| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Properties of matrix operations |
| Difficulty | Standard +0.3 This is a standard Further Maths matrices question covering routine techniques: finding when det(A)=0 for singularity, describing a geometric transformation, verifying a matrix identity with specific matrices, and proving the general result. Part (c)(ii) requires understanding of matrix inverse properties but follows a standard proof pattern (multiply by NM and show it gives the identity). Slightly above average difficulty due to the proof component, but all parts are textbook-standard for Further Maths. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03l Singular/non-singular matrices4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(k - 2 = 0 \Rightarrow k = 2\) | B1 | Finds \(k = 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Reflection in the \(y\)-axis | B1 | States correct transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{BA} = \begin{bmatrix} -1 & -2 \\ 1 & k \end{bmatrix}\) | M1 | Finds product \(\mathbf{BA}\); allow one slip |
| \((\mathbf{BA})^{-1} = \dfrac{1}{-k-(-2)}\begin{bmatrix} k & 2 \\ -1 & -1 \end{bmatrix}\) | A1F | Obtains inverse; FT 'their' \(\mathbf{BA}\) provided M1 awarded |
| \(\mathbf{A}^{-1} = \dfrac{1}{k-2}\begin{bmatrix} k & -2 \\ -1 & 1 \end{bmatrix}\), \(\mathbf{B}^{-1} = \dfrac{1}{-1}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\) | B1 | Finds \(\mathbf{A}^{-1}\) and \(\mathbf{B}^{-1}\) |
| \(\mathbf{A}^{-1}\mathbf{B}^{-1} = \dfrac{1}{(k-2)\times(-1)}\begin{bmatrix} k+0 & (-2\times-1)+0 \\ (-1\times1)+0 & 0+(-1\times1) \end{bmatrix}\), showing this equals \((\mathbf{BA})^{-1} = \dfrac{1}{2-k}\begin{bmatrix} k & 2 \\ -1 & -1 \end{bmatrix}\) | R1 | Obtains correct \(\mathbf{A}^{-1}\mathbf{B}^{-1}\) and shows \((k-2)\times(-1) = 2-k = -k-(-2)\); must clearly show \(\mathbf{A}^{-1} \times \mathbf{B}^{-1}\) method; disallow if answer simply stated |
| Answer | Marks | Guidance |
|---|---|---|
| \((\mathbf{NM}) \times \{\mathbf{M}^{-1}\mathbf{N}^{-1}\} = \mathbf{I}\) | M1 | Uses equation for identity from definition |
| \((\mathbf{NM}) \times \mathbf{M}^{-1}\mathbf{N}^{-1} = \mathbf{N}(\mathbf{M} \times \mathbf{M}^{-1})\mathbf{N}^{-1} = \mathbf{N}\mathbf{I}\mathbf{N}^{-1} = \mathbf{N}\mathbf{N}^{-1}\) | R1 | Commences argument by manipulating matrix products within the equation with clear pairing |
| \(= \mathbf{I}\) | B1 | Clearly demonstrates that \(\mathbf{M} \times \mathbf{M}^{-1} = \mathbf{I}\) used |
| Using definition of matrix inverse and associativity of matrix multiplication; hence true for all non-singular matrices \(\mathbf{N}\) and \(\mathbf{M}\) | R1 | Completes argument using rigorous reasoning with definition of matrix inverse and associativity mentioned; must see all working with correct pairing of each matrix with inverse |
## Question 4(a):
$k - 2 = 0 \Rightarrow k = 2$ | B1 | Finds $k = 2$
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## Question 4(b):
Reflection in the $y$-axis | B1 | States correct transformation
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## Question 4(c)(i):
$\mathbf{BA} = \begin{bmatrix} -1 & -2 \\ 1 & k \end{bmatrix}$ | M1 | Finds product $\mathbf{BA}$; allow one slip
$(\mathbf{BA})^{-1} = \dfrac{1}{-k-(-2)}\begin{bmatrix} k & 2 \\ -1 & -1 \end{bmatrix}$ | A1F | Obtains inverse; FT 'their' $\mathbf{BA}$ provided M1 awarded
$\mathbf{A}^{-1} = \dfrac{1}{k-2}\begin{bmatrix} k & -2 \\ -1 & 1 \end{bmatrix}$, $\mathbf{B}^{-1} = \dfrac{1}{-1}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ | B1 | Finds $\mathbf{A}^{-1}$ and $\mathbf{B}^{-1}$
$\mathbf{A}^{-1}\mathbf{B}^{-1} = \dfrac{1}{(k-2)\times(-1)}\begin{bmatrix} k+0 & (-2\times-1)+0 \\ (-1\times1)+0 & 0+(-1\times1) \end{bmatrix}$, showing this equals $(\mathbf{BA})^{-1} = \dfrac{1}{2-k}\begin{bmatrix} k & 2 \\ -1 & -1 \end{bmatrix}$ | R1 | Obtains correct $\mathbf{A}^{-1}\mathbf{B}^{-1}$ and shows $(k-2)\times(-1) = 2-k = -k-(-2)$; must clearly show $\mathbf{A}^{-1} \times \mathbf{B}^{-1}$ method; disallow if answer simply stated
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## Question 4(c)(ii):
$(\mathbf{NM}) \times \{\mathbf{M}^{-1}\mathbf{N}^{-1}\} = \mathbf{I}$ | M1 | Uses equation for identity from definition
$(\mathbf{NM}) \times \mathbf{M}^{-1}\mathbf{N}^{-1} = \mathbf{N}(\mathbf{M} \times \mathbf{M}^{-1})\mathbf{N}^{-1} = \mathbf{N}\mathbf{I}\mathbf{N}^{-1} = \mathbf{N}\mathbf{N}^{-1}$ | R1 | Commences argument by manipulating matrix products within the equation with clear pairing
$= \mathbf{I}$ | B1 | Clearly demonstrates that $\mathbf{M} \times \mathbf{M}^{-1} = \mathbf{I}$ used
Using definition of matrix inverse and associativity of matrix multiplication; hence true for all non-singular matrices $\mathbf{N}$ and $\mathbf{M}$ | R1 | Completes argument using rigorous reasoning with definition of matrix inverse and associativity mentioned; must see all working with correct pairing of each matrix with inverse
**Total: 10 marks**4
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$ for which matrix $\mathbf { A }$ is singular.
4
\item Describe the transformation represented by matrix $\mathbf { B }$.
4
\item (i) Given that $\mathbf { A }$ and $\mathbf { B }$ are both non-singular, verify that $\mathbf { A } ^ { \mathbf { - 1 } } \mathbf { B } ^ { \mathbf { - 1 } } = ( \mathbf { B A } ) ^ { \mathbf { - 1 } }$.\\[0pt]
[4 marks]\\
4 (c) (ii) Prove the result $\mathbf { M } ^ { - \mathbf { 1 } } \mathbf { N } ^ { - \mathbf { 1 } } = ( \mathbf { N M } ) ^ { - \mathbf { 1 } }$ for all non-singular square matrices $\mathbf { M }$ and $\mathbf { N }$ of the same size.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 Q4 [8]}}