| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Matrix arithmetic operations |
| Difficulty | Easy -1.8 This is a very routine introductory question on basic matrix operations (scalar multiplication, addition, multiplication) with explicit matrices given. Part (i) requires only direct computation following definitions, and part (ii) is a standard textbook exercise showing AB ≠ BA. This is easier than typical A-level questions as it tests pure recall and mechanical computation with no problem-solving element. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\mathbf{B} = \begin{pmatrix} 4 & -6 \\ 2 & 8 \end{pmatrix}\) | B1 | cao |
| \(\mathbf{A} + \mathbf{C}\) not possible — different orders | B1 | must state reason |
| \(\mathbf{CA} = \begin{pmatrix} 1 & -1 \\ 0 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 4 & 3 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 2 & 4 \\ 1 & 2 \end{pmatrix}\) | M1 A1 | M1 for attempting multiplication |
| \(\mathbf{A} - \mathbf{B} = \begin{pmatrix} 2 & 6 \\ 0 & -2 \end{pmatrix}\) | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{AB} = \begin{pmatrix} 11 & 0 \\ 4 & 5 \end{pmatrix}\), \(\mathbf{BA} = \begin{pmatrix} 5 & 0 \\ 8 & 11 \end{pmatrix}\) | M1 | attempt both products |
| \(\mathbf{AB} \neq \mathbf{BA}\), so not commutative | A1 | both correct and conclusion stated |
# Question 1:
**(i)**
$2\mathbf{B} = \begin{pmatrix} 4 & -6 \\ 2 & 8 \end{pmatrix}$ | B1 | cao
$\mathbf{A} + \mathbf{C}$ not possible — different orders | B1 | must state reason
$\mathbf{CA} = \begin{pmatrix} 1 & -1 \\ 0 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 4 & 3 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 2 & 4 \\ 1 & 2 \end{pmatrix}$ | M1 A1 | M1 for attempting multiplication
$\mathbf{A} - \mathbf{B} = \begin{pmatrix} 2 & 6 \\ 0 & -2 \end{pmatrix}$ | B1 | cao
**(ii)**
$\mathbf{AB} = \begin{pmatrix} 11 & 0 \\ 4 & 5 \end{pmatrix}$, $\mathbf{BA} = \begin{pmatrix} 5 & 0 \\ 8 & 11 \end{pmatrix}$ | M1 | attempt both products
$\mathbf{AB} \neq \mathbf{BA}$, so not commutative | A1 | both correct and conclusion stated
---1 You are given that $\mathbf { A } = \left( \begin{array} { l l } 4 & 3 \\ 1 & 2 \end{array} \right) , \mathbf { B } = \left( \begin{array} { r r } 2 & - 3 \\ 1 & 4 \end{array} \right) , \mathbf { C } = \left( \begin{array} { r r } 1 & - 1 \\ 0 & 2 \\ 0 & 1 \end{array} \right)$.\\
(i) Calculate, where possible, $2 \mathbf { B } , \mathbf { A } + \mathbf { C } , \mathbf { C A }$ and $\mathbf { A } - \mathbf { B }$.\\
(ii) Show that matrix multiplication is not commutative.
\hfill \mbox{\textit{OCR MEI FP1 2006 Q1 [7]}}