| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Non-singular matrix proof |
| Difficulty | Moderate -0.5 This is a straightforward multi-part question testing standard matrix operations (squaring, determinant, inverse) and a simple application. All parts use routine algorithms with no problem-solving insight required. Part (d) involves basic algebraic manipulation after finding P = A^(-1)Q. Slightly easier than average due to the mechanical nature and small matrix size, though it's Further Maths content. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03h Determinant 2x2: calculation4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \[\begin{pmatrix}3 & 1\\1 & -2\end{pmatrix}\begin{pmatrix}3 & 1\\1 & -2\end{pmatrix}=\begin{pmatrix}10 & 1\\1 & 5\end{pmatrix}\] | M1A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\det \mathbf{A} = -7 \neq 0\) so \(\mathbf{A}\) is non-singular | M1A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \[\mathbf{A}^{-1} = -\frac{1}{7}\begin{pmatrix}-2 & -1\\-1 & 3\end{pmatrix}\] | M1A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \[-\frac{1}{7}\begin{pmatrix}-2 & -1\\-1 & 3\end{pmatrix}\begin{pmatrix}k-1\\2-k\end{pmatrix} = -\frac{1}{7}\begin{pmatrix}-2(k-1)-1(2-k)\\-1(k-1)+3(2-k)\end{pmatrix}\] | M1 | |
| \[= \begin{pmatrix}\frac{1}{7}k\\\frac{4}{7}k-1\end{pmatrix}\] | A1, A1 | |
| (\(p\) lies on \(y = 4x - 1\)) | (3) |
| Answer | Marks |
|---|---|
| \[\begin{pmatrix}3 & 1\\1 & -2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}k-1\\2-k\end{pmatrix}\] then multiply out and attempt to solve simultaneous equations for \(x\) or \(y\) in terms of \(k\) | M1 |
| \(x = \frac{1}{7}k\) | A1 |
| \(y = \frac{4}{7}k - 1\) | A1 |
## Question 9:
### Part (a):
$$\begin{pmatrix}3 & 1\\1 & -2\end{pmatrix}\begin{pmatrix}3 & 1\\1 & -2\end{pmatrix}=\begin{pmatrix}10 & 1\\1 & 5\end{pmatrix}$$ | M1A1 | **(2)**
---
### Part (b):
$\det \mathbf{A} = -7 \neq 0$ so $\mathbf{A}$ is non-singular | M1A1 | **(2)**
---
### Part (c):
$$\mathbf{A}^{-1} = -\frac{1}{7}\begin{pmatrix}-2 & -1\\-1 & 3\end{pmatrix}$$ | M1A1 | **(2)**
---
### Part (d):
$$-\frac{1}{7}\begin{pmatrix}-2 & -1\\-1 & 3\end{pmatrix}\begin{pmatrix}k-1\\2-k\end{pmatrix} = -\frac{1}{7}\begin{pmatrix}-2(k-1)-1(2-k)\\-1(k-1)+3(2-k)\end{pmatrix}$$ | M1 |
$$= \begin{pmatrix}\frac{1}{7}k\\\frac{4}{7}k-1\end{pmatrix}$$ | A1, A1 |
($p$ lies on $y = 4x - 1$) | | **(3)**
**[9]**
---
### Part (d) Alt:
$$\begin{pmatrix}3 & 1\\1 & -2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}k-1\\2-k\end{pmatrix}$$ then multiply out and attempt to solve simultaneous equations for $x$ or $y$ in terms of $k$ | M1 |
$x = \frac{1}{7}k$ | A1 |
$y = \frac{4}{7}k - 1$ | A1 |9. With reference to a fixed origin $O$ and coordinate axes $O x$ and $O y$, a transformation from $\mathbb { R } ^ { 2 } \rightarrow \mathbb { R } ^ { 2 }$ is represented by the matrix $A$ where
$$A = \left( \begin{array} { c c }
3 & 1 \\
1 & - 2
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { A } ^ { 2 }$.
\item Show that the matrix A is non-singular.
\item Find $\mathrm { A } ^ { - 1 }$.
The transformation represented by matrix A maps the point $P$ onto the point $Q$.\\
Given that $Q$ has coordinates $( k - 1,2 - k )$, where $k$ is a constant,
\item show that $P$ lies on the line with equation $y = 4 x - 1$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q9 [9]}}