AS June 2025 Paper 1 Q3
3.
\[\mathbf{M} = \begin{pmatrix}a & b\\ -1 & -1\end{pmatrix}\]where \(a\) and \(b\) are real constants and \(a \neq b\)
Given that \(\mathbf{M} + \mathbf{M}^{-1} = \mathbf{I}\), where \(\mathbf{I}\) is the \(2 \times 2\) identity matrix,
| Scheme | Marks | AO |
|---|---|---|
| Determinant \(= -a + b\) | B1 | 1.1b |
| \(\left\{\mathbf{M}^{-1} =\right\} \dfrac{1}{b - a}\begin{pmatrix}-1 & -b\\ 1 & a\end{pmatrix}\) | M1 A1 | 1.1b 1.1b |
| (3) |
Notes
B1: Correct determinant
M1: Correct method to find the inverse matrix using their determinant
A1: Fully correct inverse matrix, isw
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}a & b\\ -1 & -1\end{pmatrix} + \dfrac{1}{b - a}\begin{pmatrix}-1 & -b\\ 1 & a\end{pmatrix} = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\) Which may include \(a - \dfrac{1}{b - a} = 1\) or \(-1 + \dfrac{1}{b - a} = 0\) or \(ab - a^2 - 1 = b - a\) or \(-b + a + 1 = 0\) \(b - \dfrac{b}{b - a} = 0\) or \(-1 + \dfrac{a}{b - a} = 1\) or \(b^2 - ab - b = 0\) or \(-b + a + a = b - a\) | M1 | 3.1a |
| Solves any two simultaneous equations to achieves a value for \(a\) and a value for \(b\) e.g., \(a - 1 = 1 \Rightarrow a = \ldots \Rightarrow b = \ldots\) | dM1 | 2.1 |
| \(a = 2,\ b = 3\) only | A1 | 1.1b |
| (3) | ||
| (6 marks) |
Notes
M1: A complete method to form two simultaneous equations in \(a\) and \(b\), using \(\mathbf{M} + \mathbf{M}^{-1} = \mathbf{I}\) and their inverse matrix
dM1: Solves their two different equations to achieve a value for \(a\) and a value for \(b\). Must come from a valid attempt at \(\mathbf{M} + \mathbf{M}^{-1} = \mathbf{I}\).
A1: Correct values for \(a\) and \(b\)
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{M}^2 + \mathbf{I} = \mathbf{M}\quad \begin{pmatrix}a^2 - b & ab - b\\ -a + 1 & -b + 1\end{pmatrix} + \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} = \begin{pmatrix}a & b\\ -1 & -1\end{pmatrix}\) Leading to any two equations which may include \(-a + 1 = -1\) or \(-b + 1 + 1 = -1\) | M1 | 3.1a |
| Solves to achieves a value for \(a\) and a value for \(b\) e.g., \(a - 1 = 1 \Rightarrow a = \ldots \Rightarrow b = \ldots\) | dM1 | 2.1 |
| \(a = 2,\ b = 3\) | A1 | 1.1b |
| (3) |
(Corrected from the printed mark scheme: the bottom-right entry of \(\mathbf{M}^2\) is printed as \(-b + a\); it is \(-b + 1\), as used in the next line.)
M1: Forms the equation \(\mathbf{M}^2 + \mathbf{I} = \mathbf{M}\) and form two equations
dM1: Solves their equations to achieve a value for \(a\) and a value for \(b\)
A1: Correct values for \(a\) and \(b\)