| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Prove eigenvalue/eigenvector properties |
| Difficulty | Challenging +1.2 This is a structured multi-part question on eigenvalues/eigenvectors that guides students through standard techniques. Part (a) is a routine proof using the definition of eigenvalues. Parts (b)-(d) involve mechanical calculations (solving (A-λI)e=0, matrix multiplication, diagonalization). Part (e) uses Cayley-Hamilton theorem, which is a standard Further Maths technique. While it's a Further Maths topic and has multiple parts, each step is well-scaffolded with no novel insights required, making it slightly above average difficulty but not challenging for the target cohort. |
| Spec | 4.03l Singular/non-singular matrices4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(Ae = \lambda e\) leading to \(e = \lambda A^{-1}e\) | M1 | Multiplies by \(A^{-1}\) on LHS |
| \(\lambda^{-1}e = A^{-1}e\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda = -1\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & 3 \\ 15 & -3 & 3 \end{vmatrix} = \begin{pmatrix} 9 \\ 36 \\ -9 \end{pmatrix} \sim \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix}\) | M1 A1 | Uses vector product (or equations) to find corresponding eigenvector |
| Answer | Marks |
|---|---|
| \(\begin{pmatrix} 2 & 0 & 3 \\ 15 & -4 & 3 \\ 3 & 0 & 2 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ -4 \\ 0 \end{pmatrix}\), \(\lambda = -4\) | B1 |
| \(\begin{pmatrix} 2 & 0 & 3 \\ 15 & -4 & 3 \\ 3 & 0 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 10 \\ 5 \end{pmatrix}\), \(\lambda = 5\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{P} = \begin{pmatrix} 1 & 0 & 1 \\ 4 & 1 & 2 \\ -1 & 0 & 1 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -\frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{5} \end{pmatrix}\) | M1 A1 | Or correctly matched permutations of columns. M1 for eigenvectors (all three non-zero) correctly matched to reciprocals of their eigenvalues |
| Answer | Marks | Guidance |
|---|---|---|
| \((\lambda+1)(\lambda+4)(\lambda-5) = \lambda^3 - 21\lambda - 20 = 0\) | B1 | Characteristic equation |
| \(A^3 - 21A - 20\mathbf{I} = 0\) | M1 | Substitutes for \(A\). Tolerate missing \(\mathbf{I}\) |
| \(20A^{-1} = A^2 - 21\mathbf{I}\) leading to \(A^{-1} = \frac{1}{20}A^2 - \frac{21}{20}\mathbf{I}\) | M1 A1 | Multiplies both sides by \(A^{-1}\), need \(\mathbf{I}\) for A1 |
## Question 7(a):
$Ae = \lambda e$ leading to $e = \lambda A^{-1}e$ | M1 | Multiplies by $A^{-1}$ on LHS |
$\lambda^{-1}e = A^{-1}e$ | A1 | |
---
## Question 7(b):
$\lambda = -1$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 0 & 3 \\ 15 & -3 & 3 \end{vmatrix} = \begin{pmatrix} 9 \\ 36 \\ -9 \end{pmatrix} \sim \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix}$ | M1 A1 | Uses vector product (or equations) to find corresponding eigenvector |
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## Question 7(c):
$\begin{pmatrix} 2 & 0 & 3 \\ 15 & -4 & 3 \\ 3 & 0 & 2 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ -4 \\ 0 \end{pmatrix}$, $\lambda = -4$ | B1 | |
$\begin{pmatrix} 2 & 0 & 3 \\ 15 & -4 & 3 \\ 3 & 0 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 10 \\ 5 \end{pmatrix}$, $\lambda = 5$ | B1 | |
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## Question 7(d):
$\mathbf{P} = \begin{pmatrix} 1 & 0 & 1 \\ 4 & 1 & 2 \\ -1 & 0 & 1 \end{pmatrix}$ and $\mathbf{D} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -\frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{5} \end{pmatrix}$ | M1 A1 | Or correctly matched permutations of columns. M1 for eigenvectors (all three non-zero) correctly matched to reciprocals of their eigenvalues |
---
## Question 7(e):
$(\lambda+1)(\lambda+4)(\lambda-5) = \lambda^3 - 21\lambda - 20 = 0$ | B1 | Characteristic equation |
$A^3 - 21A - 20\mathbf{I} = 0$ | M1 | Substitutes for $A$. Tolerate missing $\mathbf{I}$ |
$20A^{-1} = A^2 - 21\mathbf{I}$ leading to $A^{-1} = \frac{1}{20}A^2 - \frac{21}{20}\mathbf{I}$ | M1 A1 | Multiplies both sides by $A^{-1}$, need $\mathbf{I}$ for A1 |
---7
\begin{enumerate}[label=(\alph*)]
\item It is given that $\lambda$ is an eigenvalue of the non-singular square matrix $\mathbf { A }$, with corresponding eigenvector $\mathbf { e }$.
Show that $\lambda ^ { - 1 }$ is an eigenvalue of $\mathbf { A } ^ { - 1 }$ for which $\mathbf { e }$ is a corresponding eigenvector.\\
The matrix $\mathbf { A }$ is given by
$$\mathbf { A } = \left( \begin{array} { r r r }
2 & 0 & 3 \\
15 & - 4 & 3 \\
3 & 0 & 2
\end{array} \right)$$
\item Given that - 1 is an eigenvalue of $\mathbf { A }$, find a corresponding eigenvector.
\item It is also given that $\left( \begin{array} { l } 0 \\ 1 \\ 0 \end{array} \right)$ and $\left( \begin{array} { l } 1 \\ 2 \\ 1 \end{array} \right)$ are eigenvectors of $\mathbf { A }$. Find the corresponding eigenvalues.
\item Hence find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } ^ { - 1 } = \mathbf { P D P } ^ { - 1 }$.
\item Use the characteristic equation of $\mathbf { A }$ to show that $\mathbf { A } ^ { - 1 } = p \mathbf { A } ^ { 2 } + q l$, where $p$ and $q$ are rational numbers to be determined.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q7 [12]}}