| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Solving matrix equations for unknown matrix |
| Difficulty | Standard +0.3 This is a straightforward FP1 matrix question requiring finding an inverse of a 2×2 matrix using the standard formula, then solving AB = P by multiplying both sides by A^(-1). Both are routine procedures with no conceptual difficulty, making it slightly easier than average for A-level but typical for basic Further Maths content. |
| Spec | 4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A^{-1} = \frac{1}{10}\begin{pmatrix}3 & 1\\-4 & 2\end{pmatrix}\) | M1 | Either \(\frac{1}{10}\) or \(\begin{pmatrix}3 & 1\\-4 & 2\end{pmatrix}\) |
| A1 | Correct matrix seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P = AB \Rightarrow A^{-1}P = A^{-1}AB \Rightarrow B = A^{-1}P\) | ||
| \(B = \frac{1}{10}\begin{pmatrix}3 & 1\\-4 & 2\end{pmatrix}\begin{pmatrix}3 & 6\\11 & -8\end{pmatrix}\) | M1 | Multiplies their \(A^{-1}\) by \(P\) in correct order |
| \(= \begin{pmatrix}2 & 1\\1 & -4\end{pmatrix}\) | A1 | At least 2 elements correct or \(k\begin{pmatrix}20 & 10\\10 & -40\end{pmatrix}\) |
| A1 | Correct simplified matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}3 & 6\\11 & -8\end{pmatrix} = \begin{pmatrix}2 & -1\\4 & 3\end{pmatrix}\begin{pmatrix}a & b\\c & d\end{pmatrix}\) | M1 | Attempt to multiply \(A\) by \(B\) in correct order and puts equal to \(P\) |
| \(\Rightarrow a=2, c=1, b=1, d=-4\) | A1 | At least 2 elements correct |
| \(B = \begin{pmatrix}2 & 1\\1 & -4\end{pmatrix}\) | A1 | Correct matrix |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A^{-1} = \frac{1}{10}\begin{pmatrix}3 & 1\\-4 & 2\end{pmatrix}$ | M1 | Either $\frac{1}{10}$ or $\begin{pmatrix}3 & 1\\-4 & 2\end{pmatrix}$ |
| | A1 | Correct matrix seen |
## Part (b) Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = AB \Rightarrow A^{-1}P = A^{-1}AB \Rightarrow B = A^{-1}P$ | | |
| $B = \frac{1}{10}\begin{pmatrix}3 & 1\\-4 & 2\end{pmatrix}\begin{pmatrix}3 & 6\\11 & -8\end{pmatrix}$ | M1 | Multiplies their $A^{-1}$ by $P$ in correct order |
| $= \begin{pmatrix}2 & 1\\1 & -4\end{pmatrix}$ | A1 | At least 2 elements correct or $k\begin{pmatrix}20 & 10\\10 & -40\end{pmatrix}$ |
| | A1 | Correct simplified matrix |
## Part (b) Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}3 & 6\\11 & -8\end{pmatrix} = \begin{pmatrix}2 & -1\\4 & 3\end{pmatrix}\begin{pmatrix}a & b\\c & d\end{pmatrix}$ | M1 | Attempt to multiply $A$ by $B$ in correct order and puts equal to $P$ |
| $\Rightarrow a=2, c=1, b=1, d=-4$ | A1 | At least 2 elements correct |
| $B = \begin{pmatrix}2 & 1\\1 & -4\end{pmatrix}$ | A1 | Correct matrix |
---2.
$$\mathbf { A } = \left( \begin{array} { r r }
2 & - 1 \\
4 & 3
\end{array} \right) , \quad \mathbf { P } = \left( \begin{array} { r r }
3 & 6 \\
11 & - 8
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { A } ^ { - 1 }$\\
(2)
The transformation represented by the matrix $\mathbf { B }$ followed by the transformation represented by the matrix $\mathbf { A }$ is equivalent to the transformation represented by the matrix $\mathbf { P }$.
\item Find $\mathbf { B }$, giving your answer in its simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2017 Q2 [5]}}