| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | November |
| Marks | 11 |
| 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 question requires understanding of composite transformations, inverse matrices, and the relationship between eigenvalues and invariant lines. Part (c) is particularly demanding as students must recognize that invariant lines exist when eigenvalues are real, requiring them to solve a characteristic equation involving surds and interpret the discriminant condition. This goes beyond routine matrix manipulation to require conceptual understanding of the geometric meaning of complex eigenvalues. |
| Spec | 4.03a Matrix language: terminology and notation4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03o Inverse 3x3 matrix4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Shear (in the \(x\)-direction) followed by a rotation (anticlockwise about the origin through 45°) | B2 | Award B1 if given in the wrong order |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{M}^{-1} = \dfrac{1}{1}\begin{pmatrix}\frac{k+1}{\sqrt{2}} & -\frac{k-1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix} = \begin{pmatrix}\frac{k+1}{\sqrt{2}} & -\frac{k-1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix}\) | M1 A1 | Finds \(\mathbf{M}^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{M} = \begin{pmatrix}\frac{1}{2}\sqrt{2} & \frac{1}{2}(k-1)\sqrt{2} \\ \frac{1}{2}\sqrt{2} & \frac{1}{2}(k+1)\sqrt{2}\end{pmatrix}\) | B1 | |
| \(\frac{1}{2}\sqrt{2}\begin{pmatrix}1 & k-1\\1 & k+1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \frac{1}{2}\sqrt{2}\begin{pmatrix}x+(k-1)y\\x+(k+1)y\end{pmatrix}\) | B1 | Transforms \(\begin{pmatrix}x\\y\end{pmatrix}\) to \(\begin{pmatrix}X\\Y\end{pmatrix}\) |
| \(x + m(k+1)x = m(x + m(k-1)x)\) | M1 A1 | Uses \(y = mx\) and \(Y = mX\) |
| \((k-1)m^2 - km - 1 = 0\) | A1 | |
| \(k^2 + 4k - 4 < 0\) | M1 | Sets discriminant negative |
| \(2(-1-\sqrt{2}) < k < 2(-1+\sqrt{2})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{M} = \begin{pmatrix}\frac{1}{2}\sqrt{2} & \frac{1}{2}(k-1)\sqrt{2} \\ \frac{1}{2}\sqrt{2} & \frac{1}{2}(k+1)\sqrt{2}\end{pmatrix}\) | B1 | |
| \([\mathbf{M} - \lambda\mathbf{I}=]\begin{vmatrix}\frac{1}{2}\sqrt{2}-\lambda & \frac{1}{2}(k-1)\sqrt{2} \\ \frac{1}{2}\sqrt{2} & \frac{1}{2}(k+1)\sqrt{2}-\lambda\end{vmatrix} = 0\) | B1 | Sets up to find characteristic equation |
| \(\left(\frac{1}{2}\sqrt{2}-\lambda\right)\left(\frac{1}{2}(k+1)\sqrt{2}-\lambda\right) - \frac{1}{2}(k-1)\sqrt{2}\cdot\frac{1}{2}\sqrt{2} = 0\) | M1 A1 | Evaluates determinant |
| \(2\lambda^2 - \sqrt{2}(k+2)\lambda + 2 = 0\) | A1 | |
| \(k^2 + 4k - 4 < 0\) | M1 | Sets discriminant negative |
| \(2(-1-\sqrt{2}) < k < 2(-1+\sqrt{2})\) | A1 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Shear (in the $x$-direction) followed by a rotation (anticlockwise about the origin through 45°) | B2 | Award B1 if given in the wrong order |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{M}^{-1} = \dfrac{1}{1}\begin{pmatrix}\frac{k+1}{\sqrt{2}} & -\frac{k-1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix} = \begin{pmatrix}\frac{k+1}{\sqrt{2}} & -\frac{k-1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix}$ | M1 A1 | Finds $\mathbf{M}^{-1}$ |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{M} = \begin{pmatrix}\frac{1}{2}\sqrt{2} & \frac{1}{2}(k-1)\sqrt{2} \\ \frac{1}{2}\sqrt{2} & \frac{1}{2}(k+1)\sqrt{2}\end{pmatrix}$ | B1 | |
| $\frac{1}{2}\sqrt{2}\begin{pmatrix}1 & k-1\\1 & k+1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \frac{1}{2}\sqrt{2}\begin{pmatrix}x+(k-1)y\\x+(k+1)y\end{pmatrix}$ | B1 | Transforms $\begin{pmatrix}x\\y\end{pmatrix}$ to $\begin{pmatrix}X\\Y\end{pmatrix}$ |
| $x + m(k+1)x = m(x + m(k-1)x)$ | M1 A1 | Uses $y = mx$ and $Y = mX$ |
| $(k-1)m^2 - km - 1 = 0$ | A1 | |
| $k^2 + 4k - 4 < 0$ | M1 | Sets discriminant negative |
| $2(-1-\sqrt{2}) < k < 2(-1+\sqrt{2})$ | A1 | |
**Alternative method for 5(c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{M} = \begin{pmatrix}\frac{1}{2}\sqrt{2} & \frac{1}{2}(k-1)\sqrt{2} \\ \frac{1}{2}\sqrt{2} & \frac{1}{2}(k+1)\sqrt{2}\end{pmatrix}$ | B1 | |
| $[\mathbf{M} - \lambda\mathbf{I}=]\begin{vmatrix}\frac{1}{2}\sqrt{2}-\lambda & \frac{1}{2}(k-1)\sqrt{2} \\ \frac{1}{2}\sqrt{2} & \frac{1}{2}(k+1)\sqrt{2}-\lambda\end{vmatrix} = 0$ | B1 | Sets up to find characteristic equation |
| $\left(\frac{1}{2}\sqrt{2}-\lambda\right)\left(\frac{1}{2}(k+1)\sqrt{2}-\lambda\right) - \frac{1}{2}(k-1)\sqrt{2}\cdot\frac{1}{2}\sqrt{2} = 0$ | M1 A1 | Evaluates determinant |
| $2\lambda^2 - \sqrt{2}(k+2)\lambda + 2 = 0$ | A1 | |
| $k^2 + 4k - 4 < 0$ | M1 | Sets discriminant negative |
| $2(-1-\sqrt{2}) < k < 2(-1+\sqrt{2})$ | A1 | |5 The matrix $\mathbf { M }$ is given by $\mathbf { M } = \left( \begin{array} { r r } \frac { 1 } { 2 } \sqrt { 2 } & - \frac { 1 } { 2 } \sqrt { 2 } \\ \frac { 1 } { 2 } \sqrt { 2 } & \frac { 1 } { 2 } \sqrt { 2 } \end{array} \right) \left( \begin{array} { c c } 1 & k \\ 0 & 1 \end{array} \right)$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item The matrix $\mathbf { M }$ represents a sequence of two geometrical transformations.
State the type of each transformation, and make clear the order in which they are applied.
\item The triangle $A B C$ in the $x - y$ plane is transformed by $\mathbf { M }$ onto triangle $D E F$.
Find, in terms of $k$, the single matrix which transforms triangle $D E F$ onto triangle $A B C$.
\item Find the set of values of $k$ for which the transformation represented by $\mathbf { M }$ has no invariant lines through the origin.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q5 [11]}}