| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find invariant lines through origin |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question requiring determinant calculation to show non-singularity, matrix multiplication to find k, and finding invariant lines via eigenvalues. While it involves several techniques, each step is relatively standard: part (a) is routine determinant expansion, part (b) involves straightforward matrix multiplication and equation solving, and part (c) is a textbook application of eigenvalue methods for invariant lines. The question is more computational than conceptually challenging, though the multi-step nature and Further Maths context place it slightly above average difficulty. |
| Spec | 4.03g Invariant points and lines4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k\begin{vmatrix}-1 & -1\\1 & -k\end{vmatrix} + 2\begin{vmatrix}0 & -1\\1 & 1\end{vmatrix} = 0 \Rightarrow k^2+k+2=0\) | M1 A1 | |
| \(1-4(2) = -7 < 0 \Rightarrow\) Non-singular | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}-3 & -1 & 1\\1 & 1 & 2\end{pmatrix}\begin{pmatrix}k & 0 & 2\\0 & -1 & -1\\1 & 1 & -k\end{pmatrix}\begin{pmatrix}0 & -3\\-1 & 3\\0 & 0\end{pmatrix} = \begin{pmatrix}-3 & -1 & 1\\1 & 1 & 2\end{pmatrix}\begin{pmatrix}0 & -3k\\1 & -3\\-1 & 0\end{pmatrix}\) | M1 A1 | |
| \(= \begin{pmatrix}-2 & 9k+3\\-1 & -3k-3\end{pmatrix} = \begin{pmatrix}-2 & -\frac{3}{2}\\-1 & -\frac{3}{2}\end{pmatrix} \Rightarrow k = -\frac{1}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}2 & \frac{3}{2}\\1 & \frac{3}{2}\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}2x+\frac{3}{2}y\\x+\frac{3}{2}y\end{pmatrix}\) | B1 | |
| \(x + \frac{3}{2}mx = m\left(2x+\frac{3}{2}mx\right)\) | M1 A1 | |
| \(1+\frac{3}{2}m = 2m+\frac{3}{2}m^2 \Rightarrow 3m^2+m-2=0\) | A1 | |
| \(y=-x\) and \(3y-2x=0\) | A1 |
## Question 4:
**Part 4(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k\begin{vmatrix}-1 & -1\\1 & -k\end{vmatrix} + 2\begin{vmatrix}0 & -1\\1 & 1\end{vmatrix} = 0 \Rightarrow k^2+k+2=0$ | M1 A1 | |
| $1-4(2) = -7 < 0 \Rightarrow$ Non-singular | A1 | |
**Part 4(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-3 & -1 & 1\\1 & 1 & 2\end{pmatrix}\begin{pmatrix}k & 0 & 2\\0 & -1 & -1\\1 & 1 & -k\end{pmatrix}\begin{pmatrix}0 & -3\\-1 & 3\\0 & 0\end{pmatrix} = \begin{pmatrix}-3 & -1 & 1\\1 & 1 & 2\end{pmatrix}\begin{pmatrix}0 & -3k\\1 & -3\\-1 & 0\end{pmatrix}$ | M1 A1 | |
| $= \begin{pmatrix}-2 & 9k+3\\-1 & -3k-3\end{pmatrix} = \begin{pmatrix}-2 & -\frac{3}{2}\\-1 & -\frac{3}{2}\end{pmatrix} \Rightarrow k = -\frac{1}{2}$ | A1 | |
**Part 4(c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}2 & \frac{3}{2}\\1 & \frac{3}{2}\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}2x+\frac{3}{2}y\\x+\frac{3}{2}y\end{pmatrix}$ | B1 | |
| $x + \frac{3}{2}mx = m\left(2x+\frac{3}{2}mx\right)$ | M1 A1 | |
| $1+\frac{3}{2}m = 2m+\frac{3}{2}m^2 \Rightarrow 3m^2+m-2=0$ | A1 | |
| $y=-x$ and $3y-2x=0$ | A1 | |
---4 The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { r r r }
k & 0 & 2 \\
0 & - 1 & - 1 \\
1 & 1 & - k
\end{array} \right)$$
where $k$ is a real constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathbf { A }$ is non-singular.\\
The matrices $\mathbf { B }$ and $\mathbf { C }$ are given by
$$\mathbf { B } = \left( \begin{array} { r r }
0 & - 3 \\
- 1 & 3 \\
0 & 0
\end{array} \right) \text { and } \mathbf { C } = \left( \begin{array} { r r r }
- 3 & - 1 & 1 \\
1 & 1 & 2
\end{array} \right)$$
It is given that $\mathbf { C A B } = \left( \begin{array} { l l } - 2 & - \frac { 3 } { 2 } \\ - 1 & - \frac { 3 } { 2 } \end{array} \right)$.
\item Find the value of $k$.
\item Find the equations of the invariant lines, through the origin, of the transformation in the $x - y$ plane represented by $\mathbf { C A B }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q4 [11]}}