

  1. Given three vectors u = (1, 2), v = (3, 1) and b = (3, 4) in R 2 , suppose that b can be written as a linear combination of u and v as cu + dv = b (a) Write two equations in c and d corresponding to the vector equation cu + dv = b. (b) Solve the equations in part (a) for c and d. (c) Express b as a linear combination of u and v.

Solution

Given the vectors u=(1,2)u = (1, 2), v=(3,1)v = (3, 1), and b=(3,4)b = (3, 4) in R2\mathbb{R}^2, we need to solve the following:

Part (a) - Writing the Equations in cc and dd

The vector equation cu+dv=bcu + dv = b can be expressed as:

c(1,2)+d(3,1)=(3,4)c(1, 2) + d(3, 1) = (3, 4)

This gives us two equations by equating the components:

  1. For the xx-component: c(1)+d(3)=3c(1) + d(3) = 3 Simplifying, we get: c+3d=3c + 3d = 3

  2. For the yy-component: c(2)+d(1)=4c(2) + d(1) = 4 Simplifying, we get: 2c+d=42c + d = 4

So the two equations are:

c+3d=3and2c+d=4\boxed{c + 3d = 3 \quad \text{and} \quad 2c + d = 4}

Part (b) - Solving the Equations for cc and dd

We have the system of equations:

1. & \quad c + 3d = 3 \\ 2. & \quad 2c + d = 4 \end{aligned}$$ Let's solve this system: 1. Start with the second equation: $$2c + d = 4 \quad \text{(Equation 2)}$$ Solve for $$d$$: $$d = 4 - 2c$$ 2. Substitute $$d = 4 - 2c$$ into the first equation: $$c + 3(4 - 2c) = 3$$ Distribute and simplify: $$c + 12 - 6c = 3$$ Combine like terms: $$-5c + 12 = 3$$ Subtract 12 from both sides: $$-5c = -9$$ Divide by -5: $$c = \frac{9}{5}$$ 3. Now substitute $$c = \frac{9}{5}$$ back into the expression for $$d$$: $$d = 4 - 2\left(\frac{9}{5}\right)$$ Simplify: $$d = 4 - \frac{18}{5} = \frac{20}{5} - \frac{18}{5} = \frac{2}{5}$$ So, $$c = \frac{9}{5}$$ and $$d = \frac{2}{5}$$. ### Part (c) - Express $$b$$ as a Linear Combination of $$u$$ and $$v$$ Using the values of $$c$$ and $$d$$ found in part (b), we express $$b$$ as: $$b = cu + dv$$ Substitute the values: $$b = \frac{9}{5}(1, 2) + \frac{2}{5}(3, 1)$$ Distribute the scalar multiplication: $$b = \left(\frac{9}{5}, \frac{18}{5}\right) + \left(\frac{6}{5}, \frac{2}{5}\right)$$ Add the corresponding components: $$b = \left(\frac{9 + 6}{5}, \frac{18 + 2}{5}\right) = \left(\frac{15}{5}, \frac{20}{5}\right) = (3, 4)$$ So, $$b = (3, 4)$$ is expressed as the linear combination: $$\boxed{b = \frac{9}{5}u + \frac{2}{5}v}$$ Would you like more details or have any questions? --- Here are five related questions to explore: 1. How would you express a different vector $$c = (5, 7)$$ as a linear combination of $$u$$ and $$v$$? 2. What is the geometric interpretation of expressing a vector as a linear combination of two other vectors? 3. How do you determine if three vectors in $$\mathbb{R}^2$$ are linearly independent? 4. Can you find the inverse of the matrix formed by placing $$u$$ and $$v$$ as columns? 5. How do you verify if a vector $$b$$ lies within the span of $$u$$ and $$v$$? **Tip:** When solving systems of linear equations, substituting one equation into another can simplify the process and reduce the number of variables, making it easier to find the solution.

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Math Problem Analysis

Mathematical Concepts

Vectors
Linear combinations
Systems of equations

Formulas

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Theorems

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Suitable Grade Level

Advanced High School