5. 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.

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

### Part (a) - Writing the Equations in \( c \) and \( d \)

The vector equation \( cu + dv = b \) can be expressed as:

\[
c(1, 2) + d(3, 1) = (3, 4)
\]

This gives us two equations by equating the components:

1. For the \( x \)-component:
   \[
   c(1) + d(3) = 3
   \]
   Simplifying, we get:
   \[
   c + 3d = 3
   \]

2. For the \( y \)-component:
   \[
   c(2) + d(1) = 4
   \]
   Simplifying, we get:
   \[
   2c + d = 4
   \]

So the two equations are:

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

### Part (b) - Solving the Equations for \( c \) and \( d \)

We have the system of equations:

\[
\begin{aligned}
1. & \quad c + 3d = 3 \\
2. & \quad 2c + d = 4
\end{aligned}
\]

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.

Learn how to express vector b as a linear combination of vectors u and v using c and d. Solve the system of equations derived from cu + dv = b, and understand the geometric interpretation of vector combinations. Suitable for advanced high school students and beyond.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation