| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Conditions for unique solution |
| Difficulty | Standard +0.8 This is a substantial Further Maths question requiring determinant calculation for non-unique solutions, Cayley-Hamilton theorem application to find inverse of A², and diagonalization. While each technique is standard for Further Maths, the combination of three distinct matrix theory topics in one question and the need for careful algebraic manipulation (especially in part b) places it moderately above average difficulty. |
| Spec | 4.03i Determinant: area scale factor and orientation4.03r Solve simultaneous equations: using inverse matrix4.03t Plane intersection: geometric interpretation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} 3 & 1 & 1 \\ a & 6 & -1 \\ 0 & a & -2 \end{vmatrix} = 0 \Rightarrow 3(-12+a)-(-2a)+a^2=0\) | M1 | Expanding determinant and setting equal to zero |
| \(a^2+5a-36=0 \Rightarrow a=4,-9\) | M1 A1 | Solving quadratic, both values required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\lambda-3)(\lambda-6)(\lambda+2)=\lambda^3-7\lambda^2+36=0\) | B1 | Correct characteristic equation |
| \(36\mathbf{I}=7\mathbf{A}^2-\mathbf{A}^3 \Rightarrow 36(\mathbf{A}^{-1})^2=7\mathbf{I}-\mathbf{A}\) | M1 | Multiplying through by \(\mathbf{A}^{-3}\) |
| \((\mathbf{A}^{-1})^2=\dfrac{1}{36}\begin{pmatrix}4&-1&-1\\0&1&1\\0&0&9\end{pmatrix}\) | M1 A1 | Correct matrix evaluation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Eigenvalues of \(\mathbf{A}\) are \(3, 6\) and \(-2\) | B1 | All three correct |
| \(\lambda=3\): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&1&1\\0&3&-1\end{vmatrix}=\begin{pmatrix}-4\\0\\0\end{pmatrix}\sim\begin{pmatrix}1\\0\\0\end{pmatrix}\) | M1 A1 | Method for eigenvector, correct result |
| \(\lambda=6\): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-3&1&1\\0&0&-1\end{vmatrix}=\begin{pmatrix}-1\\-3\\0\end{pmatrix}\) \(\quad\lambda=-2\): \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&1&1\\0&8&-1\end{vmatrix}=\begin{pmatrix}-9\\5\\40\end{pmatrix}\) | A1 A1 | Both eigenvectors correct |
| \(\mathbf{P}=\begin{pmatrix}1&1&-9\\0&3&5\\0&0&40\end{pmatrix}\) and \(\mathbf{D}=\begin{pmatrix}243&0&0\\0&7776&0\\0&0&-32\end{pmatrix}\) or \(\begin{pmatrix}3^5&0&0\\0&6^5&0\\0&0&-2^5\end{pmatrix}\) | M1 A1 | Correct \(\mathbf{P}\) consistent with eigenvectors, correct \(\mathbf{D}\) |
## Question 8:
### Part 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} 3 & 1 & 1 \\ a & 6 & -1 \\ 0 & a & -2 \end{vmatrix} = 0 \Rightarrow 3(-12+a)-(-2a)+a^2=0$ | M1 | Expanding determinant and setting equal to zero |
| $a^2+5a-36=0 \Rightarrow a=4,-9$ | M1 A1 | Solving quadratic, both values required |
**Total: 3 marks**
---
### Part 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\lambda-3)(\lambda-6)(\lambda+2)=\lambda^3-7\lambda^2+36=0$ | B1 | Correct characteristic equation |
| $36\mathbf{I}=7\mathbf{A}^2-\mathbf{A}^3 \Rightarrow 36(\mathbf{A}^{-1})^2=7\mathbf{I}-\mathbf{A}$ | M1 | Multiplying through by $\mathbf{A}^{-3}$ |
| $(\mathbf{A}^{-1})^2=\dfrac{1}{36}\begin{pmatrix}4&-1&-1\\0&1&1\\0&0&9\end{pmatrix}$ | M1 A1 | Correct matrix evaluation |
**Total: 4 marks**
---
### Part 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Eigenvalues of $\mathbf{A}$ are $3, 6$ and $-2$ | B1 | All three correct |
| $\lambda=3$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\0&1&1\\0&3&-1\end{vmatrix}=\begin{pmatrix}-4\\0\\0\end{pmatrix}\sim\begin{pmatrix}1\\0\\0\end{pmatrix}$ | M1 A1 | Method for eigenvector, correct result |
| $\lambda=6$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-3&1&1\\0&0&-1\end{vmatrix}=\begin{pmatrix}-1\\-3\\0\end{pmatrix}$ $\quad\lambda=-2$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\5&1&1\\0&8&-1\end{vmatrix}=\begin{pmatrix}-9\\5\\40\end{pmatrix}$ | A1 A1 | Both eigenvectors correct |
| $\mathbf{P}=\begin{pmatrix}1&1&-9\\0&3&5\\0&0&40\end{pmatrix}$ and $\mathbf{D}=\begin{pmatrix}243&0&0\\0&7776&0\\0&0&-32\end{pmatrix}$ or $\begin{pmatrix}3^5&0&0\\0&6^5&0\\0&0&-2^5\end{pmatrix}$ | M1 A1 | Correct $\mathbf{P}$ consistent with eigenvectors, correct $\mathbf{D}$ |
**Total: 7 marks**8
\begin{enumerate}[label=(\alph*)]
\item Find the values of $a$ for which the system of equations
$$\begin{aligned}
3 x + y + z & = 0 \\
a x + 6 y - z & = 0 \\
a y - 2 z & = 0
\end{aligned}$$
does not have a unique solution.\\
The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { r r r }
3 & 1 & 1 \\
0 & 6 & - 1 \\
0 & 0 & - 2
\end{array} \right) .$$
\item Use the characteristic equation of $\mathbf { A }$ to find the inverse of $\mathbf { A } ^ { 2 }$.
\item Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } ^ { 5 } = \mathbf { P D P } ^ { - 1 }$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q8 [14]}}