| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Matrix equation solving (AB = C) |
| Difficulty | Standard +0.3 This is a straightforward Further Maths matrix question requiring routine matrix multiplication to verify given results, then using k=-3 to compute AB and recognize it as the identity (thus finding A^(-1)), and finally solving a matrix equation. While it involves 3×3 matrices and multiple parts, each step follows standard procedures without requiring novel insight or complex problem-solving—slightly easier than average for FP1. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = 7\times(-4)+(-1)\times(-19)+(-1)\times(-9) = 0\) | E1 | AG must see correct working |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow q = 15+2k-k^2\) | M1, A1 | AG Correct working |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{AB} = \begin{pmatrix}79&0&0\\0&79&0\\0&0&79\end{pmatrix}\) | B2 | \(-1\) each error |
| \(\mathbf{A}^{-1} = \frac{1}{79}\begin{pmatrix}-4&-5&11\\-19&-4&-7\\-9&-31&5\end{pmatrix}\) | M1, B1, A1 | Use of B; \(\frac{1}{79}\); Correct inverse |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix}x\\y\\z\end{pmatrix} = \frac{1}{79}\begin{pmatrix}-4&-5&11\\-19&-4&-7\\-9&-31&5\end{pmatrix}\begin{pmatrix}14\\-23\\9\end{pmatrix} = \begin{pmatrix}2\\-3\\8\end{pmatrix}\) | M1 | Attempt to pre-multiply by their \(\mathbf{A}^{-1}\) |
| \(\Rightarrow x=2,\ y=-3,\ z=8\) | A1, A1, A1 | SC A2 for \(x, y, z\) unspecified; sSC B1 for \(A^{-1}\) not used or incorrectly placed |
# Question 9:
## Part (i):
$p = 7\times(-4)+(-1)\times(-19)+(-1)\times(-9) = 0$ | E1 | AG must see correct working
$q = 2\times11+1\times(-7)+k\times(2-k)$
$\Rightarrow q = 15+2k-k^2$ | M1, A1 | AG Correct working
**[3]**
## Part (ii):
$\mathbf{AB} = \begin{pmatrix}79&0&0\\0&79&0\\0&0&79\end{pmatrix}$ | B2 | $-1$ each error
$\mathbf{A}^{-1} = \frac{1}{79}\begin{pmatrix}-4&-5&11\\-19&-4&-7\\-9&-31&5\end{pmatrix}$ | M1, B1, A1 | Use of **B**; $\frac{1}{79}$; Correct inverse
**[5]**
## Part (iii):
$\begin{pmatrix}x\\y\\z\end{pmatrix} = \frac{1}{79}\begin{pmatrix}-4&-5&11\\-19&-4&-7\\-9&-31&5\end{pmatrix}\begin{pmatrix}14\\-23\\9\end{pmatrix} = \begin{pmatrix}2\\-3\\8\end{pmatrix}$ | M1 | Attempt to pre-multiply by their $\mathbf{A}^{-1}$
$\Rightarrow x=2,\ y=-3,\ z=8$ | A1, A1, A1 | SC A2 for $x, y, z$ unspecified; sSC B1 for $A^{-1}$ not used or incorrectly placed
**[4]**9 You are given that $\mathbf { A } = \left( \begin{array} { r r r } - 3 & - 4 & 1 \\ 2 & 1 & k \\ 7 & - 1 & - 1 \end{array} \right) , \mathbf { B } = \left( \begin{array} { r r c } - 4 & - 5 & 11 \\ - 19 & - 4 & - 7 \\ - 9 & - 31 & 2 - k \end{array} \right)$ and $\mathbf { A B } = \left( \begin{array} { c c c } 79 & 0 & - 3 - k \\ - 9 k - 27 & - 31 k - 14 & q \\ p & 0 & 82 + k \end{array} \right)$ where $p$ and $q$ are to be determined.\\
(i) Show that $p = 0$ and $q = 15 + 2 k - k ^ { 2 }$.
It is now given that $k = - 3$.\\
(ii) Find $\mathbf { A B }$ and hence write down the inverse matrix $\mathbf { A } ^ { - 1 }$.\\
(iii) Use a matrix method to find the values of $x , y$ and $z$ that satisfy the equation $\mathbf { A } \left( \begin{array} { l } x \\ y \\ z \end{array} \right) = \left( \begin{array} { r } 14 \\ - 23 \\ 9 \end{array} \right)$.
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\hfill \mbox{\textit{OCR MEI FP1 2012 Q9 [12]}}