| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find line of invariant points |
| Difficulty | Standard +0.8 This question requires understanding of eigenvalues (determinant = 5 for area scaling), eigenvectors (line of invariant points means eigenvalue 1), and constructing a matrix from multiple constraints. Part (i) involves solving a system with underdetermined variables requiring insight about eigenvalue 1. Parts (ii-iii) are more routine eigenvalue/invariant line work, but the overall question demands sophisticated understanding of transformation properties beyond standard textbook exercises. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03g Invariant points and lines4.03i Determinant: area scale factor and orientation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\) | B1 (AO 3.1a) | \(\Rightarrow b=3, d=4\) |
| Determinant \(= ad - bc = 5\) | B1 (AO 3.1a) | \(4a - 3c = 5\). Or det \(= -5\) and follow through |
| \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\) | B1 (AO 1.2) | Understanding of invariant point seen or implied |
| \((1-a)x = by\) or \((1-d)y = cx\) | M1 (AO 2.2a) | May have \(b=3\) and/or \(d=4\) already substituted. \((1-a)x = 3y\) or \(-3y = cx\) |
| \(\frac{1-a}{c} = \frac{b}{1-d}\) or \(\frac{1-a}{b} = \frac{c}{1-d}\) | M1 (AO 1.1) | Eliminating \(x\) and \(y\) |
| \(c = a-1\) | A1 (AO 1.1) | |
| e.g. \(4a - 3(a-1) = 5\) | M1 (AO 1.1) | Attempting to solve their simultaneous equations. If no working and incorrect then M0A0. |
| \(\mathbf{A} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}\) | A1 (AO 3.2a) | Condone \(a=2\), etc as long as Matrix seen as \(\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Need \(\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3x+y \\ 2x+2y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\) | M1 (AO 1.1) | Substituting a general point into their matrix, calculating an image point and equating it to the object point. |
| \(2x + y = 0\) | A1 (AO 2.2a) | Final form must be \(y = -2x\) or \(x = -\frac{1}{2}y\) or a numerical multiple of \(2x+y=0\). Need to have considered both \(x\) and \(y\) coordinates. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x \\ x+c \end{pmatrix} = \begin{pmatrix} 3x+x+c \\ 2x+2x+2c \end{pmatrix}\) | M1* (AO 3.1a) | |
| So \(\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 4x+c \\ 4x+2c \end{pmatrix}\) ... | M1dep* (AO 2.2a) | |
| ...and \(Y = X + c\) | A1 (AO 1.1) | Could see the \(y\) component of the vector written as \(4x+c+c\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x \\ mx+c \end{pmatrix} = \begin{pmatrix} X \\ mX+c \end{pmatrix}\) | M1 | Need to eliminate \(X\), i.e. an equation in \(x\), \(m\) and \(c\). |
| \(3x+mx+c = X\) and \(2x+2mx+2c = mX+c\) leading to \(2x+2mx+2c = m(3x+mx+c)+c\) | ||
| \(x(m^2+m-2)+c(m-1)=0\) and \(x(m+2)(m-1)+c(m-1)=0\) | M1 | |
| If \(m=1\), \(c(m-1)=0\) satisfied by any \(c\) | A1 |
# Question 8:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ | B1 (AO 3.1a) | $\Rightarrow b=3, d=4$ |
| Determinant $= ad - bc = 5$ | B1 (AO 3.1a) | $4a - 3c = 5$. Or det $= -5$ and follow through |
| $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$ | B1 (AO 1.2) | Understanding of invariant point seen or implied |
| $(1-a)x = by$ or $(1-d)y = cx$ | M1 (AO 2.2a) | May have $b=3$ and/or $d=4$ already substituted. $(1-a)x = 3y$ or $-3y = cx$ |
| $\frac{1-a}{c} = \frac{b}{1-d}$ or $\frac{1-a}{b} = \frac{c}{1-d}$ | M1 (AO 1.1) | Eliminating $x$ and $y$ |
| $c = a-1$ | A1 (AO 1.1) | |
| e.g. $4a - 3(a-1) = 5$ | M1 (AO 1.1) | Attempting to solve their simultaneous equations. If no working and incorrect then M0A0. |
| $\mathbf{A} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ | A1 (AO 3.2a) | Condone $a=2$, etc as long as Matrix seen as $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ |
**[8 marks]**
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Need $\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3x+y \\ 2x+2y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$ | M1 (AO 1.1) | Substituting a general point into their matrix, calculating an image point and equating it to the object point. |
| $2x + y = 0$ | A1 (AO 2.2a) | Final form must be $y = -2x$ or $x = -\frac{1}{2}y$ or a numerical multiple of $2x+y=0$. Need to have considered both $x$ and $y$ coordinates. |
**[2 marks]**
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x \\ x+c \end{pmatrix} = \begin{pmatrix} 3x+x+c \\ 2x+2x+2c \end{pmatrix}$ | M1* (AO 3.1a) | |
| So $\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 4x+c \\ 4x+2c \end{pmatrix}$ ... | M1dep* (AO 2.2a) | |
| ...and $Y = X + c$ | A1 (AO 1.1) | Could see the $y$ component of the vector written as $4x+c+c$. |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} x \\ mx+c \end{pmatrix} = \begin{pmatrix} X \\ mX+c \end{pmatrix}$ | M1 | Need to eliminate $X$, i.e. an equation in $x$, $m$ and $c$. |
| $3x+mx+c = X$ and $2x+2mx+2c = mX+c$ leading to $2x+2mx+2c = m(3x+mx+c)+c$ | | |
| $x(m^2+m-2)+c(m-1)=0$ and $x(m+2)(m-1)+c(m-1)=0$ | M1 | |
| If $m=1$, $c(m-1)=0$ satisfied by any $c$ | A1 | |
**[3 marks]**8 The $2 \times 2$ matrix A represents a transformation T which has the following properties.
\begin{itemize}
\item The image of the point $( 0,1 )$ is the point $( 3,4 )$.
\item An object shape whose area is 7 is transformed to an image shape whose area is 35 .
\item T has a line of invariant points.\\
(i) Find a possible matrix for $\mathbf { A }$.
\end{itemize}
The transformation S is represented by the matrix $\mathbf { B }$ where $\mathbf { B } = \left( \begin{array} { l l } 3 & 1 \\ 2 & 2 \end{array} \right)$.\\
(ii) Find the equation of the line of invariant points of S .\\
(iii) Show that any line of the form $y = x + c$ is an invariant line of S .
\hfill \mbox{\textit{OCR Further Pure Core AS 2018 Q8 [13]}}