| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| 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 multi-part eigenvalue question with clear signposting. Parts (a)-(b) involve standard techniques (determinant check and characteristic equation), part (c) is routine diagonalization for A⁴, and part (d) uses Cayley-Hamilton theorem. While it requires multiple techniques and careful algebra with the parameter a, each part follows established methods without requiring novel insight. Slightly above average difficulty due to the parameter and the A⁴ calculation, but well within typical Further Maths scope. |
| Spec | 4.03l Singular/non-singular matrices4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix4.03t Plane intersection: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} a & -6a & 2a+2 \\ 0 & 1-a & 0 \\ 0 & 2-a & -1 \end{vmatrix} = a(a-1) \neq 0\) or \(ax = 1 + \frac{12a}{1-a} - \frac{2(a+1)^2}{1-a} = \frac{-2a^2+7a-1}{1-a}\), \(y = \frac{2}{1-a}\), \(z = \frac{2(2-a)}{1-a} - 3 = \frac{a+1}{1-a}\) | M1 A1 | Shows that determinant is non-zero or solves equations. |
| Three planes intersect at a *single* point. | B1 | Must be clear that there is just one point where all 3 planes intersect. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} a-\lambda & -6a & 2a+2 \\ 0 & 1-a-\lambda & 0 \\ 0 & 2-a & -1-\lambda \end{vmatrix} = 0\) | M1 | Equates determinant to zero or working to find characteristic equation. |
| \((a-\lambda)(1-a-\lambda)(-1-\lambda) = 0 \Rightarrow \lambda = a, 1-a, -1\) | A1 | AG. Factorisation must be clear. |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda = a\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1-2a & 0 \\ 0 & 2-a & -1-a \end{vmatrix} = \begin{pmatrix} 2a^2+a-1 \\ 0 \\ 0 \end{pmatrix} \sim \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors. |
| \(\lambda = -1\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a+1 & -6a & 2a+2 \\ 0 & 2-a & 0 \end{vmatrix} = \begin{pmatrix} -(2-a)(2a+2) \\ 0 \\ (a+1)(2-a) \end{pmatrix} \sim \begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}\) | A1 | |
| \(\lambda = 1-a\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2a-1 & -6a & 2a+2 \\ 0 & 2-a & -2+a \end{vmatrix} = \begin{pmatrix} -4a^2+10a-4 \\ -2a^2+5a-2 \\ -2a^2+5a-2 \end{pmatrix} \sim \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\) | A1 | |
| \(\mathbf{P} = \begin{pmatrix} 1 & -2 & 2 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} a^4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & (1-a)^4 \end{pmatrix}\) | M1 A1 | Correctly matched permutations of columns. Their eigenvectors must be non-zero for M1. |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((a-\lambda)(1-a-\lambda)(-1-\lambda) = \lambda^3 - (a^2-a+1)\lambda - (a^2-a) = 0\) | B1 | Finds characteristic equation multiplied out. |
| \(\mathbf{A}^4 = (a^2-a+1)\mathbf{A}^2 + (a^2-a)\mathbf{A}\) | M1 A1 | Multiplies through by \(\mathbf{A}\). |
| 3 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} a & -6a & 2a+2 \\ 0 & 1-a & 0 \\ 0 & 2-a & -1 \end{vmatrix} = a(a-1) \neq 0$ or $ax = 1 + \frac{12a}{1-a} - \frac{2(a+1)^2}{1-a} = \frac{-2a^2+7a-1}{1-a}$, $y = \frac{2}{1-a}$, $z = \frac{2(2-a)}{1-a} - 3 = \frac{a+1}{1-a}$ | M1 A1 | Shows that determinant is non-zero or solves equations. |
| Three planes intersect at a *single* point. | B1 | Must be clear that there is just one point where all 3 planes intersect. |
| | **3** | |
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} a-\lambda & -6a & 2a+2 \\ 0 & 1-a-\lambda & 0 \\ 0 & 2-a & -1-\lambda \end{vmatrix} = 0$ | M1 | Equates determinant to zero or working to find characteristic equation. |
| $(a-\lambda)(1-a-\lambda)(-1-\lambda) = 0 \Rightarrow \lambda = a, 1-a, -1$ | A1 | AG. Factorisation must be clear. |
| | **2** | |
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda = a$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1-2a & 0 \\ 0 & 2-a & -1-a \end{vmatrix} = \begin{pmatrix} 2a^2+a-1 \\ 0 \\ 0 \end{pmatrix} \sim \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ | M1 A1 | Uses vector product (or equations) to find corresponding eigenvectors. |
| $\lambda = -1$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a+1 & -6a & 2a+2 \\ 0 & 2-a & 0 \end{vmatrix} = \begin{pmatrix} -(2-a)(2a+2) \\ 0 \\ (a+1)(2-a) \end{pmatrix} \sim \begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}$ | A1 | |
| $\lambda = 1-a$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2a-1 & -6a & 2a+2 \\ 0 & 2-a & -2+a \end{vmatrix} = \begin{pmatrix} -4a^2+10a-4 \\ -2a^2+5a-2 \\ -2a^2+5a-2 \end{pmatrix} \sim \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$ | A1 | |
| $\mathbf{P} = \begin{pmatrix} 1 & -2 & 2 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}$ and $\mathbf{D} = \begin{pmatrix} a^4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & (1-a)^4 \end{pmatrix}$ | M1 A1 | Correctly matched permutations of columns. Their eigenvectors must be non-zero for M1. |
| | **6** | |
## Question 8(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(a-\lambda)(1-a-\lambda)(-1-\lambda) = \lambda^3 - (a^2-a+1)\lambda - (a^2-a) = 0$ | B1 | Finds characteristic equation multiplied out. |
| $\mathbf{A}^4 = (a^2-a+1)\mathbf{A}^2 + (a^2-a)\mathbf{A}$ | M1 A1 | Multiplies through by $\mathbf{A}$. |
| | **3** | |8 The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { c c c }
a & - 6 a & 2 a + 2 \\
0 & 1 - a & 0 \\
0 & 2 - a & - 1
\end{array} \right)$$
where $a$ is a constant with $a \neq 0$ and $a \neq 1$.
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathbf { A } \left( \begin{array} { c } x \\ y \\ z \end{array} \right) = \left( \begin{array} { c } 1 \\ 2 \\ 3 \end{array} \right)$ has a unique solution and interpret this situation geometrically.
\item Show that the eigenvalues of $\mathbf { A }$ are $a , 1 - a$ and - 1 .
\item Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } ^ { 4 } = \mathbf { P D P } ^ { - 1 }$.
\item Use the characteristic equation of $\mathbf { A }$ to find $\mathbf { A } ^ { 4 }$ in terms of $\mathbf { A }$ and $a$.\\
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 2023 Q8 [14]}}