| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Matrix inverse calculation |
| Difficulty | Moderate -0.5 This is a straightforward Further Maths question requiring matrix multiplication and recognizing that if AB = I, then B = A^(-1). The calculation is routine but involves more arithmetic than typical A-level questions. Being FP1 content makes it slightly harder than average A-level, but the method is direct with no problem-solving insight needed. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{AB} = \begin{pmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end{pmatrix}\) | B3 [3] | Minus 1 each error to minimum of 0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{A}^{-1} = \frac{1}{7}\begin{pmatrix} -1 & 0 & 2 \\ 14 & -14 & 7 \\ -5 & 7 & -4 \end{pmatrix}\) | M1, A1 [2] | Use of B; c.a.o. |
# Question 5:
## Question 5(i):
| $\mathbf{AB} = \begin{pmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end{pmatrix}$ | B3 [3] | Minus 1 each error to minimum of 0 |
## Question 5(ii):
| $\mathbf{A}^{-1} = \frac{1}{7}\begin{pmatrix} -1 & 0 & 2 \\ 14 & -14 & 7 \\ -5 & 7 & -4 \end{pmatrix}$ | M1, A1 [2] | Use of B; c.a.o. |
---5 You are given that $\mathbf { A } = \left( \begin{array} { l l l } 1 & 2 & 4 \\ 3 & 2 & 5 \\ 4 & 1 & 2 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { r r r } - 1 & 0 & 2 \\ 14 & - 14 & 7 \\ - 5 & 7 & - 4 \end{array} \right)$.\\
(i) Calculate AB.\\
(ii) Write down $\mathbf { A } ^ { - 1 }$.
\hfill \mbox{\textit{OCR MEI FP1 2008 Q5 [5]}}