| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Reconstruct matrix from eigenvalues and eigenvectors |
| Difficulty | Challenging +1.2 This is a standard diagonalization question requiring students to construct P from eigenvectors, form the diagonal matrix D, compute A = PDP^(-1), then use this to find A^(2n) = PD^(2n)P^(-1). While it involves multiple steps and matrix inversion, it follows a well-established algorithm taught explicitly in Further Maths with no novel insight required. The even power 2n simplifies nicely since (-1)^(2n) = 1. Slightly above average difficulty due to the computational load and being a Further Maths topic, but remains a textbook exercise. |
| Spec | 4.03a Matrix language: terminology and notation4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{P} = \begin{pmatrix} 0 & -1 & 1 \\ 1 & 0 & 1 \\ -1 & 1 & 0 \end{pmatrix}\), \(\mathbf{D} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}\) | B1B1 | Columns can be in any order but must match |
| \(\det \mathbf{P} = 2\) | B1 | |
| \(\mathbf{P}^{-1} = \frac{1}{2}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}\) | M1A1 | Adj \(\div\) Det; no working for \(\frac{1}{3}\); row operations M1A1A1 (3 errors) |
| \(\mathbf{A} = \mathbf{PDP}^{-1}\) | M1 | |
| \(\mathbf{A} = \begin{pmatrix} 0 & -1 & 2 \\ -1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix} \cdot \frac{1}{2}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}\) | M1A1 | |
| \(= \begin{pmatrix} 1.5 & 0.5 & 0.5 \\ 1.5 & 0.5 & 1.5 \\ -1 & 1 & 0 \end{pmatrix}\) | A1 | Part mark: 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{A}^{2n} = \mathbf{PD}^{2n}\mathbf{P}^{-1}\) | M1 | |
| \(= \frac{1}{2}\begin{pmatrix} 0 & -1 & 1 \\ 1 & 0 & 1 \\ -1 & 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2^{2n} \end{pmatrix}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}\) | A1 | |
| \(= \frac{1}{2}\begin{pmatrix} 0 & -1 & 2^{2n} \\ 1 & 0 & 2^{2n} \\ -1 & 1 & 0 \end{pmatrix}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}\) | M1A1 | |
| \(= \frac{1}{2}\begin{pmatrix} 2^{2n}+1 & 2^{2n}-1 & 2^{2n}-1 \\ 2^{2n}-1 & 2^{2n}+1 & 2^{2n}-1 \\ 0 & 0 & 2 \end{pmatrix}\) | A1 | Part mark: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using \(\mathbf{Ae} = \lambda\mathbf{e}\) three times to form 3 sets of linear equations | M1A1 | |
| \(b-c=0\), \(e-f=-1\), \(h-j=1\) | ||
| \(-a+c=-1\), \(-d+f=0\), \(-g+j=1\) | ||
| \(a+b=2\), \(d+e=2\), \(g+h=0\) | ||
| Solves one set, then other two sets | M1 | |
| \(\mathbf{A} = \begin{pmatrix} 1.5 & 0.5 & 0.5 \\ 1.5 & 0.5 & 1.5 \\ -1 & 1 & 0 \end{pmatrix}\) | A1 | Part mark: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{P}\) and \(\mathbf{D}\) as above | B1B1 | |
| \(\mathbf{P}^{-1} = \frac{1}{2}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}\) | B1, M1A1 | Part mark: 5 |
| \(\mathbf{A}^{2n} = \mathbf{PD}^{2n}\mathbf{P}^{-1}\) | M1 | |
| Full multiplication as above | A1, M1A1 | |
| \(= \frac{1}{2}\begin{pmatrix} 2^{2n}+1 & 2^{2n}-1 & 2^{2n}-1 \\ 2^{2n}-1 & 2^{2n}+1 & 2^{2n}-1 \\ 0 & 0 & 2 \end{pmatrix}\) | A1 | Part mark: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix} 1 & -1 & -1 & 1 \\ 2 & -1 & -4 & 3 \\ 3 & -3 & -2 & 2 \\ 5 & -4 & -6 & 5 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & -1 & -1 & 1 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}\) | M1A1 | Reduces to echelon form |
| \(r(\mathbf{A}) = 4 - 1 = 3\) | A1 | Part mark: 3 |
| \(x - y - z + t = 0\), \(y - 2z + t = 0\), \(z - t = 0\) | M1 | |
| \(\Rightarrow t = \lambda,\ z = \lambda,\ y = \lambda,\ x = \lambda\) | ||
| Basis of null space is \(\left\{\begin{pmatrix}1\\1\\1\\1\end{pmatrix}\right\}\) | A1 | |
| \(\left\{\mathbf{Ax} = \begin{pmatrix}p\\q\\r\\0\end{pmatrix}\right\} \Rightarrow \mathbf{x} = \begin{pmatrix}p\\q\\r\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\1\\1\end{pmatrix}\) (AG) | M1A1 | Part mark: 4 |
| \(p - q - r = 3\), \(2p - q - 4r = 7\), \(3p - 3q - 2r = 8\) | M1 | |
| \(p = 1,\ q = -1,\ r = -1\) | A1, A1 | B2 all correct; B1 two correct with no working. Part mark: 3 |
| \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = \frac{11}{4} \Rightarrow 4\lambda^2 - 2\lambda + \frac{1}{4} = 0\) | M1A1 | |
| \(\Rightarrow \left(2\lambda - \frac{1}{2}\right)^2 = 0 \Rightarrow \lambda = \frac{1}{4} \Rightarrow \mathbf{x} = \begin{pmatrix}1.25\\-0.75\\-0.75\\0.25\end{pmatrix}\) | M1A1 | Part mark: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Writes \(\mathbf{x} = \begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}\) and forms equations from \(\mathbf{Ax} = p\begin{pmatrix}1\\2\\3\\5\end{pmatrix} + q\begin{pmatrix}-1\\-1\\-3\\-4\end{pmatrix} + r\begin{pmatrix}-1\\-4\\-2\\-6\end{pmatrix}\) | M1A1 | |
| \(x_1 - x_2 - x_3 + x_4 = p - q - r\) etc. | M1 | |
| Obtains \(x_1 = x_4 + p\), \(x_2 = x_4 + q\), \(x_3 = x_4 + r\) | A1 | |
| Sets \(x_4 = \lambda\): \(\mathbf{x} = \begin{pmatrix}p+\lambda\\q+\lambda\\r+\lambda\\\lambda\end{pmatrix}\) | Those who verify only get M1A1 (2/4) |
# Question 11:
## EITHER (First Method)
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{P} = \begin{pmatrix} 0 & -1 & 1 \\ 1 & 0 & 1 \\ -1 & 1 & 0 \end{pmatrix}$, $\mathbf{D} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$ | B1B1 | Columns can be in any order but must match |
| $\det \mathbf{P} = 2$ | B1 | |
| $\mathbf{P}^{-1} = \frac{1}{2}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ | M1A1 | Adj $\div$ Det; no working for $\frac{1}{3}$; row operations M1A1A1 (3 errors) |
| $\mathbf{A} = \mathbf{PDP}^{-1}$ | M1 | |
| $\mathbf{A} = \begin{pmatrix} 0 & -1 & 2 \\ -1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix} \cdot \frac{1}{2}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ | M1A1 | |
| $= \begin{pmatrix} 1.5 & 0.5 & 0.5 \\ 1.5 & 0.5 & 1.5 \\ -1 & 1 & 0 \end{pmatrix}$ | A1 | **Part mark: 9** |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{A}^{2n} = \mathbf{PD}^{2n}\mathbf{P}^{-1}$ | M1 | |
| $= \frac{1}{2}\begin{pmatrix} 0 & -1 & 1 \\ 1 & 0 & 1 \\ -1 & 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2^{2n} \end{pmatrix}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ | A1 | |
| $= \frac{1}{2}\begin{pmatrix} 0 & -1 & 2^{2n} \\ 1 & 0 & 2^{2n} \\ -1 & 1 & 0 \end{pmatrix}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ | M1A1 | |
| $= \frac{1}{2}\begin{pmatrix} 2^{2n}+1 & 2^{2n}-1 & 2^{2n}-1 \\ 2^{2n}-1 & 2^{2n}+1 & 2^{2n}-1 \\ 0 & 0 & 2 \end{pmatrix}$ | A1 | **Part mark: 5** |
**Total: [14]**
---
## EITHER (Alternative Method)
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using $\mathbf{Ae} = \lambda\mathbf{e}$ three times to form 3 sets of linear equations | M1A1 | |
| $b-c=0$, $e-f=-1$, $h-j=1$ | | |
| $-a+c=-1$, $-d+f=0$, $-g+j=1$ | | |
| $a+b=2$, $d+e=2$, $g+h=0$ | | |
| Solves one set, then other two sets | M1 | |
| $\mathbf{A} = \begin{pmatrix} 1.5 & 0.5 & 0.5 \\ 1.5 & 0.5 & 1.5 \\ -1 & 1 & 0 \end{pmatrix}$ | A1 | **Part mark: 4** |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{P}$ and $\mathbf{D}$ as above | B1B1 | |
| $\mathbf{P}^{-1} = \frac{1}{2}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ | B1, M1A1 | **Part mark: 5** |
| $\mathbf{A}^{2n} = \mathbf{PD}^{2n}\mathbf{P}^{-1}$ | M1 | |
| Full multiplication as above | A1, M1A1 | |
| $= \frac{1}{2}\begin{pmatrix} 2^{2n}+1 & 2^{2n}-1 & 2^{2n}-1 \\ 2^{2n}-1 & 2^{2n}+1 & 2^{2n}-1 \\ 0 & 0 & 2 \end{pmatrix}$ | A1 | **Part mark: 5** |
**Total: [14]**
---
## OR (Third Method)
### Part 1
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix} 1 & -1 & -1 & 1 \\ 2 & -1 & -4 & 3 \\ 3 & -3 & -2 & 2 \\ 5 & -4 & -6 & 5 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & -1 & -1 & 1 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ | M1A1 | Reduces to echelon form |
| $r(\mathbf{A}) = 4 - 1 = 3$ | A1 | **Part mark: 3** |
| $x - y - z + t = 0$, $y - 2z + t = 0$, $z - t = 0$ | M1 | |
| $\Rightarrow t = \lambda,\ z = \lambda,\ y = \lambda,\ x = \lambda$ | | |
| Basis of null space is $\left\{\begin{pmatrix}1\\1\\1\\1\end{pmatrix}\right\}$ | A1 | |
| $\left\{\mathbf{Ax} = \begin{pmatrix}p\\q\\r\\0\end{pmatrix}\right\} \Rightarrow \mathbf{x} = \begin{pmatrix}p\\q\\r\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\1\\1\end{pmatrix}$ (AG) | M1A1 | **Part mark: 4** |
| $p - q - r = 3$, $2p - q - 4r = 7$, $3p - 3q - 2r = 8$ | M1 | |
| $p = 1,\ q = -1,\ r = -1$ | A1, A1 | B2 all correct; B1 two correct with no working. **Part mark: 3** |
| $\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = \frac{11}{4} \Rightarrow 4\lambda^2 - 2\lambda + \frac{1}{4} = 0$ | M1A1 | |
| $\Rightarrow \left(2\lambda - \frac{1}{2}\right)^2 = 0 \Rightarrow \lambda = \frac{1}{4} \Rightarrow \mathbf{x} = \begin{pmatrix}1.25\\-0.75\\-0.75\\0.25\end{pmatrix}$ | M1A1 | **Part mark: 4** |
**Total: [14]**
---
## OR (Alternative for 2nd part)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Writes $\mathbf{x} = \begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}$ and forms equations from $\mathbf{Ax} = p\begin{pmatrix}1\\2\\3\\5\end{pmatrix} + q\begin{pmatrix}-1\\-1\\-3\\-4\end{pmatrix} + r\begin{pmatrix}-1\\-4\\-2\\-6\end{pmatrix}$ | M1A1 | |
| $x_1 - x_2 - x_3 + x_4 = p - q - r$ etc. | M1 | |
| Obtains $x_1 = x_4 + p$, $x_2 = x_4 + q$, $x_3 = x_4 + r$ | A1 | |
| Sets $x_4 = \lambda$: $\mathbf{x} = \begin{pmatrix}p+\lambda\\q+\lambda\\r+\lambda\\\lambda\end{pmatrix}$ | | Those who verify only get M1A1 (2/4) |A $3 \times 3$ matrix $\mathbf { A }$ has eigenvalues $- 1,1,2$, with corresponding eigenvectors
$$\left( \begin{array} { r }
0 \\
1 \\
- 1
\end{array} \right) , \quad \left( \begin{array} { r }
- 1 \\
0 \\
1
\end{array} \right) , \quad \left( \begin{array} { l }
1 \\
1 \\
0
\end{array} \right) ,$$
respectively. Find\\
(i) the matrix $\mathbf { A }$,\\
(ii) $\mathbf { A } ^ { 2 n }$, where $n$ is a positive integer.
\hfill \mbox{\textit{CAIE FP1 2011 Q11 EITHER}}