| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Matrix multiplication |
| Difficulty | Moderate -0.8 This is a straightforward matrix multiplication exercise followed by a simple observation. Part (i) requires routine calculation of a 2×3 matrix times a 3×2 matrix. Part (ii) merely asks students to note that BA cannot be computed (dimension mismatch), which is immediate and requires no problem-solving. This is easier than average A-level content, though the Further Maths context slightly elevates it above trivial. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{AB} = \begin{pmatrix} 2 & -1 & 1 \\ 0 & p & -4 \end{pmatrix}\begin{pmatrix} 0 & q \\ 2 & -2 \\ 1 & -3 \end{pmatrix} = \begin{pmatrix} -1 & 2q-1 \\ 2p-4 & -2p+12 \end{pmatrix}\) | M1 | Attempt to multiply in correct order |
| Correct and simplified | A2 | -1 each error |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{BA} = \begin{pmatrix} 0 & q \\ 2 & -2 \\ 1 & -3 \end{pmatrix}\begin{pmatrix} 2 & -1 & 1 \\ 0 & p & -4 \end{pmatrix} = \begin{pmatrix} * & * & * \\ * & * & * \\ * & * & * \end{pmatrix}\) | M1 | Valid method to compare products |
| \(\mathbf{BA} \neq \mathbf{AB}\) hence not commutative | A1 | Reason for conclusion stated |
| [2] |
## Question 1:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{AB} = \begin{pmatrix} 2 & -1 & 1 \\ 0 & p & -4 \end{pmatrix}\begin{pmatrix} 0 & q \\ 2 & -2 \\ 1 & -3 \end{pmatrix} = \begin{pmatrix} -1 & 2q-1 \\ 2p-4 & -2p+12 \end{pmatrix}$ | M1 | Attempt to multiply in correct order |
| Correct and simplified | A2 | -1 each error |
| **[3]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{BA} = \begin{pmatrix} 0 & q \\ 2 & -2 \\ 1 & -3 \end{pmatrix}\begin{pmatrix} 2 & -1 & 1 \\ 0 & p & -4 \end{pmatrix} = \begin{pmatrix} * & * & * \\ * & * & * \\ * & * & * \end{pmatrix}$ | M1 | Valid method to compare products |
| $\mathbf{BA} \neq \mathbf{AB}$ hence not commutative | A1 | Reason for conclusion stated |
| **[2]** | | |
---$\mathbf { 1 }$ You are given that $\mathbf { A } = \left( \begin{array} { r r r } 2 & - 1 & 1 \\ 0 & p & - 4 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { r r } 0 & q \\ 2 & - 2 \\ 1 & - 3 \end{array} \right)$.\\
(i) Find $\mathbf { A B }$.\\
(ii) Hence prove that matrix multiplication is not commutative.
\hfill \mbox{\textit{OCR MEI FP1 2012 Q1 [5]}}