Let's tackle the problem step by step.

Part (a) - Writing the equations

Given that the vector $\mathbf{b}$ can be written as a linear combination of $\mathbf{u}$ and $\mathbf{v}$, we have:

$\mathbf{b} = c\mathbf{u} + d\mathbf{v}$

Substituting the values of the vectors:

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

This equation can be broken down into two scalar equations by equating the corresponding components:

1. For the first component (x-coordinate):

$3 = c(1) + d(3)$ $3 = c + 3d \quad \text{(Equation 1)}$

2. For the second component (y-coordinate):

$4 = c(2) + d(1)$ $4 = 2c + d \quad \text{(Equation 2)}$

So, the two equations in $c$ and $d$ are:

3 &= c + 3d \quad \text{(Equation 1)} \\ 4 &= 2c + d \quad \text{(Equation 2)} \end{aligned}$$ ### Part (b) - Solving the equations Now, let's solve the system of equations: $$\begin{aligned} 3 &= c + 3d \quad \text{(Equation 1)} \\ 4 &= 2c + d \quad \text{(Equation 2)} \end{aligned}$$ 1. **Solve Equation 1 for $$c$$:** $$c = 3 - 3d$$ 2. **Substitute this expression for $$c$$ into Equation 2:** $$4 = 2(3 - 3d) + d$$ Simplify and solve for $$d$$: $$4 = 6 - 6d + d$$ $$4 = 6 - 5d$$ $$5d = 6 - 4$$ $$5d = 2$$ $$d = \frac{2}{5}$$ 3. **Substitute $$d = \frac{2}{5}$$ back into the expression for $$c$$:** $$c = 3 - 3\left(\frac{2}{5}\right)$$ $$c = 3 - \frac{6}{5}$$ $$c = \frac{15}{5} - \frac{6}{5}$$ $$c = \frac{9}{5}$$ So the solution is: $$c = \frac{9}{5}, \quad d = \frac{2}{5}$$ ### Part (c) - Expressing $$\mathbf{b}$$ as a linear combination of $$\mathbf{u}$$ and $$\mathbf{v}$$ Finally, substitute the values of $$c$$ and $$d$$ back into the linear combination equation: $$\mathbf{b} = c\mathbf{u} + d\mathbf{v}$$ Substituting $$c = \frac{9}{5}$$ and $$d = \frac{2}{5}$$: $$\mathbf{b} = \frac{9}{5}(1, 2) + \frac{2}{5}(3, 1)$$ Calculate the vectors: $$\mathbf{b} = \left(\frac{9}{5}, \frac{18}{5}\right) + \left(\frac{6}{5}, \frac{2}{5}\right)$$ Add the vectors component-wise: $$\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)$$ $$\mathbf{b} = (3, 4)$$ So, indeed: $$\mathbf{b} = \frac{9}{5} \mathbf{u} + \frac{2}{5} \mathbf{v}$$ This confirms that $$\mathbf{b}$$ can be expressed as a linear combination of $$\mathbf{u}$$ and $$\mathbf{v}$$. --- Would you like more details, or do you have any questions? Here are some related questions you might find helpful: 1. How would the solution change if $$\mathbf{b} = (2, 3)$$? 2. Can every vector in $$\mathbb{R}^2$$ be expressed as a linear combination of $$\mathbf{u}$$ and $$\mathbf{v}$$? 3. What is the geometric interpretation of $$c$$ and $$d$$ in the context of linear combinations? 4. How would you determine if $$\mathbf{u}$$ and $$\mathbf{v}$$ are linearly independent? 5. What happens if $$\mathbf{u}$$ and $$\mathbf{v}$$ are parallel vectors? **Tip:** When solving linear systems, always check if the equations are independent to avoid unnecessary calculations.