| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | November |
| 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 +0.8 This is a multi-part Further Maths question requiring understanding of eigenvalues from triangular matrices, using Cayley-Hamilton theorem to find inverse, and constructing A = PDP⁻¹. Part (a) is routine, part (b) requires non-standard technique (Cayley-Hamilton), and part (c) requires careful matrix multiplication with the diagonalization formula. The combination of techniques and computational demand places this above average difficulty. |
| Spec | 4.03a Matrix language: terminology and notation4.03h Determinant 2x2: calculation4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1, -1, 2\) | B1 | |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{P}^3 - 2\mathbf{P}^2 - \mathbf{P} + 2\mathbf{I} = 0\) | B1 | States that P satisfies its characteristic equation |
| \(2\mathbf{P}^{-1} = \mathbf{I} + 2\mathbf{P} - \mathbf{P}^2\) | M1 | Multiplies through by \(\mathbf{P}^{-1}\) |
| \(\mathbf{P}^2 = \begin{pmatrix} 1 & 0 & 10 \\ 0 & 1 & 1 \\ 0 & 0 & 4 \end{pmatrix} \Rightarrow \mathbf{P}^{-1} = \begin{pmatrix} 1 & 4 & -3 \\ 0 & -1 & \frac{1}{2} \\ 0 & 0 & \frac{1}{2} \end{pmatrix}\) | M1 A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{A} = \mathbf{P}\begin{pmatrix} b & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\mathbf{P}^{-1}\) | M1 | Applies \(\mathbf{A} = \mathbf{PDP}^{-1}\) |
| \(= \begin{pmatrix} b & -4 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & 4 & -3 \\ 0 & -1 & \frac{1}{2} \\ 0 & 0 & \frac{1}{2} \end{pmatrix}\) | M1 A1 | Multiplies two adjacent matrices |
| \(\begin{pmatrix} b & 4b+4 & -3b-1 \\ 0 & -1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\) | A1 | |
| 4 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1, -1, 2$ | **B1** | |
| | **1** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{P}^3 - 2\mathbf{P}^2 - \mathbf{P} + 2\mathbf{I} = 0$ | **B1** | States that **P** satisfies its characteristic equation |
| $2\mathbf{P}^{-1} = \mathbf{I} + 2\mathbf{P} - \mathbf{P}^2$ | **M1** | Multiplies through by $\mathbf{P}^{-1}$ |
| $\mathbf{P}^2 = \begin{pmatrix} 1 & 0 & 10 \\ 0 & 1 & 1 \\ 0 & 0 & 4 \end{pmatrix} \Rightarrow \mathbf{P}^{-1} = \begin{pmatrix} 1 & 4 & -3 \\ 0 & -1 & \frac{1}{2} \\ 0 & 0 & \frac{1}{2} \end{pmatrix}$ | **M1 A1** | |
| | **4** | |
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A} = \mathbf{P}\begin{pmatrix} b & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\mathbf{P}^{-1}$ | **M1** | Applies $\mathbf{A} = \mathbf{PDP}^{-1}$ |
| $= \begin{pmatrix} b & -4 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & 4 & -3 \\ 0 & -1 & \frac{1}{2} \\ 0 & 0 & \frac{1}{2} \end{pmatrix}$ | **M1 A1** | Multiplies two adjacent matrices |
| $\begin{pmatrix} b & 4b+4 & -3b-1 \\ 0 & -1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ | **A1** | |
| | **4** | |7 The matrix $\mathbf { P }$ is given by
$$\mathbf { P } = \left( \begin{array} { r r r }
1 & 4 & 2 \\
0 & - 1 & 1 \\
0 & 0 & 2
\end{array} \right) .$$
\begin{enumerate}[label=(\alph*)]
\item State the eigenvalues of $\mathbf { P }$.
\item Use the characteristic equation of $\mathbf { P }$ to find $\mathbf { P } ^ { - 1 }$.\\
The $3 \times 3$ matrix $\mathbf { A }$ has distinct eigenvalues $b , - 1,1$ with corresponding eigenvectors
$$\left( \begin{array} { l }
1 \\
0 \\
0
\end{array} \right) , \quad \left( \begin{array} { r }
4 \\
- 1 \\
0
\end{array} \right) , \quad \left( \begin{array} { l }
2 \\
1 \\
2
\end{array} \right)$$
respectively.
\item Find $\mathbf { A }$ in terms of b.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q7 [9]}}