AS June 2023 Paper 1 Q1
1.
\[\begin{pmatrix}x & 9\\ y & z\end{pmatrix} - 3\begin{pmatrix}z & y\\ z & y\end{pmatrix} = k\mathbf{I}\]where \(x\), \(y\), \(z\) and \(k\) are constants.
Determine the value of \(x\), the value of \(y\) and the value of \(z\). (4)
| Scheme | Marks | AO |
|---|---|---|
| \(y = 3\) | B1 | 2.2a |
| \(z = \dfrac{\text{their } y}{3} = \ldots\{1\}\) | B1ft | 1.1b |
| Uses \(z - 3y = k \Rightarrow k = -8\) and \(x - 3z = k \Rightarrow x = k + 3z = \text{their } k + 3 \times \text{their } z\) leading to a value for \(x\) Alternatively uses \(x - 3z = k = z - 3y\) with values for \(y\) and \(z\) to find a value for \(x\). | M1 | 3.1a |
| \(x = -5\) | A1 | 1.1b |
| (4) | ||
| (4 marks) |
Notes
B1: \(y = 3\)
B1ft: Follow through on the value of \(z\) which comes from their \(y\) divided by 3
M1: A complete method to find the value of \(x\). Uses \(z - 3y = k\) to find a value for \(k\) then finds a value for \(x\) using \(x - 3z = k\) and their values for \(z\) and \(k\). Condone a slip with the coefficients if the intention is clear but must have the correct letters.
Alternatively uses \(x - 3z = k = z - 3y\) with values for \(y\) and \(z\) to find a value for \(x\).
A1: \(x = -5\)
Correct answers only scores full marks.