| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find P and D for A = PDP⁻¹ |
| Difficulty | Standard +0.3 This is a standard Further Maths linear algebra question requiring eigenvalue/eigenvector calculation (routine characteristic equation with fractional entries), recognition that perpendicular eigenvectors indicate orthogonal invariant lines, and identification of a stretch transformation. While it involves multiple steps and Further Maths content, the techniques are algorithmic and well-practiced, making it slightly easier than average for Further Maths students. |
| Spec | 4.03a Matrix language: terminology and notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = \left(\frac{5}{2}-\lambda\right)\left(\frac{13}{2}-\lambda\right) - \frac{9}{4}\); \(0 = \lambda^2 - 9\lambda + 14\) | M1 | Forms correct characteristic equation and solves. Condone one error |
| \(\lambda = 2\) and \(\lambda = 7\) | A1 | Correct eigenvalues |
| \(\lambda = 2\): \(\mathbf{0} = \begin{bmatrix}\frac{1}{2} & \frac{-3}{2}\\ \frac{-3}{2} & \frac{9}{2}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\), eigenvector \(\begin{bmatrix}3\\1\end{bmatrix}\) | M1 | Uses correct equation to find eigenvector for one of their two eigenvalues |
| \(\lambda = 7\): \(\mathbf{0} = \begin{bmatrix}\frac{-9}{2} & \frac{-3}{2}\\ \frac{-3}{2} & \frac{-1}{2}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\), eigenvector \(\begin{bmatrix}-1\\3\end{bmatrix}\) | A1F | Correct eigenvector for one eigenvalue. Allow any scalar multiple |
| Both eigenvectors correctly paired with corresponding eigenvalue | A1 | Allow any scalar multiple |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{bmatrix}3\\1\end{bmatrix}\cdot\begin{bmatrix}-1\\3\end{bmatrix} = 0\), so invariant lines are perpendicular | M1 | Compares directions of eigenvectors, e.g. considers gradients or obtains scalar product |
| Deduces invariant lines are perpendicular | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| It is a two-way stretch: stretch parallel to \(y = \frac{1}{3}x\), SF = 2; stretch parallel to \(y = -3x\), SF = 7 | M1 | Deduces it is a two-way stretch |
| Full description following through their eigenvalues and corresponding eigenvectors | A1F | Describes the transformation fully |
## Question 11(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = \left(\frac{5}{2}-\lambda\right)\left(\frac{13}{2}-\lambda\right) - \frac{9}{4}$; $0 = \lambda^2 - 9\lambda + 14$ | M1 | Forms correct characteristic equation and solves. Condone one error |
| $\lambda = 2$ and $\lambda = 7$ | A1 | Correct eigenvalues |
| $\lambda = 2$: $\mathbf{0} = \begin{bmatrix}\frac{1}{2} & \frac{-3}{2}\\ \frac{-3}{2} & \frac{9}{2}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$, eigenvector $\begin{bmatrix}3\\1\end{bmatrix}$ | M1 | Uses correct equation to find eigenvector for one of their two eigenvalues |
| $\lambda = 7$: $\mathbf{0} = \begin{bmatrix}\frac{-9}{2} & \frac{-3}{2}\\ \frac{-3}{2} & \frac{-1}{2}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$, eigenvector $\begin{bmatrix}-1\\3\end{bmatrix}$ | A1F | Correct eigenvector for one eigenvalue. Allow any scalar multiple |
| Both eigenvectors correctly paired with corresponding eigenvalue | A1 | Allow any scalar multiple |
---
## Question 11(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{bmatrix}3\\1\end{bmatrix}\cdot\begin{bmatrix}-1\\3\end{bmatrix} = 0$, so invariant lines are perpendicular | M1 | Compares directions of eigenvectors, e.g. considers gradients or obtains scalar product |
| Deduces invariant lines are perpendicular | R1 | |
---
## Question 11(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| It is a two-way stretch: stretch parallel to $y = \frac{1}{3}x$, SF = 2; stretch parallel to $y = -3x$, SF = 7 | M1 | Deduces it is a two-way stretch |
| Full description following through their eigenvalues and corresponding eigenvectors | A1F | Describes the transformation fully |
---11
\begin{enumerate}[label=(\alph*)]
\item Find the eigenvalues and corresponding eigenvectors of the matrix
$$\mathbf { M } = \left[ \begin{array} { c c }
\frac { 5 } { 2 } & - \frac { 3 } { 2 } \\
- \frac { 3 } { 2 } & \frac { 13 } { 2 }
\end{array} \right]$$
11
\item (i) Describe how the directions of the invariant lines of the transformation represented by $\mathbf { M }$ are related to each other.
Fully justify your answer.\\[0pt]
[2 marks]\\
11 (b) (ii) Describe fully the transformation represented by $\mathbf { M }$
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2022 Q11 [9]}}