| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Singular matrix conditions |
| Difficulty | Moderate -0.8 This is a straightforward Further Maths question testing basic matrix operations: finding when determinant equals zero (routine), computing a 2×2 inverse (standard formula), and applying inverse transformation (algebraic manipulation). All parts are direct applications of standard techniques with no problem-solving insight required, making it easier than average even for Further Maths. |
| Spec | 4.03l Singular/non-singular matrices4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Use \(4a - (-2 \times -1) = 0 \Rightarrow a = \frac{1}{2}\) | M1, A1 | Allow sign slips for first M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Determinant: \((3 \times 4) - (-2 \times -1) = 10\) \((\Delta)\) | M1 | Allow sign slip for determinant for first M1 (may be awarded for 1/10 appearing in inverse matrix) |
| \(\mathbf{B}^{-1} = \frac{1}{10}\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\) | M1, A1cso | Second M1 for correctly treating the \(2 \times 2\) matrix, i.e. for \(\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{1}{10}\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} k-6 \\ 3k+12 \end{pmatrix} = \frac{1}{10}\begin{pmatrix} 4(k-6)+2(3k+12) \\ (k-6)+3(3k+12) \end{pmatrix}\) | M1, A1ft | M1 for multiplying matrix by appropriate column vector; A1 correct work (ft wrong determinant) |
| \(\begin{pmatrix} k \\ k+3 \end{pmatrix}\) Lies on \(y = x + 3\) | A1 | A1 for conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\begin{pmatrix} 3 & -2 \\ -1 & 4 \end{pmatrix}\begin{pmatrix} x \\ x+3 \end{pmatrix} = \begin{pmatrix} 3x-2(x+3) \\ -x+4(x+3) \end{pmatrix} = \begin{pmatrix} x-6 \\ 3x+12 \end{pmatrix}\), which was of the form \((k-6,\ 3k+12)\) | M1, A1, A1 | |
| Or \(\begin{pmatrix} 3 & -2 \\ -1 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3x-2y \\ -x+4y \end{pmatrix} = \begin{pmatrix} k-6 \\ 3k+12 \end{pmatrix}\) and solves simultaneous equations | M1 | |
| Both equations correct and eliminate one letter to get \(x = k\) or \(y = k+3\) or \(10x - 10y = -30\) or equivalent | A1 | |
| Completely correct work (to \(x = k\) and \(y = k+3\)), and conclusion lies on \(y = x+3\) | A1 |
# Question 7:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| Use $4a - (-2 \times -1) = 0 \Rightarrow a = \frac{1}{2}$ | M1, A1 | Allow sign slips for first M1 |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| Determinant: $(3 \times 4) - (-2 \times -1) = 10$ $(\Delta)$ | M1 | Allow sign slip for determinant for first M1 (may be awarded for 1/10 appearing in inverse matrix) |
| $\mathbf{B}^{-1} = \frac{1}{10}\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}$ | M1, A1cso | Second M1 for correctly treating the $2 \times 2$ matrix, i.e. for $\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}$ |
## Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{1}{10}\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} k-6 \\ 3k+12 \end{pmatrix} = \frac{1}{10}\begin{pmatrix} 4(k-6)+2(3k+12) \\ (k-6)+3(3k+12) \end{pmatrix}$ | M1, A1ft | M1 for multiplying matrix by appropriate column vector; A1 correct work (ft wrong determinant) |
| $\begin{pmatrix} k \\ k+3 \end{pmatrix}$ Lies on $y = x + 3$ | A1 | A1 for conclusion |
**Alternative (c):**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\begin{pmatrix} 3 & -2 \\ -1 & 4 \end{pmatrix}\begin{pmatrix} x \\ x+3 \end{pmatrix} = \begin{pmatrix} 3x-2(x+3) \\ -x+4(x+3) \end{pmatrix} = \begin{pmatrix} x-6 \\ 3x+12 \end{pmatrix}$, which was of the form $(k-6,\ 3k+12)$ | M1, A1, A1 | |
| Or $\begin{pmatrix} 3 & -2 \\ -1 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3x-2y \\ -x+4y \end{pmatrix} = \begin{pmatrix} k-6 \\ 3k+12 \end{pmatrix}$ and solves simultaneous equations | M1 | |
| Both equations correct and eliminate one letter to get $x = k$ or $y = k+3$ or $10x - 10y = -30$ or equivalent | A1 | |
| Completely correct work (to $x = k$ and $y = k+3$), and conclusion lies on $y = x+3$ | A1 | |
---7. $\mathbf { A } = \left( \begin{array} { r r } a & - 2 \\ - 1 & 4 \end{array} \right)$, where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ for which the matrix $\mathbf { A }$ is singular.
$$\mathbf { B } = \left( \begin{array} { r r }
3 & - 2 \\
- 1 & 4
\end{array} \right)$$
\item Find $\mathbf { B } ^ { - 1 }$.
The transformation represented by $\mathbf { B }$ maps the point $P$ onto the point $Q$.\\
Given that $Q$ has coordinates $( k - 6,3 k + 12 )$, where $k$ is a constant,
\item show that $P$ lies on the line with equation $y = x + 3$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2009 Q7 [8]}}