| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Matrix satisfying given equation |
| Difficulty | Standard +0.3 This is a standard FP1 matrix question requiring matrix multiplication (routine), algebraic manipulation to find the inverse, and applying the inverse transformation. All steps are straightforward applications of learned techniques with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 4.03b Matrix operations: addition, multiplication, scalar4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{A}^2 = \begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix} = \begin{pmatrix} 44 & -14 \\ -28 & 9 \end{pmatrix}\) | M1A1 | M1: Attempt both \(\mathbf{A}^2\) and \(7\mathbf{A} + 2\mathbf{I}\). A1: Both matrices correct. |
| Answer | Marks | Guidance |
|---|---|---|
| OR \(\mathbf{A}^2 - 7\mathbf{A} = \mathbf{A}(\mathbf{A} - 7\mathbf{I})\) | M1 | M1 for expression and attempt to substitute and multiply \((2\times2)(2\times2)=2\times2\) |
| \(\mathbf{A}(\mathbf{A}-7\mathbf{I}) = \begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} -1 & -2 \\ -4 & -6 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = 2\mathbf{I}\) | A1 cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{A}^2 = 7\mathbf{A} + 2\mathbf{I} \Rightarrow \mathbf{A} = 7\mathbf{I} + 2\mathbf{A}^{-1}\) | M1 | Require one correct line using accurate expressions involving \(\mathbf{A}^{-1}\) and identity matrix to be clearly stated as \(\mathbf{I}\). |
| \(\mathbf{A}^{-1} = \frac{1}{2}(\mathbf{A} - 7\mathbf{I})^*\) | A1* cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{A}^{-1} = \frac{1}{2}\begin{pmatrix} -1 & -2 \\ -4 & -6 \end{pmatrix}\) | B1 | Correct inverse matrix or equivalent. |
| \(\frac{1}{2}\begin{pmatrix} -1 & -2 \\ -4 & -6 \end{pmatrix}\begin{pmatrix} 2k+8 \\ -2k-5 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} -2k-8+4k+10 \\ -8k-32+12k+30 \end{pmatrix}\) | M1 | Matrix multiplication involving their inverse and \(k\): \((2\times2)(2\times1)=2\times1\). N.B. \(\begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} 2k+8 \\ -2k-5 \end{pmatrix}\) is M0. |
| \(\begin{pmatrix} k+1 \\ 2k-1 \end{pmatrix}\) or \((k+1, 2k-1)\) | A1, A1 | \((k+1)\) first A1, \((2k-1)\) second A1. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2k+8 \\ -2k-5 \end{pmatrix}\) | B1 | Correct matrix equation. |
| Answer | Marks | Guidance |
|---|---|---|
| \(-4x + y = -2k-5 \Rightarrow x = \ldots \text{ or } y = \ldots\) | M1 | Multiply out and attempt to solve simultaneous equations for \(x\) or \(y\) in terms of \(k\). |
| \(\begin{pmatrix} k+1 \\ 2k-1 \end{pmatrix}\) or \((k+1, 2k-1)\) | A1, A1 | \((k+1)\) first A1, \((2k-1)\) second A1. |
# Question 8:
## Part (a):
$\mathbf{A}^2 = \begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix} = \begin{pmatrix} 44 & -14 \\ -28 & 9 \end{pmatrix}$ | M1A1 | M1: Attempt both $\mathbf{A}^2$ and $7\mathbf{A} + 2\mathbf{I}$. A1: Both matrices correct.
$7\mathbf{A} + 2\mathbf{I} = \begin{pmatrix} 42 & -14 \\ -28 & 7 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 44 & -14 \\ -28 & 9 \end{pmatrix}$
OR $\mathbf{A}^2 - 7\mathbf{A} = \mathbf{A}(\mathbf{A} - 7\mathbf{I})$ | M1 | M1 for expression and attempt to substitute and multiply $(2\times2)(2\times2)=2\times2$
$\mathbf{A}(\mathbf{A}-7\mathbf{I}) = \begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} -1 & -2 \\ -4 & -6 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = 2\mathbf{I}$ | A1 cso |
**(2 marks)**
## Part (b):
$\mathbf{A}^2 = 7\mathbf{A} + 2\mathbf{I} \Rightarrow \mathbf{A} = 7\mathbf{I} + 2\mathbf{A}^{-1}$ | M1 | Require one correct line using accurate expressions involving $\mathbf{A}^{-1}$ and identity matrix to be clearly stated as $\mathbf{I}$.
$\mathbf{A}^{-1} = \frac{1}{2}(\mathbf{A} - 7\mathbf{I})^*$ | A1* cso |
Numerical approach award 0/2.
**(2 marks)**
## Part (c):
$\mathbf{A}^{-1} = \frac{1}{2}\begin{pmatrix} -1 & -2 \\ -4 & -6 \end{pmatrix}$ | B1 | Correct inverse matrix or equivalent.
$\frac{1}{2}\begin{pmatrix} -1 & -2 \\ -4 & -6 \end{pmatrix}\begin{pmatrix} 2k+8 \\ -2k-5 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} -2k-8+4k+10 \\ -8k-32+12k+30 \end{pmatrix}$ | M1 | Matrix multiplication involving their inverse and $k$: $(2\times2)(2\times1)=2\times1$. N.B. $\begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} 2k+8 \\ -2k-5 \end{pmatrix}$ is M0.
$\begin{pmatrix} k+1 \\ 2k-1 \end{pmatrix}$ or $(k+1, 2k-1)$ | A1, A1 | $(k+1)$ first A1, $(2k-1)$ second A1.
**Or:**
$\begin{pmatrix} 6 & -2 \\ -4 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2k+8 \\ -2k-5 \end{pmatrix}$ | B1 | Correct matrix equation.
$6x - 2y = 2k+8$
$-4x + y = -2k-5 \Rightarrow x = \ldots \text{ or } y = \ldots$ | M1 | Multiply out and attempt to solve simultaneous equations for $x$ or $y$ in terms of $k$.
$\begin{pmatrix} k+1 \\ 2k-1 \end{pmatrix}$ or $(k+1, 2k-1)$ | A1, A1 | $(k+1)$ first A1, $(2k-1)$ second A1.
**(4 marks)**
**Total: 8 marks**
---8.
$$\mathbf { A } = \left( \begin{array} { c c }
6 & - 2 \\
- 4 & 1
\end{array} \right)$$
and $\mathbf { I }$ is the $2 \times 2$ identity matrix.
\begin{enumerate}[label=(\alph*)]
\item Prove that
$$\mathbf { A } ^ { 2 } = 7 \mathbf { A } + 2 \mathbf { I }$$
\item Hence show that
$$\mathbf { A } ^ { - 1 } = \frac { 1 } { 2 } ( \mathbf { A } - 7 \mathbf { I } )$$
The transformation represented by $\mathbf { A }$ maps the point $P$ onto the point $Q$.\\
Given that $Q$ has coordinates $( 2 k + 8 , - 2 k - 5 )$, where $k$ is a constant,
\item find, in terms of $k$, the coordinates of $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q8 [8]}}