Additional Mathematics 4037/12 — October/November 2012

Walkthrough

To add vectors written as columns, we simply add the top components together and the bottom components together. This gives $4 + (-1) + 21 = 24$ on top and $3 + 2 + 2 = 7$ below, so the sum is $\begin{pmatrix} 24 \\ 7 \end{pmatrix}$.

The magnitude (or modulus) of a vector measures its length. For a vector $\begin{pmatrix} x \\ y \end{pmatrix}$ it is $\sqrt{x^2 + y^2}$, which is just Pythagoras' theorem applied to the horizontal and vertical components. Here that gives $\sqrt{24^2 + 7^2} = \sqrt{625} = 25$ — a perfect square, so no calculator is needed.

Key Takeaways

• Vector addition is done component by component; order does not matter.
• The magnitude of $\begin{pmatrix} x \\ y \end{pmatrix}$ is $\sqrt{x^2 + y^2}$.
• Recognising Pythagorean triples (here a scaled 3–4–5: $24$, $7$, $25$) speeds up non-calculator work.

Common Mistakes

• Subtracting instead of adding one of the components, especially with the negative entry in $\mathbf{b}$.
• Forgetting to square both components before adding under the root.
• Giving $\sqrt{625}$ without evaluating it — the exact value $25$ is required.

Things to Be Careful About

• The answer must be fully evaluated ($25$), not left as $\sqrt{625}$.
• Show the summed vector before taking the modulus — this is where the method mark sits.

Walkthrough

The equation $\lambda \mathbf{a} + \mu \mathbf{b} = \mathbf{c}$ says that the vector $\mathbf{c}$ can be built from multiples of $\mathbf{a}$ and $\mathbf{b}$. Multiplying out gives $\begin{pmatrix} 4\lambda - \mu \\ 3\lambda + 2\mu \end{pmatrix}$.

Two column vectors are equal only when their corresponding components are equal, so we get one equation from the top row ($4\lambda - \mu = 21$) and one from the bottom row ($3\lambda + 2\mu = 2$). These are ordinary simultaneous equations.

Eliminating $\mu$ is easiest here: doubling the first equation makes its $\mu$ coefficient $-2$, which cancels against the $+2\mu$ in the second equation when added. That leaves $11\lambda = 44$, so $\lambda = 4$. Substituting back into $4\lambda - \mu = 21$ gives $16 - \mu = 21$, hence $\mu = -5$.

Key Takeaways

• Equal column vectors give one equation per component.
• Scalar multiples of vectors distribute over each component.
• The elimination method removes one variable cleanly when coefficients match after scaling.

Common Mistakes

• Sign slips when multiplying $\mathbf{b}$ by $\mu$: the top component is $-\mu$, not $+\mu$.
• Equating unlike components (top of one side with bottom of the other).
• Arithmetic errors in the elimination step; always check by substituting both values back into the second equation: $3(4) + 2(-5) = 12 - 10 = 2$ ✓.

Things to Be Careful About

• Both values must be found for full marks; the dependent M1 for solving requires correct simultaneous-equation work following the equating step.
• Verify answers in both original equations before moving on.