| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | November |
| Marks | 11 |
| 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 | Challenging +1.2 This is a Further Maths question on diagonalization with an upper triangular matrix (eigenvalues immediately visible on diagonal: 2, 5, -2). Part (a) requires finding eigenvectors and constructing P and D for A⁵ = PDP⁻¹, which is routine for Further Maths students. Part (b) uses the Cayley-Hamilton theorem to express A⁴ in terms of lower powers, which is a standard technique. The upper triangular structure simplifies the eigenvalue work significantly, making this easier than a general 3×3 diagonalization problem. Moderately above average difficulty due to the 3×3 size and two-part structure, but well within standard Further Maths scope. |
| Spec | 4.03j Determinant 3x3: calculation4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Eigenvalues of A are 2, 5 and \(-2\) | B1 | Lower diagonal matrix or characteristic equation |
| \(\lambda = 2\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 3 & 7 \\ 0 & 0 & -4 \end{vmatrix} = \begin{pmatrix}-12\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}\) | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| \(\lambda = 5\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & -3 & -7 \\ 0 & 0 & 7 \end{vmatrix} = \begin{pmatrix}-21\\21\\0\end{pmatrix} \sim \begin{pmatrix}1\\-1\\0\end{pmatrix}\) | A1 | |
| \(\lambda = -2\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -3 & -7 \\ 0 & 7 & 7 \end{vmatrix} = \begin{pmatrix}28\\-28\\28\end{pmatrix} \sim \begin{pmatrix}1\\-1\\1\end{pmatrix}\) | A1 | |
| \(\mathbf{P} = \begin{pmatrix}1&1&1\\0&-1&-1\\0&0&1\end{pmatrix}\), \(\mathbf{D} = \begin{pmatrix}32&0&0\\0&3125&0\\0&0&-32\end{pmatrix}\) | M1 A1 | Or correctly matched permutations of columns. Eigenvectors must be non-zero and correctly matched to eigenvalues raised to fifth power for M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\lambda-2)(\lambda-5)(\lambda+2) = \lambda^3 - 5\lambda^2 - 4\lambda + 20 = 0\) | B1 | Characteristic equation |
| \(\mathbf{A}^3 = 5\mathbf{A}^2 + 4\mathbf{A} - 20\mathbf{I}\) | M1 | Substitutes for A and makes \(\mathbf{A}^3\) the subject. Tolerate I missing |
| \(\mathbf{A}^4 = 5\mathbf{A}^3 + 4\mathbf{A}^2 - 20\mathbf{A} = 5(5\mathbf{A}^2 + 4\mathbf{A} - 20\mathbf{I}) + 4\mathbf{A}^2 - 20\mathbf{A}\) | M1 | Multiplies by A and substitutes for \(\mathbf{A}^3\) |
| \(\mathbf{A}^4 = 29\mathbf{A}^2 - 100\mathbf{I}\) | A1 | Has I in answer or states values of \(a\) and \(b\) |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Eigenvalues of **A** are 2, 5 and $-2$ | B1 | Lower diagonal matrix or characteristic equation |
| $\lambda = 2$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 3 & 7 \\ 0 & 0 & -4 \end{vmatrix} = \begin{pmatrix}-12\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}$ | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| $\lambda = 5$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & -3 & -7 \\ 0 & 0 & 7 \end{vmatrix} = \begin{pmatrix}-21\\21\\0\end{pmatrix} \sim \begin{pmatrix}1\\-1\\0\end{pmatrix}$ | A1 | |
| $\lambda = -2$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -3 & -7 \\ 0 & 7 & 7 \end{vmatrix} = \begin{pmatrix}28\\-28\\28\end{pmatrix} \sim \begin{pmatrix}1\\-1\\1\end{pmatrix}$ | A1 | |
| $\mathbf{P} = \begin{pmatrix}1&1&1\\0&-1&-1\\0&0&1\end{pmatrix}$, $\mathbf{D} = \begin{pmatrix}32&0&0\\0&3125&0\\0&0&-32\end{pmatrix}$ | M1 A1 | Or correctly matched permutations of columns. Eigenvectors must be non-zero and correctly matched to eigenvalues raised to fifth power for M1 |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\lambda-2)(\lambda-5)(\lambda+2) = \lambda^3 - 5\lambda^2 - 4\lambda + 20 = 0$ | B1 | Characteristic equation |
| $\mathbf{A}^3 = 5\mathbf{A}^2 + 4\mathbf{A} - 20\mathbf{I}$ | M1 | Substitutes for **A** and makes $\mathbf{A}^3$ the subject. Tolerate **I** missing |
| $\mathbf{A}^4 = 5\mathbf{A}^3 + 4\mathbf{A}^2 - 20\mathbf{A} = 5(5\mathbf{A}^2 + 4\mathbf{A} - 20\mathbf{I}) + 4\mathbf{A}^2 - 20\mathbf{A}$ | M1 | Multiplies by **A** and substitutes for $\mathbf{A}^3$ |
| $\mathbf{A}^4 = 29\mathbf{A}^2 - 100\mathbf{I}$ | A1 | Has **I** in answer or states values of $a$ and $b$ |
---6 The matrix $\mathbf { A }$ is given by
$$A = \left( \begin{array} { r r r }
2 & - 3 & - 7 \\
0 & 5 & 7 \\
0 & 0 & - 2
\end{array} \right) .$$
\begin{enumerate}[label=(\alph*)]
\item Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } ^ { 5 } = \mathbf { P D P } ^ { - 1 }$.
\item Use the characteristic equation of $\mathbf { A }$ to show that
$$\mathbf { A } ^ { 4 } = a \mathbf { A } ^ { 2 } + b \mathbf { I } ,$$
where $a$ and $b$ are integers to be determined.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q6 [11]}}