. 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.

Let's solve this step by step.

### (a) Write two equations in \( c \) and \( d \) corresponding to the vector equation \( cu + dv = b \).

Given:
- \( \mathbf{u} = (1, 2) \)
- \( \mathbf{v} = (3, 1) \)
- \( \mathbf{b} = (3, 4) \)

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

\[
c \mathbf{u} + d \mathbf{v} = \mathbf{b}
\]

Substituting the given vectors:

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

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

\[
\begin{aligned}
1. & \quad c \cdot 1 + d \cdot 3 = 3 \quad \text{(from the first component)} \\
2. & \quad c \cdot 2 + d \cdot 1 = 4 \quad \text{(from the second component)}
\end{aligned}
\]

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.

Learn how to express a vector as a linear combination of given vectors u = (1, 2) and v = (3, 1). Step-by-step solution includes solving equations for coefficients c and d. Suitable for undergraduate students studying linear algebra.

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