A2 June 2025 Paper 1 Q1
1.
\[\mathbf{M} = \begin{pmatrix}3 & 6 & 0\\ a & 3 & 1\\ 2 & a & a\end{pmatrix} \qquad \text{where } a \text{ is a constant}\]Given that matrix \(\mathbf{M}\) is non-singular and that
- \(\det(\mathbf{M}) = -108\)
- \(a \lt 0\)
| Scheme | Marks | AO |
|---|---|---|
| \(\det(\mathbf{M}) = 3(3a - a) - 6(a^2 - 2) = 0\) | M1 | 1.2 |
| \(6a^2 - 6a - 12 = 0 \Rightarrow a = \ldots\) | dM1 | 1.1b |
| \(a = -1,\ 2\) | A1 | 1.1b |
| (3) |
Notes
M1: Finds the determinant of the matrix \(\mathbf{M}\) and sets \(= 0\)
Allow for sight of \(\pm 3(3a - a) \pm 6(a^2 - 2) \pm 0(a^2 - 6) = 0\) and do not allow sign slips in their minors.
Need not be simplified
dM1: Solves their 3TQ. Usual rules apply for solving a quadratic by any means including using a calculator.
A1: Both correct values for \(a\).
Note: Correct answers without a method can be awarded M1dM1A1
Alternatively, the determinant can be found using the vector product, for example:
![Handwritten example: 0 = (3, a, 2) dot [(6, 3, a) cross (0, 1, a)] = (3, a, 2) dot (2a, -6a, 6), giving 0 = 6a - 6a squared + 12](https://www.westiesworkshop.com/wp-content/uploads/question-bank/fm-edexcel/edexcel-9fm001-jun25-q1-ms-fig1.webp)
If you are uncertain, please send to review
| Scheme | Marks | AO |
|---|---|---|
| Sets determinant \(= -108\), solves a 3TQ to find a negative value of \(a\) \(-6a^2 + 6a + 12 = -108 \Rightarrow a = \ldots\) or \(6a^2 - 6a - 12 = 108 \Rightarrow a = \ldots\) | M1 | 3.1a |
| \(a = -4\) | A1 | 2.2a |
| \(\begin{pmatrix}\dfrac{2}{27} & -\dfrac{2}{9} & -\dfrac{1}{18}\\[6pt] \dfrac{7}{54} & \dfrac{1}{9} & \dfrac{1}{36}\\[6pt] -\dfrac{5}{54} & -\dfrac{2}{9} & -\dfrac{11}{36}\end{pmatrix}\) | A1 | 1.1b |
| (3) | ||
| (6 marks) |
Notes
M1: A complete method to find a value for \(a\). Sets their determinant equal to \(-108\), solves a 3TQ and proceeds to find at least one negative value of \(a\).
They must use their determinant equal to \(-108\) and not for example a changed quadratic from (a) equal to \(-108\). Usual rules apply for solving a quadratic by any means including using a calculator.
A1: Deduces \(a = -4\) only, which could be implied by a correct matrix.
A1: Correct matrix. Accept exact equivalents but not answers rounded to decimal places. Isw once a correct matrix is seen.
e.g. Accept \(\dfrac{1}{108}\begin{pmatrix}8 & -24 & -6\\ 14 & 12 & 3\\ -10 & -24 & -33\end{pmatrix}\) or \(-\dfrac{1}{108}\begin{pmatrix}-8 & 24 & 6\\ -14 & -12 & -3\\ 10 & 24 & 33\end{pmatrix}\)