| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find matrix A given eigenvalues and eigenvectors |
| Difficulty | Challenging +1.2 This is a structured Further Maths question testing standard eigenvalue theory. Part (a) requires reading eigenvalues from a triangular matrix (trivial). Part (b) uses Cayley-Hamilton theorem to find the inverse (standard technique). Part (c) applies the diagonalization formula A = PDP^(-1) to find A^(-1) = PD^(-1)P^(-1), which is a bookwork method with straightforward matrix multiplication. While it requires knowledge of multiple techniques, each step follows a well-established procedure with no novel insight needed. |
| Spec | 4.03i Determinant: area scale factor and orientation4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1,\ 2,\ -1\) | B1 | |
| Total: 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 \(\mathbf{P}^{-1}\) |
| \(\mathbf{P}^2 = \begin{pmatrix}1 & -3 & -1\\0 & 4 & 1\\0 & 0 & 1\end{pmatrix} \Rightarrow \mathbf{P}^{-1} = \begin{pmatrix}1 & \frac{1}{2} & \frac{3}{2}\\0 & \frac{1}{2} & \frac{1}{2}\\0 & 0 & -1\end{pmatrix}\) | M1 A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{D} = \begin{pmatrix}\frac{1}{a} & 0 & 0\\0 & 2 & 0\\0 & 0 & \frac{1}{2}\end{pmatrix}\) | B1 | |
| \(\mathbf{A}^{-1} = \mathbf{P}\begin{pmatrix}\frac{1}{a} & 0 & 0\\0 & 2 & 0\\0 & 0 & \frac{1}{2}\end{pmatrix}\mathbf{P}^{-1}\) | M1 | Applies \(\mathbf{A}^{-1} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) |
| \(= \begin{pmatrix}\frac{1}{a} & -2 & \frac{1}{2}\\0 & 4 & \frac{1}{2}\\0 & 0 & -\frac{1}{2}\end{pmatrix}\begin{pmatrix}1 & \frac{1}{2} & \frac{3}{2}\\0 & \frac{1}{2} & \frac{1}{2}\\0 & 0 & -1\end{pmatrix}\) | M1 A1 | Multiplies two adjacent matrices |
| \(\begin{pmatrix}\frac{1}{a} & \frac{1-2a}{2a} & \frac{3-3a}{2a}\\0 & 2 & \frac{3}{2}\\0 & 0 & \frac{1}{2}\end{pmatrix}\) | A1 | |
| Total: 5 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1,\ 2,\ -1$ | B1 | |
| **Total: 1** | | |
## Question 6(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 $\mathbf{P}^{-1}$ |
| $\mathbf{P}^2 = \begin{pmatrix}1 & -3 & -1\\0 & 4 & 1\\0 & 0 & 1\end{pmatrix} \Rightarrow \mathbf{P}^{-1} = \begin{pmatrix}1 & \frac{1}{2} & \frac{3}{2}\\0 & \frac{1}{2} & \frac{1}{2}\\0 & 0 & -1\end{pmatrix}$ | M1 A1 | |
| **Total: 4** | | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{D} = \begin{pmatrix}\frac{1}{a} & 0 & 0\\0 & 2 & 0\\0 & 0 & \frac{1}{2}\end{pmatrix}$ | B1 | |
| $\mathbf{A}^{-1} = \mathbf{P}\begin{pmatrix}\frac{1}{a} & 0 & 0\\0 & 2 & 0\\0 & 0 & \frac{1}{2}\end{pmatrix}\mathbf{P}^{-1}$ | M1 | Applies $\mathbf{A}^{-1} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$ |
| $= \begin{pmatrix}\frac{1}{a} & -2 & \frac{1}{2}\\0 & 4 & \frac{1}{2}\\0 & 0 & -\frac{1}{2}\end{pmatrix}\begin{pmatrix}1 & \frac{1}{2} & \frac{3}{2}\\0 & \frac{1}{2} & \frac{1}{2}\\0 & 0 & -1\end{pmatrix}$ | M1 A1 | Multiplies two adjacent matrices |
| $\begin{pmatrix}\frac{1}{a} & \frac{1-2a}{2a} & \frac{3-3a}{2a}\\0 & 2 & \frac{3}{2}\\0 & 0 & \frac{1}{2}\end{pmatrix}$ | A1 | |
| **Total: 5** | | |6 The matrix $\mathbf { P }$ is given by
$$\mathbf { P } = \left( \begin{array} { r r r }
1 & - 1 & 1 \\
0 & 2 & 1 \\
0 & 0 & - 1
\end{array} \right) .$$
\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 non-zero eigenvalues $a , \frac { 1 } { 2 } , 2$ with corresponding eigenvectors
$$\left( \begin{array} { l }
1 \\
0 \\
0
\end{array} \right) , \quad \left( \begin{array} { r }
- 1 \\
2 \\
0
\end{array} \right) , \quad \left( \begin{array} { r }
1 \\
1 \\
- 1
\end{array} \right) ,$$
respectively.
\item Find $\mathbf { A } ^ { - 1 }$ in terms of $a$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q6 [10]}}