Math Problem Statement

. 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

Let's solve this step by step.

(a) Write two equations in cc and dd corresponding to the vector equation cu+dv=bcu + dv = b.

Given:

  • u=(1,2)\mathbf{u} = (1, 2)
  • v=(3,1)\mathbf{v} = (3, 1)
  • b=(3,4)\mathbf{b} = (3, 4)

We want to express b\mathbf{b} as a linear combination of u\mathbf{u} and v\mathbf{v}, i.e.,

cu+dv=bc \mathbf{u} + d \mathbf{v} = \mathbf{b}

Substituting the given vectors:

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

This gives us two separate equations, one for each component of the vectors:

1. & \quad c \cdot 1 + d \cdot 3 = 3 \quad \text{(from the first component)} \\ 2. & \quad c \cdot 2 + d \cdot 1 = 4 \quad \text{(from the second component)} \end{aligned}$$ So, the equations in $$c$$ and $$d$$ are: $$\begin{aligned} c + 3d &= 3 \quad \text{(Equation 1)} \\ 2c + d &= 4 \quad \text{(Equation 2)} \end{aligned}$$ ### (b) Solve the equations in part (a) for $$c$$ and $$d$$. We can solve this system of equations using substitution or elimination. Let's use substitution. From Equation 1: $$c = 3 - 3d$$ Substitute $$c = 3 - 3d$$ into Equation 2: $$2(3 - 3d) + d = 4$$ Simplify and solve for $$d$$: $$6 - 6d + d = 4$$ $$6 - 5d = 4$$ $$-5d = 4 - 6 = -2$$ $$d = \frac{2}{5}$$ Now, substitute $$d = \frac{2}{5}$$ back into $$c = 3 - 3d$$: $$c = 3 - 3\left(\frac{2}{5}\right) = 3 - \frac{6}{5} = \frac{15}{5} - \frac{6}{5} = \frac{9}{5}$$ Thus, $$c = \frac{9}{5}$$ and $$d = \frac{2}{5}$$. ### (c) Express $$\mathbf{b}$$ as a linear combination of $$\mathbf{u}$$ and $$\mathbf{v}$$. Finally, we express $$\mathbf{b}$$ as: $$\mathbf{b} = c \mathbf{u} + d \mathbf{v} = \frac{9}{5}(1, 2) + \frac{2}{5}(3, 1)$$ Let's confirm this calculation: $$\mathbf{b} = \left(\frac{9}{5} \times 1 + \frac{2}{5} \times 3, \frac{9}{5} \times 2 + \frac{2}{5} \times 1\right)$$ $$\mathbf{b} = \left(\frac{9}{5} + \frac{6}{5}, \frac{18}{5} + \frac{2}{5}\right)$$ $$\mathbf{b} = \left(\frac{15}{5}, \frac{20}{5}\right) = (3, 4)$$ Thus, $$\mathbf{b} = \frac{9}{5} \mathbf{u} + \frac{2}{5} \mathbf{v}$$. --- Do you want details or have any questions about the solution? Here are 5 related questions: 1. How would the solution change if $$\mathbf{b}$$ was a different vector? 2. Can you find another pair of scalars $$c$$ and $$d$$ that satisfy the equation? 3. How do we know that $$\mathbf{u}$$ and $$\mathbf{v}$$ span $$\mathbb{R}^2$$? 4. What is the geometric interpretation of the coefficients $$c$$ and $$d$$? 5. What happens if $$\mathbf{u}$$ and $$\mathbf{v}$$ are parallel? **Tip:** In linear algebra, expressing a vector as a linear combination of others helps determine if they span a space.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Linear Combinations

Formulas

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Theorems

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Suitable Grade Level

Undergraduate Level