AS June 2025 Paper 1 Q4

EdexcelCurrent spec10 marksMatrices

(i)\[\mathbf{P} = \begin{pmatrix}2 & 0\\ 0 & 1\end{pmatrix}\]

(a) Describe fully the single geometrical transformation \(P\) represented by the matrix \(\mathbf{P}\). (2)

(b) State the equation of one invariant line under the transformation \(P\). (1)

(ii)\[\mathbf{Q} = \begin{pmatrix}\cos 2\theta & 0\\ 1 & \tan 2\theta\end{pmatrix} \qquad \text{where } 0 \leqslant \theta \lt 360^\circ\]

The matrix \(\mathbf{Q}\) represents the transformation \(Q\).

Triangle \(T\) is transformed to triangle \(T^{\prime}\) by the transformation \(Q\).

Given that

the coordinates of the vertices of \(T\) are (2, 3), (3, 6) and (8, 3)
the area of \(T^{\prime}\) is 4.5

determine the possible values of \(\theta\)

(Solutions relying entirely on calculator technology are not acceptable.) (7)

Mark scheme (i)(a)

Scheme	Marks	AO
Stretch	B1	1.1a
Scale factor 2 parallel to \(x\)-axis	B1	2.5
	(2)

Notes

B1: See scheme, enlargement is B0

B1: See scheme, if they mention of \(y\)-axis must be correct e.g. scale factor 1, stays the same.
Condone defining the stretch as from/about the origin and in the \(x\)- axis, along the \(x\)-axis

Scheme	Marks	AO
Any line parallel to the \(x\)-axis or \(y = k\) or the \(y\)-axis or \(x = 0\)	B1	1.1b
	(1)

Mark scheme (ii)

Scheme	Marks	AO
Area triangle \(= \dfrac{1}{2} \times (8 - 2)(6 - 3) = 9\) Area triangle for example \(\dfrac{1}{2}\begin{vmatrix}2 & 3 & 8 & 2\\ 3 & 6 & 3 & 3\end{vmatrix} = \dfrac{1}{2}\left[2 \times 6 + 3 \times 3 + 8 \times 3 - (3 \times 3 + 8 \times 6 + 2 \times 3)\right]\) Area triangle \(= \dfrac{1}{2}(6)\left(\sqrt{10}\right)\sin 71.57\) from a valid attempt to find the angle and sides	M1	3.1a
\(\begin{vmatrix}\cos 2\theta & 0\\ 1 & \tan 2\theta\end{vmatrix} = \cos 2\theta \tan 2\theta\)	B1	1.1b
their ‘\(\cos 2\theta \tan 2\theta\)’ \(= \dfrac{4.5}{\text{their ‘9’}}\)	M1	1.1b
\(\cos 2\theta \tan 2\theta = \dfrac{4.5}{\text{their } 9} \Rightarrow \cos 2\theta \times \dfrac{\sin 2\theta}{\cos 2\theta} = \dfrac{4.5}{\text{their } 9} \Rightarrow \sin 2\theta = k\) where \(-1 \lt k \lt 1\)	dM1	3.1a
\(\sin 2\theta = \dfrac{1}{2} \Rightarrow \theta = \ldots\) or \(\sin 2\theta = -\dfrac{1}{2} \Rightarrow \theta = \ldots\)	ddM1	1.1b
Any two of Coming from \(\sin 2\theta = \dfrac{1}{2}\) 15°, 75°, 195°, 255° or \(\dfrac{\pi}{12}, \dfrac{5\pi}{12}, \dfrac{13\pi}{12}, \dfrac{17\pi}{12}\) or awrt 0.26, 1.31, 3.40, 4.45 Coming from \(\sin 2\theta = -\dfrac{1}{2}\) 105°, 165°, 285°, 345° or \(\dfrac{7\pi}{12}, \dfrac{11\pi}{12}, \dfrac{19\pi}{12}, \dfrac{23\pi}{12}\) or awrt , 1.83, 2.88, 4.97, 6.02	A1	1.1b
All eight of 15°, 75°, 195°, 255°, 105°, 165°, 285°, 345°	A1	2.3
	(7)
	(10 marks)

Notes

M1: Correct method to find the area of the triangle \(T\)

B1: Correct determinant of the matrix

M1: Sets their determinant = 4.5 divided by their area of the triangle \(T\)

dM1: Dependent on previous method mark. Uses the identity \(\tan 2\theta = \dfrac{\sin 2\theta}{\cos 2\theta}\) to achieve \(\sin 2\theta = k\) where \(-1 \lt k \lt 1\)

ddM1: Dependent on previous method marks. Solves \(\sin 2\theta = k\) where \(-1 \lt k \lt 1\) to find a value of \(\theta\)

A1: Any two correct values in degrees or radians

A1: All eight correct values and no others must be in degrees