Solve for Vectors a and b: r = a - 2b, s = a + b

To solve for vectors $\mathbf{a}$ and $\mathbf{b}$ in terms of $\mathbf{r}$ and $\mathbf{s}$, we treat the given equations as a system of linear equations.

Answer

The vectors are expressed as $\mathbf{a} = \frac{1}{3}(\mathbf{r} + 2\mathbf{s})$ and $\mathbf{b} = \frac{1}{3}(\mathbf{s} - \mathbf{r})$. These results are derived by using elimination to isolate each variable.

Explanation

1. Set up the system of equations We are given: (1) $\mathbf{r} = \mathbf{a} - 2\mathbf{b}$ (2) $\mathbf{s} = \mathbf{a} + \mathbf{b}$ We aim to solve for $\mathbf{a}$ and $\mathbf{b}$ using basic algebraic manipulation. System of equations using vectors $\mathbf{a}$ and $\mathbf{b}$.

2. Solve for $\mathbf{b}$ by elimination Subtract equation (1) from equation (2) to eliminate $\mathbf{a}$: $\mathbf{s} - \mathbf{r} = (\mathbf{a} + \mathbf{b}) - (\mathbf{a} - 2\mathbf{b})$ $\mathbf{s} - \mathbf{r} = 3\mathbf{b}$ Divide by 3: $\mathbf{b} = \frac{1}{3}(\mathbf{s} - \mathbf{r})$ Subtracting the equations isolates vector $\mathbf{b}$.

3. Solve for $\mathbf{a}$ by substitution Rearrange equation (2) to get $\mathbf{a} = \mathbf{s} - \mathbf{b}$. Substitute our expression for $\mathbf{b}$ into this: $\mathbf{a} = \mathbf{s} - \frac{1}{3}(\mathbf{s} - \mathbf{r})$ $\mathbf{a} = \frac{3\mathbf{s} - \mathbf{s} + \mathbf{r}}{3} = \frac{1}{3}(\mathbf{r} + 2\mathbf{s})$ Substituting $\mathbf{b}$ back into the simpler equation yields $\mathbf{a}$.

Final Answer

The vectors are: $\mathbf{a} = \frac{1}{3}\mathbf{r} + \frac{2}{3}\mathbf{s}, \quad \mathbf{b} = \frac{1}{3}\mathbf{s} - \frac{1}{3}\mathbf{r}$ Final expressions for vectors $\mathbf{a}$ and $\mathbf{b}$ in term of $\mathbf{r}$ and $\mathbf{s}$.

Common Mistakes

• Sign Errors: Students often fail to distribute the negative sign correctly when subtracting vector equations (e.g., forgetting that $-(-2\mathbf{b}) = +2\mathbf{b}$).
• Mixing up variables: Accidentally solving the system for the wrong variables or failing to fully simplify the final fraction.