| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Properties of matrix operations |
| Difficulty | Moderate -0.5 This is a straightforward Further Maths question testing basic matrix determinant properties. Part (i) requires simple determinant calculation with a parameter, (ii) is direct verification of the multiplicative property (a standard result), and (iii) involves solving a quadratic equation. While it's Further Maths content, the techniques are routine and require no problem-solving insight—just methodical application of determinant formulas. |
| Spec | 4.03h Determinant 2x2: calculation4.03m det(AB) = det(A)*det(B) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \( | \mathbf{A} | = -2t - 6t\) or \( |
| \( | \mathbf{B} | = -8t = |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix} t & 6 \\ t & -2 \end{pmatrix}\begin{pmatrix} 2t & 4 \\ t & -2 \end{pmatrix} = \begin{pmatrix} 2t^2+6t & 4t-12 \\ 2t^2-2t & 4t+4 \end{pmatrix}\) | M1 (AO 3.1a) | Must be attempt at proper matrix multiplication (columns into rows). Condone one error |
| \( | \mathbf{AB} | = \begin{vmatrix} 2t^2+6t & 4t-12 \\ 2t^2-2t & 4t+4 \end{vmatrix}\) \((2t^2+6t)(4t+4)-(2t^2-2t)(4t-12)\) |
| \(= 8t^3+8t^2+24t^2+24t-(8t^3-24t^2-8t^2+24t)\) \(= 64t^2 = (-8t)(-8t) = | \mathbf{A} |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Set their \( | \mathbf{AB} | =-1\) or \( |
| \(\Rightarrow 64t^2 = -1\) so \(t\) must be complex/imaginary/not real | A1ft (AO 3.2a) | Accept \(t = -i/8\) or \(t = i/8\). Allow follow through if their \( |
# Question 6:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $|\mathbf{A}| = -2t - 6t$ or $|\mathbf{B}| = -4t - 4t$ | M1 (AO 1.1a) | Correct expression for either seen or implied |
| $|\mathbf{B}| = -8t = |\mathbf{A}|$ | A1 (AO 2.2a) | Both correct and statement of equality. Need indication candidate understands they have shown these are equal. Could be done by re-writing $|\mathbf{A}|= -8t$ immediately next to $|\mathbf{B}|= -8t$. $|\mathbf{A}|=|\mathbf{B}|$ is fine after having shown both are equal to $-8t$, but $8t = -8t$ is not ok for the A mark. |
**[2 marks]**
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix} t & 6 \\ t & -2 \end{pmatrix}\begin{pmatrix} 2t & 4 \\ t & -2 \end{pmatrix} = \begin{pmatrix} 2t^2+6t & 4t-12 \\ 2t^2-2t & 4t+4 \end{pmatrix}$ | M1 (AO 3.1a) | Must be attempt at proper matrix multiplication (columns into rows). Condone one error |
| $|\mathbf{AB}| = \begin{vmatrix} 2t^2+6t & 4t-12 \\ 2t^2-2t & 4t+4 \end{vmatrix}$ $(2t^2+6t)(4t+4)-(2t^2-2t)(4t-12)$ | M1 (AO 2.1) | Correct expression for determinant of their matrix. Condone one error |
| $= 8t^3+8t^2+24t^2+24t-(8t^3-24t^2-8t^2+24t)$ $= 64t^2 = (-8t)(-8t) = |\mathbf{A}||\mathbf{B}|$ | A1 (AO 2.1) | Convincing expansion, correct answer and conclusion. Need candidate to conclude that $|\mathbf{AB}| = |\mathbf{A}||\mathbf{B}|$. Condone not seeing $(-8t)(-8t)$ explicitly |
**[3 marks]**
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Set their $|\mathbf{AB}|=-1$ or $|\mathbf{A}||\mathbf{A}|=|\mathbf{A}|^2=-1$ | M1 (AO 3.1a) | Seen or implied. $64t^2 = -1$ or $(-8t)^2 = -1$ |
| $\Rightarrow 64t^2 = -1$ so $t$ must be complex/imaginary/not real | A1ft (AO 3.2a) | Accept $t = -i/8$ or $t = i/8$. Allow follow through if their $|\mathbf{AB}|$ is of the form $kt^2$. |
**[2 marks]**
---6 The matrices $\mathbf { A }$ and $\mathbf { B }$ are given by $\mathbf { A } = \left( \begin{array} { r r } t & 6 \\ t & - 2 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { r r } 2 t & 4 \\ t & - 2 \end{array} \right)$ where $t$ is a constant.\\
(i) Show that $| \mathbf { A } | = | \mathbf { B } |$.\\
(ii) Verify that $| \mathbf { A B } | = | \mathbf { A } \| \mathbf { B } |$.\\
(iii) Given that $| \mathbf { A B } | = - 1$ explain what this means about the constant $t$.
\hfill \mbox{\textit{OCR Further Pure Core AS 2018 Q6 [7]}}