AS June 2025 Paper 1 Q8
8 Matrix \(\mathbf{A}\) is given by \(\mathbf{A} = \begin{pmatrix} 5 + a & 5 & 1 \\ 3 & 13 + a & 6 \\ a - 4 & -20 & -9 \end{pmatrix}\) where \(a\) is a constant and all entries of \(\mathbf{A}\) are integers.
The transformation represented by \(\mathbf{A}\) is applied to a shape of volume 8 units.
The image shape has volume 40 units and the orientation of the image is reversed.
Determine the image under \(\mathbf{A}\) of the point \((1, 2, 3)\). [7]
| Scheme | Marks | AO |
|---|---|---|
| Vol SF \(= 40/8 = 5\) & reverse of orientation \(\Rightarrow \det\mathbf{A} = -5\) | B1 | 3.1a |
| \(\begin{vmatrix} 5 + a & 5 & 1 \\ 3 & 13 + a & 6 \\ a - 4 & -20 & -9 \end{vmatrix} = (5 + a)((13 + a) \times -9 - 6 \times -20)\) \(-5(3 \times -9 - 6(a - 4)) + 1(3 \times -20 - (13 + a)(a - 4))\) | M1 | 1.1 |
| \(= (5 + a)(-117 - 9a + 120) - 5(-27 - 6a + 24))\) \(+(-60 - (13a - 52 + a^2 - 4a))\) \(= (5 + a)(3 - 9a) - 5(-3 - 6a) - 8 - 9a - a^2\) \(= 15 - 45a + 3a - 9a^2 + 15 + 30a - 8 - 9a - a^2\) \(= -10a^2 - 21a + 22\) | M1 | 1.1 |
| \(= -5 \Rightarrow 10a^2 + 21a - 27 = 0\) | A1 | 1.1 |
| \(a = -3\) since \(a = 9/10\) leads to non-integer entries in \(\mathbf{A}\). | A1 | 3.2a |
| \(\begin{pmatrix} 2 & 5 & 1 \\ 3 & 10 & 6 \\ -7 & -20 & -9 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) | M1 | 1.1 |
| \(= \begin{pmatrix} 15 \\ 41 \\ -74 \end{pmatrix}\) so image is \((15, 41, -74)\) | A1 | 1.1 |
| [7] |
Notes
B1: Could see embedded in equation below
Can allow recovery in equation
M1: Expanding the determinant
Allow sign errors
Allow if two of the three minor determinants are correct
M1: Reduction of determinant to three term quadratic form (could be done in conjunction with “\(= -5\)” or “\(= 5\)”).
Allow sign errors
Must have used correct form of determinant (apart from sign errors)
A1: Arriving at correct 3 term quadratic equation in \(a\).
BOD omission of “\(= 0\)” if correct \(a\) values appears
A1: Must see both correct roots and correct reason for rejection
M1: Multiplying their \(\mathbf{A}\) by the position vector of the given point.
Must be correct order of multiplication
Must have used a numerical value of \(a\)
A1: Condone position vector as answer.
Must have come from correct equation found correctly. Allow if reason for rejection of \(a = 9/10\) is incorrect or omitted. (Corrected from the printed mark scheme: the printed guidance says \(a = -9/10\); the rejected root is \(a = 9/10\).)