| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Singular matrix conditions |
| Difficulty | Moderate -0.8 This is a straightforward application of the singular matrix condition (determinant = 0). Students need to compute det(A) = (1+k)(1-k) - k², set it equal to zero, and solve the resulting quadratic equation. It's a routine FP1 exercise requiring only basic algebraic manipulation with no problem-solving insight needed. |
| Spec | 4.03h Determinant 2x2: calculation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Determinant of \(\mathbf{A} = (1-k)(1+k) - k^2 = 0\) | M1 | For attempting \(ad - bc = 0\) with \(= 0\) seen or used later |
| \(1 - k + k - k^2 - k^2 (= 0)\), i.e. \(1 - 2k^2 (= 0)\) | A1 | Correct (unsimplified) expression on LHS or correct equation after brackets expanded |
| \(k = \dfrac{\pm\sqrt{2}}{2}\) | A1 | Accept \(\pm\dfrac{\sqrt{2}}{2}\), \(\pm\dfrac{1}{\sqrt{2}}\), \(\pm\sqrt{\dfrac{1}{2}}\), \(\pm\sqrt{0.5}\). Must have \(\pm\) for mark |
## Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Determinant of $\mathbf{A} = (1-k)(1+k) - k^2 = 0$ | M1 | For attempting $ad - bc = 0$ with $= 0$ seen or **used** later |
| $1 - k + k - k^2 - k^2 (= 0)$, i.e. $1 - 2k^2 (= 0)$ | A1 | Correct (unsimplified) expression on LHS or correct equation after brackets expanded |
| $k = \dfrac{\pm\sqrt{2}}{2}$ | A1 | Accept $\pm\dfrac{\sqrt{2}}{2}$, $\pm\dfrac{1}{\sqrt{2}}$, $\pm\sqrt{\dfrac{1}{2}}$, $\pm\sqrt{0.5}$. Must have $\pm$ for mark |
---\begin{enumerate}
\item Given that $k$ is a real number and that
\end{enumerate}
$$\mathbf { A } = \left( \begin{array} { c c }
1 + k & k \\
k & 1 - k
\end{array} \right)$$
find the exact values of $k$ for which $\mathbf { A }$ is a singular matrix. Give your answers in their simplest form.\\
(3)\\
\hfill \mbox{\textit{Edexcel FP1 2016 Q1 [3]}}