| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 13 |
| 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 structured Further Maths question requiring eigenvalue computation, diagonalization of A², and manipulation of the characteristic equation. Part (a) is routine (determinant = 0), part (b) is standard diagonalization with the twist of A² rather than A (eigenvalues are squared), and part (c) requires algebraic manipulation using the characteristic equation. While multi-step, each component follows established techniques without requiring novel insight, making it moderately above average difficulty for Further Maths. |
| Spec | 4.03a Matrix language: terminology and notation4.03l Singular/non-singular matrices4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} 3 & a & 0 \\ 5 & -1 & 0 \\ 1 & 3 & 2 \end{vmatrix} = 0 \Rightarrow -10a - 6 = 0\) | M1 | Sets determinant equal to zero |
| \(a = -\frac{3}{5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Eigenvalues of A are \(3,\ -1\) and \(2\) | B1 | Lower diagonal matrix or characteristic equation |
| \(\lambda = 3\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & -4 & 0 \\ 1 & 3 & -1 \end{vmatrix} = \begin{pmatrix} 4 \\ 5 \\ 19 \end{pmatrix}\) | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| \(\lambda = -1\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 0 & 0 \\ 1 & 3 & 3 \end{vmatrix} = \begin{pmatrix} 0 \\ -12 \\ 12 \end{pmatrix} \sim \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}\) | A1 | |
| \(\lambda = 2\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 0 \\ 1 & 3 & 0 \end{vmatrix} = \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} \sim \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\) | A1 | |
| Thus \(\mathbf{P} = \begin{pmatrix} 4 & 0 & 0 \\ 5 & -1 & 0 \\ 19 & 1 & 1 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} 9 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix}\) | M1 A1 | Or correctly matched permutations of columns |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\lambda-3)(\lambda+1)(\lambda-2) = \lambda^3 - 4\lambda^2 + \lambda + 6 = 0\) | B1 | Characteristic equation |
| \(\mathbf{A} + 6\mathbf{I} = -\mathbf{A}^3 + 4\mathbf{A}^2\) | M1 | Substitutes for A and makes \(\mathbf{A} + 6\mathbf{I}\) the subject |
| \((\mathbf{A}+6\mathbf{I})^2 = \left(-\mathbf{A}^3+4\mathbf{A}^2\right)^2 = \mathbf{A}^4(\mathbf{A}-4\mathbf{I})^2\) | M1 A1 | Squares and factorises |
## Question 8:
### Part 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} 3 & a & 0 \\ 5 & -1 & 0 \\ 1 & 3 & 2 \end{vmatrix} = 0 \Rightarrow -10a - 6 = 0$ | M1 | Sets determinant equal to zero |
| $a = -\frac{3}{5}$ | A1 | |
---
### Part 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Eigenvalues of **A** are $3,\ -1$ and $2$ | B1 | Lower diagonal matrix or characteristic equation |
| $\lambda = 3$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & -4 & 0 \\ 1 & 3 & -1 \end{vmatrix} = \begin{pmatrix} 4 \\ 5 \\ 19 \end{pmatrix}$ | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors |
| $\lambda = -1$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 0 & 0 \\ 1 & 3 & 3 \end{vmatrix} = \begin{pmatrix} 0 \\ -12 \\ 12 \end{pmatrix} \sim \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}$ | A1 | |
| $\lambda = 2$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 0 \\ 1 & 3 & 0 \end{vmatrix} = \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} \sim \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ | A1 | |
| Thus $\mathbf{P} = \begin{pmatrix} 4 & 0 & 0 \\ 5 & -1 & 0 \\ 19 & 1 & 1 \end{pmatrix}$ and $\mathbf{D} = \begin{pmatrix} 9 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix}$ | M1 A1 | Or correctly matched permutations of columns |
---
### Part 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\lambda-3)(\lambda+1)(\lambda-2) = \lambda^3 - 4\lambda^2 + \lambda + 6 = 0$ | B1 | Characteristic equation |
| $\mathbf{A} + 6\mathbf{I} = -\mathbf{A}^3 + 4\mathbf{A}^2$ | M1 | Substitutes for **A** and makes $\mathbf{A} + 6\mathbf{I}$ the subject |
| $(\mathbf{A}+6\mathbf{I})^2 = \left(-\mathbf{A}^3+4\mathbf{A}^2\right)^2 = \mathbf{A}^4(\mathbf{A}-4\mathbf{I})^2$ | M1 A1 | Squares and factorises |8
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ for which the system of equations
$$\begin{gathered}
3 x + a y = 0 \\
5 x - y = 0 \\
x + 3 y + 2 z = 0
\end{gathered}$$
does not have a unique solution.\\
The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { r r r }
3 & 0 & 0 \\
5 & - 1 & 0 \\
1 & 3 & 2
\end{array} \right)$$
\item Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } ^ { 2 } = \mathbf { P D P } ^ { - 1 }$.
\item Use the characteristic equation of $\mathbf { A }$ to show that
$$( \mathbf { A } + 6 \mathbf { I } ) ^ { 2 } = \mathbf { A } ^ { 4 } ( \mathbf { A } + b \mathbf { I } ) ^ { 2 }$$
where $b$ is an integer to be determined.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q8 [13]}}