| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Properties of matrix operations |
| Difficulty | Standard +0.3 Part (a) is a standard matrix algebra proof requiring application of a given result to show a transpose-inverse identity—straightforward for Further Maths students. Parts (b)(i) and (b)(ii) involve routine matrix inversion and stating the non-singularity condition. This is textbook-level material with no novel insight required, making it slightly easier than average even for Further Maths. |
| Spec | 4.03n Inverse 2x2 matrix4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1) |
| Answer | Marks |
|---|---|
| 8(a) | Uses the result ( AB )T =B T A T |
| Answer | Marks | Guidance |
|---|---|---|
| acceptable here. | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| result. | 2.1 | R1 |
| Subtotal | 3 | |
| Q | Marking Instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 8(b)(i) | -1 | |
| Obtains A or A | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| outside matrix. | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking Instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 8(b)(ii) | Obtains correct restriction on k | |
| FT their det(A) from (b)(i). | 1.1b | B1F |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Question total | 6 | |
| Q | Marking Instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Uses the result ( AB )T =B T A T
in a statement involving A-1
Use of the notation A-T is not
acceptable here. | 3.1a | M1 | ( A T)( A −1)T = ( A −1 A )T =IT =I
( T)−1 ( −1)T
∴ A = A
Uses the fact that the identity
matrix is its own transpose.
PI | 1.1a | M1
Completes a reasoned
argument to show the required
result. | 2.1 | R1
Subtotal | 3
Q | Marking Instructions | AO | Marks | Typical solution
--- 8(b)(i) ---
8(b)(i) | -1
Obtains A or A | 1.1a | M1 | 1 k −5
−1=
A
4k+51 4
k 1
( −1)T 4k+5 4k+5
A =
−5 4
4k+5 4k+5
T
T
-1
Obtains A
Allow answer with factor
� �
outside matrix. | 1.1b | A1
Subtotal | 2
Q | Marking Instructions | AO | Marks | Typical solution
--- 8(b)(ii) ---
8(b)(ii) | Obtains correct restriction on k
FT their det(A) from (b)(i). | 1.1b | B1F | 5
k ≠−
4
Subtotal | 1
Question total | 6
Q | Marking Instructions | AO | Marks | Typical solution$\mathbf{A}$ is a non-singular $2 \times 2$ matrix and $\mathbf{A}^T$ is the transpose of $\mathbf{A}$
\begin{enumerate}[label=(\alph*)]
\item Using the result
$$(\mathbf{AB})^T = \mathbf{B}^T\mathbf{A}^T$$
show that
$$(\mathbf{A}^{-1})^T = (\mathbf{A}^T)^{-1}$$
[3 marks]
\item It is given that $\mathbf{A} = \begin{pmatrix} 4 & 5 \\ -1 & k \end{pmatrix}$, where $k$ is a real constant.
\begin{enumerate}[label=(\roman*)]
\item Find $(\mathbf{A}^{-1})^T$, giving your answer in terms of $k$
[2 marks]
\item State the restriction on the possible values of $k$
[1 mark]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2023 Q8 [6]}}