OCR Further Pure Core AS 2023 June — Question 9 10 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2023
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear transformations
TypeMatrix powers and repeated transformations
DifficultyStandard +0.8 This is a substantial multi-part question requiring matrix multiplication with parameters, algebraic manipulation of nested surds to simplify expressions, recognition of patterns in matrix powers, and geometric interpretation of 3D transformations. The surd simplification in part (b) and identifying the rotation angle in part (d) require non-routine insight beyond standard textbook exercises, though the matrix multiplication itself is mechanical.
Spec1.02b Surds: manipulation and rationalising denominators4.03b Matrix operations: addition, multiplication, scalar4.03d Linear transformations 2D: reflection, rotation, enlargement, shear

9 Matrix \(\mathbf { R }\) is given by \(\mathbf { R } = \left( \begin{array} { c c c } a & 0 & - b \\ 0 & 1 & 0 \\ b & 0 & a \end{array} \right)\) where \(a\) and \(b\) are constants.
  1. Find \(\mathbf { R } ^ { 2 }\) in terms of \(a\) and \(b\). The constants \(a\) and \(b\) are given by \(a = \frac { \sqrt { 2 } } { 4 } ( \sqrt { 3 } + 1 )\) and \(b = \frac { \sqrt { 2 } } { 4 } ( \sqrt { 3 } - 1 )\).
  2. By determining exact expressions for \(a b\) and \(a ^ { 2 } - b ^ { 2 }\) and using the result from part (a), show that \(\mathbf { R } ^ { 2 } = k \left( \begin{array} { c c c } \sqrt { 3 } & 0 & - 1 \\ 0 & 2 & 0 \\ 1 & 0 & \sqrt { 3 } \end{array} \right)\) where \(k\) is a real number whose value is to be determined.
  3. Find \(\mathbf { R } ^ { 6 } , \mathbf { R } ^ { 12 }\) and \(\mathbf { R } ^ { 24 }\).
  4. Describe fully the transformation represented by \(\mathbf { R }\). \section*{END OF QUESTION PAPER}

Question 9:
Part (a):
\[\begin{pmatrix} a & 0 & -b \\ 0 & 1 & 0 \\ b & 0 & a \end{pmatrix}\begin{pmatrix} a & 0 & -b \\ 0 & 1 & 0 \\ b & 0 & a \end{pmatrix} = \begin{pmatrix} a^2-b^2 & 0 & -2ab \\ 0 & 1 & 0 \\ 2ab & 0 & a^2-b^2 \end{pmatrix}\]
AnswerMarks
M1Some indication of knowledge of how to multiply 3 by 3 matrices. Can be implied by 3 non-zero correct terms.
A1Final solutions must have all terms simplified.
[2 marks]
Part (b):
\(ab = \frac{\sqrt{2}}{4}(\sqrt{3}+1)\frac{\sqrt{2}}{4}(\sqrt{3}-1) = \frac{2}{16}(3-1) = \frac{1}{4}\)
\(a^2 - b^2 = (a-b)(a+b) = \frac{2\sqrt{2}}{4} \times \frac{2\sqrt{2}\sqrt{3}}{4} = \frac{1}{2}\sqrt{3}\)
\[R^2 = \begin{pmatrix} \frac{1}{2}\sqrt{3} & 0 & -\frac{1}{2} \\ 0 & 1 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2}\sqrt{3} \end{pmatrix} = \frac{1}{2}\begin{pmatrix} \sqrt{3} & 0 & -1 \\ 0 & 2 & 0 \\ 1 & 0 & \sqrt{3} \end{pmatrix}\]
so \(k = \frac{1}{2}\)
AnswerMarks
B1Explicitly finding an expression for either \(ab\) (or \(2ab\)) or \(a^2 - b^2\). AO 3.1a
B1AG. Explicitly finding the expression for the other, substituting into expression for \(R^2\) (or carrying out the matrix multiplication again). \(k\) can be embedded. If \(k\) embedded need to see \(\frac{1}{2} \times 2\) or \(\frac{2}{2}\). AO 2.2a
[2 marks]
Part (c):
\[R^4 = \begin{pmatrix} \frac{1}{2} & 0 & -\frac{1}{2}\sqrt{3} \\ 0 & 1 & 0 \\ \frac{1}{2}\sqrt{3} & 0 & \frac{1}{2} \end{pmatrix}\]
\[R^6 = \begin{pmatrix} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\]
AnswerMarks
B1For correct \(R^6\). By calculator expected, so intermediate steps might not be shown. AO 1.1
\[R^{12} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\]
AnswerMarks
B1For correct \(R^{12}\). AO 1.1
\[R^{24} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\]
AnswerMarks
B1For correct \(R^{24}\). AO 1.1
SC – if answers left in terms of \(k\) then award B1 if one or two correct and B2 if all three correct.
[3 marks]
Part (d):
AnswerMarks Guidance
RotationB1 AO 3.1a
\((360°/24 =)\ 15°\)B1 Also allow \(345°\) (rotation in opposite sense). Also allow in radians \(\frac{\pi}{12}\). AO 2.2a
Clockwise about the \(y\)-axis.B1 Both sense and axis must be correct. Could be a rotation of \(345°\) anticlockwise about \(y\)-axis. If correct transformation is combined with an incorrect one (such as correct rotation combined with a reflection) then maximum mark is B2. AO 3.2a
[3 marks]
# Question 9:

## Part (a):

$$\begin{pmatrix} a & 0 & -b \\ 0 & 1 & 0 \\ b & 0 & a \end{pmatrix}\begin{pmatrix} a & 0 & -b \\ 0 & 1 & 0 \\ b & 0 & a \end{pmatrix} = \begin{pmatrix} a^2-b^2 & 0 & -2ab \\ 0 & 1 & 0 \\ 2ab & 0 & a^2-b^2 \end{pmatrix}$$

| **M1** | Some indication of knowledge of how to multiply 3 by 3 matrices. Can be implied by 3 non-zero correct terms. |

| **A1** | Final solutions must have all terms simplified. |

**[2 marks]**

---

## Part (b):

$ab = \frac{\sqrt{2}}{4}(\sqrt{3}+1)\frac{\sqrt{2}}{4}(\sqrt{3}-1) = \frac{2}{16}(3-1) = \frac{1}{4}$

$a^2 - b^2 = (a-b)(a+b) = \frac{2\sqrt{2}}{4} \times \frac{2\sqrt{2}\sqrt{3}}{4} = \frac{1}{2}\sqrt{3}$

$$R^2 = \begin{pmatrix} \frac{1}{2}\sqrt{3} & 0 & -\frac{1}{2} \\ 0 & 1 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2}\sqrt{3} \end{pmatrix} = \frac{1}{2}\begin{pmatrix} \sqrt{3} & 0 & -1 \\ 0 & 2 & 0 \\ 1 & 0 & \sqrt{3} \end{pmatrix}$$

so $k = \frac{1}{2}$

| **B1** | Explicitly finding an expression for either $ab$ (or $2ab$) or $a^2 - b^2$. AO 3.1a |

| **B1** | AG. Explicitly finding the expression for the other, substituting into expression for $R^2$ (or carrying out the matrix multiplication again). $k$ can be embedded. If $k$ embedded need to see $\frac{1}{2} \times 2$ or $\frac{2}{2}$. AO 2.2a |

**[2 marks]**

---

## Part (c):

$$R^4 = \begin{pmatrix} \frac{1}{2} & 0 & -\frac{1}{2}\sqrt{3} \\ 0 & 1 & 0 \\ \frac{1}{2}\sqrt{3} & 0 & \frac{1}{2} \end{pmatrix}$$

$$R^6 = \begin{pmatrix} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$$

| **B1** | For correct $R^6$. By calculator expected, so intermediate steps might not be shown. AO 1.1 |

$$R^{12} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$

| **B1** | For correct $R^{12}$. AO 1.1 |

$$R^{24} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

| **B1** | For correct $R^{24}$. AO 1.1 |

SC – if answers left in terms of $k$ then award B1 if one or two correct and B2 if all three correct.

**[3 marks]**

---

## Part (d):

Rotation | **B1** | AO 3.1a |

$(360°/24 =)\ 15°$ | **B1** | Also allow $345°$ (rotation in opposite sense). Also allow in radians $\frac{\pi}{12}$. AO 2.2a |

Clockwise about the $y$-axis. | **B1** | Both sense and axis must be correct. Could be a rotation of $345°$ anticlockwise about $y$-axis. If correct transformation is combined with an incorrect one (such as correct rotation combined with a reflection) then maximum mark is B2. AO 3.2a |

**[3 marks]**
9 Matrix $\mathbf { R }$ is given by $\mathbf { R } = \left( \begin{array} { c c c } a & 0 & - b \\ 0 & 1 & 0 \\ b & 0 & a \end{array} \right)$ where $a$ and $b$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { R } ^ { 2 }$ in terms of $a$ and $b$.

The constants $a$ and $b$ are given by $a = \frac { \sqrt { 2 } } { 4 } ( \sqrt { 3 } + 1 )$ and $b = \frac { \sqrt { 2 } } { 4 } ( \sqrt { 3 } - 1 )$.
\item By determining exact expressions for $a b$ and $a ^ { 2 } - b ^ { 2 }$ and using the result from part (a), show that $\mathbf { R } ^ { 2 } = k \left( \begin{array} { c c c } \sqrt { 3 } & 0 & - 1 \\ 0 & 2 & 0 \\ 1 & 0 & \sqrt { 3 } \end{array} \right)$ where $k$ is a real number whose value is to be determined.
\item Find $\mathbf { R } ^ { 6 } , \mathbf { R } ^ { 12 }$ and $\mathbf { R } ^ { 24 }$.
\item Describe fully the transformation represented by $\mathbf { R }$.

\section*{END OF QUESTION PAPER}
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core AS 2023 Q9 [10]}}