| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 10 |
| 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 +0.8 This is a Further Maths diagonalization problem with a 3×3 upper triangular matrix. While eigenvalues are easily read from the diagonal, finding three linearly independent eigenvectors (especially for the repeated eigenvalue when m=2 is excluded) and computing P⁻¹ requires careful algebraic manipulation. Part (ii) adds a computational layer using the diagonalization. This is moderately challenging for Further Maths, above average difficulty overall. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Eigenvalues of upper diagonal matrix \(\mathbf{A}\) are \(2, m\) and \(1\). (Or from characteristic equation: \((\lambda-2)(\lambda-m)(\lambda-1)=0\)) | B1 | |
| \(\lambda=2\): \(\mathbf{e}_1 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & m-2 & 1 \\ 0 & 0 & -1 \end{vmatrix} = \begin{pmatrix}2-m\\0\\0\end{pmatrix} = t\begin{pmatrix}1\\0\\0\end{pmatrix}\) | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| \(\lambda=m\): \(\mathbf{e}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2-m & m & 1 \\ 0 & 0 & 7 \end{vmatrix} = \begin{pmatrix}7m\\7(m-2)\\0\end{pmatrix} = t\begin{pmatrix}m\\m-2\\0\end{pmatrix}\) | A1 | |
| \(\lambda=1\): \(\mathbf{e}_3 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & m & 1 \\ 0 & m-1 & 7 \end{vmatrix} = \begin{pmatrix}6m+1\\-7\\m-1\end{pmatrix} = t\begin{pmatrix}6m+1\\-7\\m-1\end{pmatrix}\) | A1 | |
| Thus \(\mathbf{P} = \begin{pmatrix}1 & m & 6m+1\\0 & m-2 & -7\\0 & 0 & m-1\end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix}2 & 0 & 0\\0 & m & 0\\0 & 0 & 1\end{pmatrix}\) | M1 A1 FT | Or correctly matched permutations of columns. No follow through on two or more zero eigenvectors |
| 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{M}^7\mathbf{P} = \mathbf{P}\mathbf{D}^7\mathbf{P}^{-1}\mathbf{P} = \mathbf{P}\mathbf{D}^7 = \begin{pmatrix}1 & m & 6m+1\\0 & m-2 & -7\\0 & 0 & m-1\end{pmatrix}\begin{pmatrix}2^7 & 0 & 0\\0 & m^7 & 0\\0 & 0 & 1\end{pmatrix}\) | M1 A1 FT | Applies \(\mathbf{M}^7 = \mathbf{P}\mathbf{D}^7\mathbf{P}^{-1}\) |
| \(= \begin{pmatrix}2^7 & m^8 & 6m+1\\0 & m^8-2m^7 & -7\\0 & 0 & m-1\end{pmatrix}\) | A1 | Order of columns might be swapped depending on \(\mathbf{P}\) |
| 3 |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Eigenvalues of upper diagonal matrix $\mathbf{A}$ are $2, m$ and $1$. (Or from characteristic equation: $(\lambda-2)(\lambda-m)(\lambda-1)=0$) | B1 | |
| $\lambda=2$: $\mathbf{e}_1 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & m-2 & 1 \\ 0 & 0 & -1 \end{vmatrix} = \begin{pmatrix}2-m\\0\\0\end{pmatrix} = t\begin{pmatrix}1\\0\\0\end{pmatrix}$ | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| $\lambda=m$: $\mathbf{e}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2-m & m & 1 \\ 0 & 0 & 7 \end{vmatrix} = \begin{pmatrix}7m\\7(m-2)\\0\end{pmatrix} = t\begin{pmatrix}m\\m-2\\0\end{pmatrix}$ | A1 | |
| $\lambda=1$: $\mathbf{e}_3 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & m & 1 \\ 0 & m-1 & 7 \end{vmatrix} = \begin{pmatrix}6m+1\\-7\\m-1\end{pmatrix} = t\begin{pmatrix}6m+1\\-7\\m-1\end{pmatrix}$ | A1 | |
| Thus $\mathbf{P} = \begin{pmatrix}1 & m & 6m+1\\0 & m-2 & -7\\0 & 0 & m-1\end{pmatrix}$ and $\mathbf{D} = \begin{pmatrix}2 & 0 & 0\\0 & m & 0\\0 & 0 & 1\end{pmatrix}$ | M1 A1 FT | Or correctly matched permutations of columns. No follow through on two or more zero eigenvectors |
| | **7** | |
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{M}^7\mathbf{P} = \mathbf{P}\mathbf{D}^7\mathbf{P}^{-1}\mathbf{P} = \mathbf{P}\mathbf{D}^7 = \begin{pmatrix}1 & m & 6m+1\\0 & m-2 & -7\\0 & 0 & m-1\end{pmatrix}\begin{pmatrix}2^7 & 0 & 0\\0 & m^7 & 0\\0 & 0 & 1\end{pmatrix}$ | M1 A1 FT | Applies $\mathbf{M}^7 = \mathbf{P}\mathbf{D}^7\mathbf{P}^{-1}$ |
| $= \begin{pmatrix}2^7 & m^8 & 6m+1\\0 & m^8-2m^7 & -7\\0 & 0 & m-1\end{pmatrix}$ | A1 | Order of columns might be swapped depending on $\mathbf{P}$ |
| | **3** | |8 The matrix $\mathbf { M }$ is defined by
$$\mathbf { M } = \left( \begin{array} { c c c }
2 & m & 1 \\
0 & m & 7 \\
0 & 0 & 1
\end{array} \right) ,$$
where $m \neq 0,1,2$.\\
(i) Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { M } = \mathbf { P D P } ^ { - 1 }$.\\
(ii) Find $\mathbf { M } ^ { 7 } \mathbf { P }$.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q8 [10]}}