| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | January |
| Marks | 6 |
| 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 matrices question requiring (a) finding the inverse of a 2×2 matrix using the standard formula, and (b) solving MQ = PQ by pre-multiplying by P^(-1). Both parts use routine techniques with algebraic simplification, making it slightly easier than average for A-level but typical for early Further Maths content. |
| Spec | 4.03o Inverse 3x3 matrix4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{P}^{-1} = \frac{1}{4ab}\begin{pmatrix} 2b & 2a \\ b & 3a \end{pmatrix}\) | B1, M1, A1 | \(\frac{1}{4ab}\); Two out of four elements correct; Correct matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{M} = \mathbf{PQ} \Rightarrow \mathbf{P}^{-1}\mathbf{M} = \mathbf{P}^{-1}\mathbf{PQ} \Rightarrow \mathbf{Q} = \mathbf{P}^{-1}\mathbf{M}\) | ||
| \(\mathbf{Q} = \frac{1}{4ab}\begin{pmatrix} 2b & 2a \\ b & 3a \end{pmatrix}\begin{pmatrix} -6a & 7a \\ 2b & -b \end{pmatrix}\) | M1 | Multiplies their \(\mathbf{P}^{-1}\) by M |
| \(= \frac{1}{4ab}\begin{pmatrix} -8ab & 12ab \\ 0 & 4ab \end{pmatrix}\) | ||
| \(= \begin{pmatrix} -2 & 3 \\ 0 & 1 \end{pmatrix}\) | A1, A1 | Two out of four elements correct; Correct matrix |
**(a)**
| $\mathbf{P}^{-1} = \frac{1}{4ab}\begin{pmatrix} 2b & 2a \\ b & 3a \end{pmatrix}$ | B1, M1, A1 | $\frac{1}{4ab}$; Two out of four elements correct; Correct matrix |
**(b)**
| $\mathbf{M} = \mathbf{PQ} \Rightarrow \mathbf{P}^{-1}\mathbf{M} = \mathbf{P}^{-1}\mathbf{PQ} \Rightarrow \mathbf{Q} = \mathbf{P}^{-1}\mathbf{M}$ | | |
| $\mathbf{Q} = \frac{1}{4ab}\begin{pmatrix} 2b & 2a \\ b & 3a \end{pmatrix}\begin{pmatrix} -6a & 7a \\ 2b & -b \end{pmatrix}$ | M1 | Multiplies their $\mathbf{P}^{-1}$ by M |
| $= \frac{1}{4ab}\begin{pmatrix} -8ab & 12ab \\ 0 & 4ab \end{pmatrix}$ | | |
| $= \begin{pmatrix} -2 & 3 \\ 0 & 1 \end{pmatrix}$ | A1, A1 | Two out of four elements correct; Correct matrix |
---7.
$$\mathbf { P } = \left( \begin{array} { c c }
3 a & - 2 a \\
- b & 2 b
\end{array} \right) , \quad \mathbf { M } = \left( \begin{array} { c c }
- 6 a & 7 a \\
2 b & - b
\end{array} \right)$$
where $a$ and $b$ are non-zero constants.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { P } ^ { - 1 }$, leaving your answer in terms of $a$ and $b$.
Given that
$$\mathbf { M } = \mathbf { P Q }$$
\item find the matrix $\mathbf { Q }$, giving your answer in its simplest form.\\
\includegraphics[max width=\textwidth, alt={}, center]{9093bb1d-4f32-44e7-b0e7-b8c4f8a844e1-19_95_77_2617_1804}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2014 Q7 [6]}}