| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear transformations |
| Type | Commutativity of transformations |
| Difficulty | Moderate -0.3 This is a straightforward Further Maths question testing basic matrix operations and transformations. Part (a) requires simple matrix multiplication to show non-commutativity, (b) recognizes a rotation matrix, (c) uses geometric reasoning to find the inverse, and (d) applies the determinant-area relationship. All parts are standard textbook exercises requiring recall and routine application rather than problem-solving or insight. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{AB} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \frac{5}{13} & \frac{12}{13} \\ \frac{12}{13} & \frac{5}{13} \end{pmatrix} = \begin{pmatrix} -\frac{5}{13} & -\frac{12}{13} \\ \frac{12}{13} & \frac{5}{13} \end{pmatrix}\) | M1 | BC. AB or BA correct. Could see \(\frac{1}{13}\begin{pmatrix}-1&0\\0&1\end{pmatrix}\begin{pmatrix}5&-12\\12&5\end{pmatrix} = \frac{1}{13}\begin{pmatrix}-5&12\\12&5\end{pmatrix}\) |
| \(\mathbf{BA} = \begin{pmatrix} \frac{5}{13} & -\frac{12}{13} \\ \frac{12}{13} & \frac{5}{13} \end{pmatrix}\begin{pmatrix}-1&0\\0&1\end{pmatrix} = \begin{pmatrix}-\frac{5}{13} & -\frac{12}{13} \\ -\frac{12}{13} & \frac{5}{13}\end{pmatrix} \neq \mathbf{AB}\), so matrix multiplication is not commutative | A1 | 2.2a — BC. Other multiplication correct and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Rotation about \(O\), \(67.4°\) anticlockwise | M1, A1 | 1.2, 1.1 — or \(1.18\) rads |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((T_B)^{-1}\) is a rotation about \(O\) by \(-67.4°\) anticlockwise (or \(67.4°\) clockwise). So \(\mathbf{B}^{-1} = \begin{pmatrix}\cos(-67.4°) & -\sin(-67.4°) \\ \sin(-67.4°) & \cos(-67.4°)\end{pmatrix}\) | M1 | 3.1a — Correct inverse of their rotation \(T_B\). Or \(\mathbf{B}^{-1} = \begin{pmatrix}0.385 & 0.923 \\ -0.923 & 0.385\end{pmatrix}\) (allow \(0.384\) for \(0.385\)). Could also be rotation of \(292.6°\) anticlockwise. NB: Question states "by considering the inverse transformation". SC1 for correct inverse by other method. |
| \(= \begin{pmatrix}\frac{5}{13} & \frac{12}{13} \\ -\frac{12}{13} & \frac{5}{13}\end{pmatrix}\) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\det \mathbf{B} = 1\) and \(\det \mathbf{C} = -3\) | M1 | 3.1a — Could find BC and then find \(\det(\mathbf{BC}) = -3\) |
| So area of \(N = | 1 \times -3 | \times 5 = 15\) |
# Question 5:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{AB} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \frac{5}{13} & \frac{12}{13} \\ \frac{12}{13} & \frac{5}{13} \end{pmatrix} = \begin{pmatrix} -\frac{5}{13} & -\frac{12}{13} \\ \frac{12}{13} & \frac{5}{13} \end{pmatrix}$ | M1 | BC. **AB** or **BA** correct. Could see $\frac{1}{13}\begin{pmatrix}-1&0\\0&1\end{pmatrix}\begin{pmatrix}5&-12\\12&5\end{pmatrix} = \frac{1}{13}\begin{pmatrix}-5&12\\12&5\end{pmatrix}$ |
| $\mathbf{BA} = \begin{pmatrix} \frac{5}{13} & -\frac{12}{13} \\ \frac{12}{13} & \frac{5}{13} \end{pmatrix}\begin{pmatrix}-1&0\\0&1\end{pmatrix} = \begin{pmatrix}-\frac{5}{13} & -\frac{12}{13} \\ -\frac{12}{13} & \frac{5}{13}\end{pmatrix} \neq \mathbf{AB}$, so matrix multiplication is not commutative | A1 | 2.2a — BC. Other multiplication correct and conclusion |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Rotation about $O$, $67.4°$ anticlockwise | M1, A1 | 1.2, 1.1 — or $1.18$ rads |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(T_B)^{-1}$ is a rotation about $O$ by $-67.4°$ anticlockwise (or $67.4°$ clockwise). So $\mathbf{B}^{-1} = \begin{pmatrix}\cos(-67.4°) & -\sin(-67.4°) \\ \sin(-67.4°) & \cos(-67.4°)\end{pmatrix}$ | M1 | 3.1a — Correct inverse of their rotation $T_B$. Or $\mathbf{B}^{-1} = \begin{pmatrix}0.385 & 0.923 \\ -0.923 & 0.385\end{pmatrix}$ (allow $0.384$ for $0.385$). Could also be rotation of $292.6°$ anticlockwise. NB: Question states "by considering the inverse transformation". SC1 for correct inverse by other method. |
| $= \begin{pmatrix}\frac{5}{13} & \frac{12}{13} \\ -\frac{12}{13} & \frac{5}{13}\end{pmatrix}$ | A1 | 1.1 |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\det \mathbf{B} = 1$ and $\det \mathbf{C} = -3$ | M1 | 3.1a — Could find **BC** and then find $\det(\mathbf{BC}) = -3$ |
| So area of $N = |1 \times -3| \times 5 = 15$ | A1 | 3.2a — Area must be $15$, do not allow $-15$ or $\pm 15$ |
---5 Matrices $\mathbf { A }$ and $\mathbf { B }$ are given by $\mathbf { A } = \left( \begin{array} { r l } - 1 & 0 \\ 0 & 1 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { c c } \frac { 5 } { 13 } & - \frac { 12 } { 13 } \\ \frac { 12 } { 13 } & \frac { 5 } { 13 } \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item Use $\mathbf { A }$ and $\mathbf { B }$ to disprove the proposition: "Matrix multiplication is commutative".
Matrix $\mathbf { B }$ represents the transformation $\mathrm { T } _ { \mathrm { B } }$.
\item Describe the transformation $\mathrm { T } _ { \mathrm { B } }$.
\item By considering the inverse transformation of $\mathrm { T } _ { \mathrm { B } }$, determine $\mathbf { B } ^ { - 1 }$.
Matrix $\mathbf { C }$ is given by $\mathbf { C } = \left( \begin{array} { r r } 1 & 0 \\ 0 & - 3 \end{array} \right)$ and represents the transformation $\mathrm { T } _ { \mathrm { C } }$.\\
The transformation $\mathrm { T } _ { \mathrm { BC } }$ is transformation $\mathrm { T } _ { \mathrm { C } }$ followed by transformation $\mathrm { T } _ { \mathrm { B } }$.\\
An object shape of area 5 is transformed by $\mathrm { T } _ { \mathrm { BC } }$ to an image shape $N$.
\item Determine the area of $N$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core AS 2021 Q5 [8]}}