| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| 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 structured multi-part question on matrix multiplication and inverse matrices. Parts (i)-(ii) involve routine matrix multiplication to find parameters, part (iii) requires recognizing that B is a scalar multiple of A^{-1} (standard FP1 technique), and part (iv) is straightforward application. While it requires multiple steps, each part follows standard procedures without requiring novel insight or complex problem-solving. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\beta = (-1)(3\alpha - 1) + 5\alpha + (-1)(2\alpha + 1)\) | M1 | Multiply second row of A with first column of B |
| \(= -3\alpha + 1 + 5\alpha - 2\alpha - 1 = 0\) | A1 | Correct |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\gamma = (1)(3\alpha - 1) + 15 + (-1)(2\alpha + 1)\) | M1 | Attempt to multiply relevant row of A with relevant column of B. Condone use of BA instead |
| \(= \alpha + 13\) | A1 | Correct |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(\alpha = 2\), \(\gamma = 15\) | M1 | Multiplication of B by \(\dfrac{1}{\gamma}\), \((\gamma \neq 1)\) using \(\alpha = 2\) in both |
| \(\mathbf{A}^{-1} = \dfrac{1}{15}\begin{pmatrix} 5 & -8 & -1 \\ 5 & 1 & 2 \\ 5 & -5 & 5 \end{pmatrix}\) | A1 | Correct elements in matrix and correct \(\gamma\) |
| \(\mathbf{A}^{-1}\) does not exist when \(\alpha = -13\) | B1ft | ft their \(\gamma = 0\). Condone "\(\alpha \neq -13\)" |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{1}{15}\begin{pmatrix}5 & -8 & -1\\5 & 1 & 2\\5 & -5 & 5\end{pmatrix}\begin{pmatrix}25\\11\\-23\end{pmatrix} = \begin{pmatrix}x\\y\\z\end{pmatrix}\) | M1 | Set-up of pre-multiplication by their \(3\times3\) \(\mathbf{A}^{-1}\), or by B (using \(\alpha = 2\)) |
| \(= \dfrac{1}{15}\begin{pmatrix}60\\90\\-45\end{pmatrix} = \begin{pmatrix}4\\6\\-3\end{pmatrix}\) | B1 | \((60\ \ 90\ \ {-45})^T\) need not be fully evaluated |
| \(\Rightarrow x = 4,\ y = 6,\ z = -3\) | A3 | cao. A1 for each explicit identification of \(x, y, z\) in a vector or list. (−1 unidentified). Answers only or solution by other method: M0A0 |
| [5] |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\beta = (-1)(3\alpha - 1) + 5\alpha + (-1)(2\alpha + 1)$ | M1 | Multiply second row of **A** with first column of **B** |
| $= -3\alpha + 1 + 5\alpha - 2\alpha - 1 = 0$ | A1 | Correct |
| | **[2]** | |
---
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\gamma = (1)(3\alpha - 1) + 15 + (-1)(2\alpha + 1)$ | M1 | Attempt to multiply relevant row of **A** with relevant column of **B**. Condone use of **BA** instead |
| $= \alpha + 13$ | A1 | Correct |
| | **[2]** | |
---
## Question 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $\alpha = 2$, $\gamma = 15$ | M1 | Multiplication of **B** by $\dfrac{1}{\gamma}$, $(\gamma \neq 1)$ using $\alpha = 2$ in both |
| $\mathbf{A}^{-1} = \dfrac{1}{15}\begin{pmatrix} 5 & -8 & -1 \\ 5 & 1 & 2 \\ 5 & -5 & 5 \end{pmatrix}$ | A1 | Correct elements in matrix and correct $\gamma$ |
| $\mathbf{A}^{-1}$ does not exist when $\alpha = -13$ | B1ft | ft their $\gamma = 0$. Condone "$\alpha \neq -13$" |
| | **[3]** | |
---
## Question 9(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{1}{15}\begin{pmatrix}5 & -8 & -1\\5 & 1 & 2\\5 & -5 & 5\end{pmatrix}\begin{pmatrix}25\\11\\-23\end{pmatrix} = \begin{pmatrix}x\\y\\z\end{pmatrix}$ | M1 | Set-up of pre-multiplication by their $3\times3$ $\mathbf{A}^{-1}$, or by **B** (using $\alpha = 2$) |
| $= \dfrac{1}{15}\begin{pmatrix}60\\90\\-45\end{pmatrix} = \begin{pmatrix}4\\6\\-3\end{pmatrix}$ | B1 | $(60\ \ 90\ \ {-45})^T$ need not be fully evaluated |
| $\Rightarrow x = 4,\ y = 6,\ z = -3$ | A3 | cao. A1 for each explicit identification of $x, y, z$ in a vector or list. (−1 unidentified). Answers only or solution by other method: M0A0 |
| | **[5]** | |9 You are given that $\mathbf { A } = \left( \begin{array} { r r r } 1 & 3 & - 1 \\ - 1 & \alpha & - 1 \\ - 2 & - 1 & 3 \end{array} \right) , \mathbf { B } = \left( \begin{array} { c c c } 3 \alpha - 1 & - 8 & \alpha - 3 \\ 5 & 1 & 2 \\ 2 \alpha + 1 & - 5 & \alpha + 3 \end{array} \right)$ and $\mathbf { A B } = \left( \begin{array} { c c c } \gamma & 0 & 0 \\ \beta & \gamma & 0 \\ 0 & 0 & \gamma \end{array} \right)$.\\
(i) Show that $\beta = 0$.\\
(ii) Find $\gamma$ in terms of $\alpha$.\\
(iii) Write down $\mathbf { A } ^ { - 1 }$ for the case when $\alpha = 2$. State the value of $\alpha$ for which $\mathbf { A } ^ { - 1 }$ does not exist.\\
(iv) Use your answer to part (iii) to solve the following simultaneous equations.
$$\begin{aligned}
x + 3 y - z & = 25 \\
- x + 2 y - z & = 11 \\
- 2 x - y + 3 z & = - 23
\end{aligned}$$
\hfill \mbox{\textit{OCR MEI FP1 2014 Q9 [12]}}