| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Singular matrix conditions |
| Difficulty | Standard +0.3 This is a standard Further Maths question testing routine matrix concepts: finding when a matrix is singular (det=0), computing a 2×2 matrix inverse, applying inverse transformations, and using the area scale factor property. All parts follow textbook procedures with no novel insight required, making it slightly easier than average even for FP1. |
| Spec | 4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(5k(k+1) - (-3(3k-1))=0\) | M1 | Puts determinant equal to zero |
| \(5k^2+5k+9k-3=0\) | A1 | cao as three or four term quadratic |
| \((5k-1)(k+3)=0\) | M1 | Solve their quadratic |
| \(k=\frac{1}{5}\) or \(k=-3\) | A1 | cao – need both correct answers |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\mathbf{B}^{-1}=\frac{1}{45}\begin{pmatrix}3&-5\\3&10\end{pmatrix}\) | M1 A1 | M1: Uses correct method for inverse with fraction \(\frac{1}{45}\) or \(\frac{1}{\text{their det}}\). A1: All correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{1}{45}\begin{pmatrix}3&-5\\3&10\end{pmatrix}\begin{pmatrix}0&-20&10c\\0&6&6c\end{pmatrix}\) | M1 | Post-multiplies inverse by 2×3 matrix excluding origin |
| \(= \frac{1}{45}\begin{pmatrix}0&-90&0\\0&0&90c\end{pmatrix}\) | ||
| Vertices at \((0,0)\), \((-2,0)\), \((0,2c)\) | A1, A1 | A1: \((-2,0)\) and \((0,2c)\). A1: \((0,0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Area of \(T\) is \(\frac{1}{2}\times2\times2c = 2c\) | B1 | Area of \(T\) given as \(2c\) or area of \(T'=90c\). Accept \(\pm\) |
| Area of \(T\times\text{determinant}=135\) | M1 | Either method using area of \(T\) and det, or area of \(T'\) |
| So \(c=\frac{3}{2}\) | A1 | cao |
# Question 7:
## Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $5k(k+1) - (-3(3k-1))=0$ | M1 | Puts determinant equal to zero |
| $5k^2+5k+9k-3=0$ | A1 | cao as three or four term quadratic |
| $(5k-1)(k+3)=0$ | M1 | Solve their quadratic |
| $k=\frac{1}{5}$ or $k=-3$ | A1 | cao – need both correct answers |
## Part (ii)(a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\mathbf{B}^{-1}=\frac{1}{45}\begin{pmatrix}3&-5\\3&10\end{pmatrix}$ | M1 A1 | M1: Uses correct method for inverse with fraction $\frac{1}{45}$ or $\frac{1}{\text{their det}}$. A1: All correct |
## Part (ii)(b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{1}{45}\begin{pmatrix}3&-5\\3&10\end{pmatrix}\begin{pmatrix}0&-20&10c\\0&6&6c\end{pmatrix}$ | M1 | Post-multiplies inverse by 2×3 matrix excluding origin |
| $= \frac{1}{45}\begin{pmatrix}0&-90&0\\0&0&90c\end{pmatrix}$ | | |
| Vertices at $(0,0)$, $(-2,0)$, $(0,2c)$ | A1, A1 | A1: $(-2,0)$ and $(0,2c)$. A1: $(0,0)$ |
## Part (ii)(c):
| Working | Mark | Guidance |
|---------|------|----------|
| Area of $T$ is $\frac{1}{2}\times2\times2c = 2c$ | B1 | Area of $T$ given as $2c$ or area of $T'=90c$. Accept $\pm$ |
| Area of $T\times\text{determinant}=135$ | M1 | Either method using area of $T$ and det, or area of $T'$ |
| So $c=\frac{3}{2}$ | A1 | cao |
---\begin{enumerate}
\item (i)
\end{enumerate}
$$\mathbf { A } = \left( \begin{array} { r r }
5 k & 3 k - 1 \\
- 3 & k + 1
\end{array} \right) , \text { where } k \text { is a real constant. }$$
Given that $\mathbf { A }$ is a singular matrix, find the possible values of $k$.\\
(ii)
$$\mathbf { B } = \left( \begin{array} { l l }
10 & 5 \\
- 3 & 3
\end{array} \right)$$
A triangle $T$ is transformed onto a triangle $T ^ { \prime }$ by the transformation represented by the matrix $\mathbf { B }$.
The vertices of triangle $T ^ { \prime }$ have coordinates $( 0,0 ) , ( - 20,6 )$ and $( 10 c , 6 c )$, where $c$ is a positive constant.
The area of triangle $T ^ { \prime }$ is 135 square units.\\
(a) Find the matrix $\mathbf { B } ^ { - 1 }$\\
(b) Find the coordinates of the vertices of the triangle $T$, in terms of $c$ where necessary.\\
(c) Find the value of $c$.\\
\hfill \mbox{\textit{Edexcel FP1 2015 Q7 [12]}}