| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find invariant lines through origin |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring matrix composition (rotation + stretch), determinant-area relationship, matrix inversion, and finding invariant lines via eigenvalues. While each component uses standard techniques, the combination of transformations, the area constraint proof, and the invariant line calculation (requiring eigenvalue analysis or direct algebraic approach) makes this moderately challenging for Further Maths students. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03g Invariant points and lines4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}\cos60 & -\sin60\\ \sin60 & \cos60\end{pmatrix} = \begin{pmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix}\) | B1 | Rotation 60° anticlockwise about the origin |
| \(\begin{pmatrix}d & 0\\ 0 & 1\end{pmatrix}\) | B1 | One-way stretch in the \(x\)-direction, scale factor \(d\) |
| \(\mathbf{M} = \begin{pmatrix}d & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix} = \begin{pmatrix}\frac{d}{2} & -\frac{d\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix}\) | M1 A1 | Correct order |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(d = \frac{1}{2}d^2\) | M1 | Uses value of \(\det\mathbf{M}\) |
| \(d \neq 0 \Rightarrow d = 2\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{M}^{-1} = \frac{1}{2}\begin{pmatrix}\frac{1}{2} & \sqrt{3}\\ -\frac{\sqrt{3}}{2} & 1\end{pmatrix}\) | B1 FT | Inverse of their \(\mathbf{M}\) |
| \(\mathbf{N} = \mathbf{M}^{-1}\begin{pmatrix}1 & 1\\ \frac{1}{2} & \frac{1}{2}\end{pmatrix} = \frac{1}{2}\begin{pmatrix}\frac{1}{2} & \sqrt{3}\\ -\frac{\sqrt{3}}{2} & 1\end{pmatrix}\begin{pmatrix}1 & 1\\ \frac{1}{2} & \frac{1}{2}\end{pmatrix}\) | M1 | Multiplies on the left by the inverse of their \(\mathbf{M}\) |
| \(= \frac{1}{4}\begin{pmatrix}1+\sqrt{3} & 1+\sqrt{3}\\ 1-\sqrt{3} & 1-\sqrt{3}\end{pmatrix}\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}1 & 1\\ \frac{1}{2} & \frac{1}{2}\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}x+y\\ \frac{1}{2}x+\frac{1}{2}y\end{pmatrix}\) | B1 | Transforms \(\begin{pmatrix}x\\ y\end{pmatrix}\) to \(\begin{pmatrix}X\\ Y\end{pmatrix}\) |
| \(\frac{1}{2}x + \frac{1}{2}mx = m(x+mx)\) | M1 A1 | Uses \(y = mx\) and \(Y = mX\) |
| \(1 + m = 2m + 2m^2 \Rightarrow 2m^2 + m - 1 = 0\) | A1 | |
| \(y = \frac{1}{2}x\) and \(y = -x\) | A1 | WWW |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}\cos60 & -\sin60\\ \sin60 & \cos60\end{pmatrix} = \begin{pmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix}$ | B1 | Rotation 60° anticlockwise about the origin |
| $\begin{pmatrix}d & 0\\ 0 & 1\end{pmatrix}$ | B1 | One-way stretch in the $x$-direction, scale factor $d$ |
| $\mathbf{M} = \begin{pmatrix}d & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix} = \begin{pmatrix}\frac{d}{2} & -\frac{d\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix}$ | M1 A1 | Correct order |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $d = \frac{1}{2}d^2$ | M1 | Uses value of $\det\mathbf{M}$ |
| $d \neq 0 \Rightarrow d = 2$ | A1 | AG |
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{M}^{-1} = \frac{1}{2}\begin{pmatrix}\frac{1}{2} & \sqrt{3}\\ -\frac{\sqrt{3}}{2} & 1\end{pmatrix}$ | B1 FT | Inverse of their $\mathbf{M}$ |
| $\mathbf{N} = \mathbf{M}^{-1}\begin{pmatrix}1 & 1\\ \frac{1}{2} & \frac{1}{2}\end{pmatrix} = \frac{1}{2}\begin{pmatrix}\frac{1}{2} & \sqrt{3}\\ -\frac{\sqrt{3}}{2} & 1\end{pmatrix}\begin{pmatrix}1 & 1\\ \frac{1}{2} & \frac{1}{2}\end{pmatrix}$ | M1 | Multiplies on the left by the inverse of their $\mathbf{M}$ |
| $= \frac{1}{4}\begin{pmatrix}1+\sqrt{3} & 1+\sqrt{3}\\ 1-\sqrt{3} & 1-\sqrt{3}\end{pmatrix}$ | A1 | CAO |
## Question 4(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}1 & 1\\ \frac{1}{2} & \frac{1}{2}\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}x+y\\ \frac{1}{2}x+\frac{1}{2}y\end{pmatrix}$ | B1 | Transforms $\begin{pmatrix}x\\ y\end{pmatrix}$ to $\begin{pmatrix}X\\ Y\end{pmatrix}$ |
| $\frac{1}{2}x + \frac{1}{2}mx = m(x+mx)$ | M1 A1 | Uses $y = mx$ and $Y = mX$ |
| $1 + m = 2m + 2m^2 \Rightarrow 2m^2 + m - 1 = 0$ | A1 | |
| $y = \frac{1}{2}x$ and $y = -x$ | A1 | WWW |4 The matrix $\mathbf { M }$ represents the sequence of two transformations in the $x - y$ plane given by a rotation of $60 ^ { \circ }$ anticlockwise about the origin followed by a one-way stretch in the $x$-direction, scale factor $d ( d \neq 0 )$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { M }$ in terms of $d$.
\item The unit square in the $x - y$ plane is transformed by $\mathbf { M }$ onto a parallelogram of area $\frac { 1 } { 2 } d ^ { 2 }$ units ${ } ^ { 2 }$. Show that $d = 2$.\\
The matrix $\mathbf { N }$ is such that $\mathbf { M N } = \left( \begin{array} { l l } 1 & 1 \\ \frac { 1 } { 2 } & \frac { 1 } { 2 } \end{array} \right)$.
\item Find $\mathbf { N }$.
\item Find the equations of the invariant lines, through the origin, of the transformation in the $x - y$ plane represented by $\mathbf { M N }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q4 [14]}}