| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear transformations |
| Type | Extract enlargement and rotation parameters |
| Difficulty | Standard +0.3 This is a standard Further Pure 1 question on matrix transformations requiring routine techniques: applying rotation matrix formula with standard angle, finding scale factor (determinant) and rotation angle (arctan), and using matrix powers with pattern recognition. While it has multiple parts, each step follows textbook methods without requiring novel insight. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03i Determinant: area scale factor and orientation |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | \(\begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}\) | M1, A1 |
| 6(b)(i) | \(M = \begin{bmatrix} -1 & -1 \\ 1 & -1 \end{bmatrix} = \sqrt{2} \times \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \sqrt{2} & 0 \\ 0 & \sqrt{2} \end{bmatrix}\) | M1 |
| 6(b)(i) | Scale factor of enlargement is \(\sqrt{2}\) | A1 |
| 6(b)(i) | Angle of rotation is 135 (degrees anticlockwise) | A1 |
| 6(b)(ii) | For \(M^2\), SF of enlargement = 2 | B1F |
| 6(b)(ii) | Angle of rotation is 270 (degrees anticlockwise) | B1F |
| 6(b)(iii) | \(M^2 = \begin{bmatrix} -1 & -1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\) | M1 |
| 6(b)(iii) | \(M^4 = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix}\) | |
| 6(b)(iii) | \(M^4 = -4\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -4I\) | A1 |
| 6(b)(iv) | \(M^{2012} = (M^4)^{503} = (-4I)^{503} = -(2^2)^{503}I = -2^{1006}I\) | E1 |
| 6(b)(iv) | \(M^{2012} = -2^{1006}I\) | B1 |
| 6(b)(iv) | (Geometrically: \(M^{2012}\) represents an enlargement with SF 2^1006 followed by a rotation of angle 2012×135° is 754.5 revolutions, being equivalent to rotation of 180° ie matrix is \(-I\) so \(M^{2012} = -2^{1006}I\)) |
6(a) | $\begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$ | M1, A1 | If A1 not scored, award M1A0 for all correct entries expressed in trig form eg $\begin{bmatrix} \cos135 & -\sin135 \\ \sin135 & \cos135 \end{bmatrix}$
6(b)(i) | $M = \begin{bmatrix} -1 & -1 \\ 1 & -1 \end{bmatrix} = \sqrt{2} \times \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \sqrt{2} & 0 \\ 0 & \sqrt{2} \end{bmatrix}$ | M1 | Or better: PI by cand. having both a correct scale factor of enlargement and a correct corresponding angle of rotation
6(b)(i) | Scale factor of enlargement is $\sqrt{2}$ | A1 | SF = $\sqrt{2}$ OE surd form. Angle = 135 OE eg −225. If M0 give B1 for SF= $\sqrt{2}$ OE surd and B1 for angle = 135 OE
6(b)(i) | Angle of rotation is 135 (degrees anticlockwise) | A1 |
6(b)(ii) | For $M^2$, SF of enlargement = 2 | B1F | OE If incorrect, ft on [c's SF in (b)(i)]². OE, eg −90(degrees), eg 90 (degrees) clockwise. If incorrect, ft on 2×c's angle in (b)(i) (neither B1F B1 nor B1 B1F is possible)
6(b)(ii) | Angle of rotation is 270 (degrees anticlockwise) | B1F |
6(b)(iii) | $M^2 = \begin{bmatrix} -1 & -1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$ | M1 | For complete method (matrix calculation or geometrical reasoning). Matrix for $M^2$ could be seen earlier (M0 if >1 independent error in matrix multiplication)
6(b)(iii) | $M^4 = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix}$ | | Geometrically SF = 4, rotation angle= 540. OE scores M1 and completion scores A1
6(b)(iii) | $M^4 = -4\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -4I$ | A1 | Either of these two forms convincingly shown
6(b)(iv) | $M^{2012} = (M^4)^{503} = (-4I)^{503} = -(2^2)^{503}I = -2^{1006}I$ | E1 | OE Fully explained, algebraically from $(-4I)^{503}$, or geometrically. $M^{2012} = -2^{1006}I (n = 1006)$ (B0 if FiW)
6(b)(iv) | $M^{2012} = -2^{1006}I$ | B1 |
6(b)(iv) | (Geometrically: $M^{2012}$ represents an enlargement with SF 2^1006 followed by a rotation of angle 2012×135° is 754.5 revolutions, being equivalent to rotation of 180° ie matrix is $-I$ so $M^{2012} = -2^{1006}I$) | |6
\begin{enumerate}[label=(\alph*)]
\item Using surd forms, find the matrix of a rotation about the origin through $135 ^ { \circ }$ anticlockwise.
\item The matrix $\mathbf { M }$ is defined by $\mathbf { M } = \left[ \begin{array} { r r } - 1 & - 1 \\ 1 & - 1 \end{array} \right]$.
\begin{enumerate}[label=(\roman*)]
\item Given that $\mathbf { M }$ represents an enlargement followed by a rotation, find the scale factor of the enlargement and the angle of the rotation.
\item The matrix $\mathbf { M } ^ { 2 }$ also represents an enlargement followed by a rotation. State the scale factor of the enlargement and the angle of the rotation.
\item Show that $\mathbf { M } ^ { 4 } = k \mathbf { I }$, where $k$ is an integer and $\mathbf { I }$ is the $2 \times 2$ identity matrix.
\item Deduce that $\mathbf { M } ^ { 2012 } = - 2 ^ { n } \mathbf { I }$ for some positive integer $n$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2012 Q6 [11]}}