| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find eigenvalues of 3×3 matrix |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring eigenvalue calculation via a 3×3 determinant and application of the Cayley-Hamilton theorem to find the inverse. While these are standard Further Maths techniques, the 3×3 characteristic polynomial and algebraic manipulation elevate it above average A-level difficulty, though it remains a textbook-style exercise without requiring novel insight. |
| Spec | 4.03j Determinant 3x3: calculation4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix |
| Answer | Marks |
|---|---|
| 3(a) | 5−λ −1 7 |
| Answer | Marks |
|---|---|
| 7 7 5−λ | B1 |
| −λ3+16λ2−36λ−144=0(λ+2)(λ−6)(λ−12)=0 | M1 |
| λ= −2,6,12 | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 3(b) | −A3 +16A2 −36A −144I = 0 | B1 |
| 144A−1 = −A2 +16A −36I | M1 |
| Answer | Marks |
|---|---|
| | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 3:
--- 3(a) ---
3(a) | 5−λ −1 7
0 6−λ 0 =0
7 7 5−λ | B1
−λ3+16λ2−36λ−144=0(λ+2)(λ−6)(λ−12)=0 | M1
λ= −2,6,12 | A1 A1
4
Question | Answer | Marks
--- 3(b) ---
3(b) | −A3 +16A2 −36A −144I = 0 | B1
144A−1 = −A2 +16A −36I | M1
74 38 70 −5 −9 7
1
A2 = 0 36 0 A −1 = 0 4 0
24
70 70 74 7 7 −5
| M1 A1
4
Question | Answer | MarksThe matrix $\mathbf{A}$ is given by
$$\mathbf{A} = \begin{pmatrix} 5 & -1 & 7 \\ 0 & 6 & 0 \\ 7 & 7 & 5 \end{pmatrix}.$$
\begin{enumerate}[label=(\alph*)]
\item Find the eigenvalues of $\mathbf{A}$. [4]
\item Use the characteristic equation of $\mathbf{A}$ to find $\mathbf{A}^{-1}$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q3 [8]}}