| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find P and D for A² = PDP⁻¹ or A⁻¹ = PDP⁻¹ |
| Difficulty | Standard +0.8 This is a standard Further Maths eigenvalue question requiring characteristic equation verification, finding eigenvalues (including factoring a cubic), finding three eigenvectors for a 3×3 matrix, and constructing the diagonalization. The A^5 aspect adds minimal difficulty as D^5 is trivial. More computational work than typical A-level but follows a well-rehearsed algorithm without requiring novel insight. |
| Spec | 4.03a Matrix language: terminology and notation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{vmatrix} 18-\lambda & 5 & -11 \\ 8 & 6-\lambda & -4 \\ 32 & 10 & -20-\lambda \end{vmatrix} = 0\) | M1 | Sets determinant equal to zero |
| \((18-\lambda)\begin{vmatrix}6-\lambda & -4 \\ 10 & -20-\lambda\end{vmatrix} - 5\begin{vmatrix}8 & -4 \\ 32 & -20-\lambda\end{vmatrix} - 11\begin{vmatrix}8 & 6-\lambda \\ 32 & 10\end{vmatrix}\) | A1 | Expands determinant. Can use other rows/columns |
| \(\lambda^3 - 4\lambda^2 - 20\lambda + 48 = 0\) | A1 | AG |
| \((\lambda-2)(\lambda+4)(\lambda-6) = 0\) leading to \(\lambda = 2, -4, 6\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda=2\): \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 16 & 5 & -11 \\ 8 & 4 & -4\end{vmatrix} = \begin{pmatrix}24\\-24\\24\end{pmatrix} \sim \begin{pmatrix}1\\-1\\1\end{pmatrix}\) | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors. Must attempt to solve equations for M1 |
| \(\lambda=-4\): \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 8 & 10 & -4 \\ 32 & 10 & -16\end{vmatrix} = \begin{pmatrix}-120\\0\\-240\end{pmatrix} \sim \begin{pmatrix}1\\0\\2\end{pmatrix}\) | A1 A1 | |
| \(\lambda=6\): \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 12 & 5 & -11 \\ 8 & 0 & -4\end{vmatrix} = \begin{pmatrix}-20\\-40\\-40\end{pmatrix} \sim \begin{pmatrix}1\\2\\2\end{pmatrix}\) | ||
| Thus \(\mathbf{P} = \begin{pmatrix}1&1&1\\-1&0&2\\1&2&2\end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix}32&0&0\\0&-1024&0\\0&0&7776\end{pmatrix}\) | M1 A1 | Or correctly matched permutations of columns. M0 if a column of zeros appears in P |
## Question 5(a):
| $\begin{vmatrix} 18-\lambda & 5 & -11 \\ 8 & 6-\lambda & -4 \\ 32 & 10 & -20-\lambda \end{vmatrix} = 0$ | M1 | Sets determinant equal to zero |
|---|---|---|
| $(18-\lambda)\begin{vmatrix}6-\lambda & -4 \\ 10 & -20-\lambda\end{vmatrix} - 5\begin{vmatrix}8 & -4 \\ 32 & -20-\lambda\end{vmatrix} - 11\begin{vmatrix}8 & 6-\lambda \\ 32 & 10\end{vmatrix}$ | A1 | Expands determinant. Can use other rows/columns |
| $\lambda^3 - 4\lambda^2 - 20\lambda + 48 = 0$ | A1 | AG |
| $(\lambda-2)(\lambda+4)(\lambda-6) = 0$ leading to $\lambda = 2, -4, 6$ | B1 | |
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## Question 5(b):
| $\lambda=2$: $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 16 & 5 & -11 \\ 8 & 4 & -4\end{vmatrix} = \begin{pmatrix}24\\-24\\24\end{pmatrix} \sim \begin{pmatrix}1\\-1\\1\end{pmatrix}$ | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors. Must attempt to solve equations for M1 |
|---|---|---|
| $\lambda=-4$: $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 8 & 10 & -4 \\ 32 & 10 & -16\end{vmatrix} = \begin{pmatrix}-120\\0\\-240\end{pmatrix} \sim \begin{pmatrix}1\\0\\2\end{pmatrix}$ | A1 A1 | |
| $\lambda=6$: $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 12 & 5 & -11 \\ 8 & 0 & -4\end{vmatrix} = \begin{pmatrix}-20\\-40\\-40\end{pmatrix} \sim \begin{pmatrix}1\\2\\2\end{pmatrix}$ | | |
| Thus $\mathbf{P} = \begin{pmatrix}1&1&1\\-1&0&2\\1&2&2\end{pmatrix}$ and $\mathbf{D} = \begin{pmatrix}32&0&0\\0&-1024&0\\0&0&7776\end{pmatrix}$ | M1 A1 | Or correctly matched permutations of columns. M0 if a column of zeros appears in **P** |5 The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { r r r }
18 & 5 & - 11 \\
8 & 6 & - 4 \\
32 & 10 & - 20
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Show that the characteristic equation of $\mathbf { A }$ is $\lambda ^ { 3 } - 4 \lambda ^ { 2 } - 20 \lambda + 48 = 0$ and hence find the eigenvalues of $\mathbf { A }$.
\item Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } ^ { 5 } = \mathbf { P D P } ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q5 [10]}}