| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Solving matrix equations for unknown matrix |
| Difficulty | Moderate -0.3 This is a straightforward Further Maths FP1 matrix question requiring calculation of determinant (routine check for non-singularity), then solving BA² = A by multiplying both sides by A⁻¹ to get B = A⁻³. While it involves matrix inversion and multiplication, these are standard algorithmic procedures with no conceptual difficulty or novel insight required. Slightly easier than average due to its mechanical nature, though the Further Maths context keeps it from being trivial. |
| Spec | 4.03l Singular/non-singular matrices4.03n Inverse 2x2 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\det(\mathbf{A}) = 3\times0 - 2\times1 = -2\) | M1 | Correct attempt at the determinant |
| \(\det(\mathbf{A}) \neq 0\) so \(\mathbf{A}\) is non singular | A1 | \(\det(A) = -2\) and some reference to zero |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\mathbf{B}\mathbf{A}^2 = \mathbf{A} \Rightarrow \mathbf{B}\mathbf{A} = \mathbf{I} \Rightarrow \mathbf{B} = \mathbf{A}^{-1}\) | M1 | Recognising that \(\mathbf{A}^{-1}\) is required |
| \(\mathbf{B} = -\frac{1}{2}\begin{pmatrix} 3 & -1 \\ -2 & 0 \end{pmatrix}\) | M1 | At least 3 correct terms in \(\begin{pmatrix} 3 & -1 \\ -2 & 0 \end{pmatrix}\) |
| B1ft | \(\frac{1}{\text{their det}(\mathbf{A})}\begin{pmatrix} * & * \\ * & * \end{pmatrix}\) | |
| Fully correct answer | A1 |
## Question 8(a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\det(\mathbf{A}) = 3\times0 - 2\times1 = -2$ | M1 | Correct attempt at the determinant |
| $\det(\mathbf{A}) \neq 0$ so $\mathbf{A}$ is non singular | A1 | $\det(A) = -2$ and some reference to zero |
Note: $\frac{1}{\det(\mathbf{A})}$ scores M0. **(2 marks)**
## Question 8(b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\mathbf{B}\mathbf{A}^2 = \mathbf{A} \Rightarrow \mathbf{B}\mathbf{A} = \mathbf{I} \Rightarrow \mathbf{B} = \mathbf{A}^{-1}$ | M1 | Recognising that $\mathbf{A}^{-1}$ is required |
| $\mathbf{B} = -\frac{1}{2}\begin{pmatrix} 3 & -1 \\ -2 & 0 \end{pmatrix}$ | M1 | At least 3 correct terms in $\begin{pmatrix} 3 & -1 \\ -2 & 0 \end{pmatrix}$ |
| | B1ft | $\frac{1}{\text{their det}(\mathbf{A})}\begin{pmatrix} * & * \\ * & * \end{pmatrix}$ |
| Fully correct answer | A1 | |
**Correct answer only scores 4/4. Ignore poor matrix algebra notation if intention is clear. (4 marks) Total: 6**
---8.
$$\mathbf { A } = \left( \begin{array} { l l }
0 & 1 \\
2 & 3
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathbf { A }$ is non-singular.
\item Find $\mathbf { B }$ such that $\mathbf { B A } ^ { 2 } = \mathbf { A }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2012 Q8 [6]}}