| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Volume/area scale factors |
| Difficulty | Standard +0.3 This is a straightforward application of the determinant-volume relationship for 3×3 matrices. Part (a) requires computing a determinant with numerical values and multiplying by the original volume, plus checking the sign. Part (b) requires setting the determinant equal to zero and solving a cubic equation. While it involves 3×3 matrices (Further Maths content), the techniques are routine and the question is well-scaffolded with clear steps. |
| Spec | 4.03j Determinant 3x3: calculation4.03k Determinant 3x3: volume scale factor |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 5 \Rightarrow \det\mathbf{A} = 2\times5^3 - 4\times5^2 - 58\times5 + 60 = 250 - 100 - 290 + 60 = -80\) | B1 | AO 1.1 — Could expand as \(-1(-60+25) - 5(20-5) + 2(-50+30) = 35 - 75 - 40 = -80\) |
| Volume of \(H' = 6\times80 = 480\) cao | B1 | AO 1.1 — not \(-480\) |
| \(\mathbf{A}\) does not preserve orientation because \(\det\mathbf{A} < 0\) | E1 ft | AO 2.4 — Follow through on sign of their determinant. If positive: "\(\mathbf{A}\) does preserve orientation because \(\det\mathbf{A} > 0\)" |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Image coplanar \(\Rightarrow \det\mathbf{A} = 0\) | B1 | AO 2.2a |
| \(\det\mathbf{A} = -1(-6\times2x+5x) - x(2(7-x)-5) + 2(-5x(7-x)+6\times5)\) | M1 | AO 3.1a — Attempt to expand determinant |
| \(2x^3 - 4x^2 - 58x + 60\) | A1 | AO 1.1 — Don't need "\(=0\)" here |
| \(1,\ 6,\ -5\) | A1 | AO 1.1 — BC. All three. No working required for roots |
| [4] |
## Question 7:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 5 \Rightarrow \det\mathbf{A} = 2\times5^3 - 4\times5^2 - 58\times5 + 60 = 250 - 100 - 290 + 60 = -80$ | **B1** | AO 1.1 — Could expand as $-1(-60+25) - 5(20-5) + 2(-50+30) = 35 - 75 - 40 = -80$ |
| Volume of $H' = 6\times80 = 480$ cao | **B1** | AO 1.1 — not $-480$ |
| $\mathbf{A}$ does not preserve orientation because $\det\mathbf{A} < 0$ | **E1 ft** | AO 2.4 — Follow through on sign of their determinant. If positive: "$\mathbf{A}$ does preserve orientation because $\det\mathbf{A} > 0$" |
| **[3]** | | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Image coplanar $\Rightarrow \det\mathbf{A} = 0$ | **B1** | AO 2.2a |
| $\det\mathbf{A} = -1(-6\times2x+5x) - x(2(7-x)-5) + 2(-5x(7-x)+6\times5)$ | **M1** | AO 3.1a — Attempt to expand determinant |
| $2x^3 - 4x^2 - 58x + 60$ | **A1** | AO 1.1 — Don't need "$=0$" here |
| $1,\ 6,\ -5$ | **A1** | AO 1.1 — **BC**. All three. No working required for roots |
| **[4]** | | |
---7 A transformation A is represented by the matrix $\mathbf { A }$ where $\mathbf { A } = \left( \begin{array} { c c c } - 1 & x & 2 \\ 7 - x & - 6 & 1 \\ 5 & - 5 x & 2 x \end{array} \right)$.\\
The tetrahedron $H$ has vertices at $O , P , Q$ and $R$. The volume of $H$ is 6 units.\\
$P ^ { \prime } , Q ^ { \prime } , R ^ { \prime }$ and $H ^ { \prime }$ are the images of $P , Q , R$ and $H$ under A .
\begin{enumerate}[label=(\alph*)]
\item In the case where $x = 5$
\begin{itemize}
\item find the volume of $H ^ { \prime }$,
\item determine whether A preserves the orientation of $H$.
\item Find the values of $x$ for which $O , P ^ { \prime } , Q ^ { \prime }$ and $R ^ { \prime }$ are coplanar (i.e. the four points lie in the same plane).
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core AS 2019 Q7 [7]}}