| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Solving matrix equations for unknown matrix |
| Difficulty | Moderate -0.3 This is a straightforward two-part question testing standard matrix inversion (using the 2×2 formula) and basic matrix equation manipulation (Z = YX^(-1)). While it involves algebraic manipulation with parameters a and b, the techniques are routine for FP1 students with no conceptual challenges or novel problem-solving required. Slightly easier than average due to the mechanical nature of 2×2 matrix operations. |
| Spec | 4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Determinant \(= 5ab\) | B1 | |
| \(\mathbf{X}^{-1} = \dfrac{1}{5ab}\begin{pmatrix} 3b & -2b \\ a & a \end{pmatrix}\) | M1 A1 | (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{Z} = \mathbf{Y}\mathbf{X}^{-1}\) | M1 | |
| \(= \dfrac{1}{5ab}\begin{pmatrix} 10ab & -5ab \\ 5ab & 0 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}\) | M1 A1 ft, A1 cao | (4 marks) |
| (7 marks total) |
# Question 5:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Determinant $= 5ab$ | B1 | |
| $\mathbf{X}^{-1} = \dfrac{1}{5ab}\begin{pmatrix} 3b & -2b \\ a & a \end{pmatrix}$ | M1 A1 | (3 marks total) |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{Z} = \mathbf{Y}\mathbf{X}^{-1}$ | M1 | |
| $= \dfrac{1}{5ab}\begin{pmatrix} 10ab & -5ab \\ 5ab & 0 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}$ | M1 A1 ft, A1 cao | (4 marks) |
| **(7 marks total)** | | |
---5. Given that $a$ and $b$ are non-zero constants and that
$$\mathbf { X } = \left( \begin{array} { r r }
a & 2 b \\
- a & 3 b
\end{array} \right) ,$$
\begin{enumerate}[label=(\alph*)]
\item find $\mathbf { X } ^ { - 1 }$, giving your answer in terms of $a$ and $b$.
Given also that $\mathbf { Z X } = \mathbf { Y }$, where $\mathbf { Y } = \left( \begin{array} { c c } 3 a & b \\ a & 2 b \end{array} \right)$,
\item find $\mathbf { Z }$, simplifying your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 Q5 [7]}}