AQA Further Paper 2 Specimen — Question 9 6 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMatrices
TypeProperties of matrix operations
DifficultyChallenging +1.2 This is a Further Maths question testing matrix inverse properties. Part (a) requires recognizing that non-invertible matrices exist (simple counter-example). Part (b) needs adding the condition that matrices must be invertible. Part (c) is a standard proof using associativity and the definition of inverse. While it requires proof-writing skills beyond Core modules, this is a well-known result that students explicitly learn, making it moderately above average but not exceptionally challenging for Further Maths students.
Spec1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1)
A student claims: "Given any two non-zero square matrices, A and B, then \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\)"
  1. Explain why the student's claim is incorrect giving a counter example. [2 marks]
  2. Refine the student's claim to make it fully correct. [1 mark]
  3. Prove that your answer to part (b) is correct. [3 marks]
Question 9:
AnswerMarks
9(a)Explains that the claim is incorrect
as singular square matrices do not
AnswerMarks Guidance
have inverses.AO2.3 E1
matrix is singular/has determinant
equal to zero as the inverse will not
exist
2 2
Eg is singular
 
3 3
Correctly gives an example of a
AnswerMarks Guidance
singular matrix.AO1.1b B1
(b)Correctly refines the statement
using ‘non-singular’ or equivalent
AnswerMarks Guidance
wordingAO2.3 B1
square matrices, A and B, then
(AB)–1 = B–1A–1
AnswerMarks
(c)Correctly recalls the inverse
property for matrices A and B
AnswerMarks Guidance
(seen at least once)AO1.2 B1
inverses exist hence
A and B are non-singular so
inverses exist hence
(AB)(B1A1)A(BB1)A1
AIA1
AA1
 I
Since
(AB)(B1A1) I
Then
(AB)1 (B1A1)
Correctly uses associativity by
AnswerMarks Guidance
regrouping (seen at least once)AO2.5 B1
Correctly applies the identity
property throughout and concludes
their rigorous mathematical
argument with no errors or
AnswerMarks Guidance
omissionsAO2.1 R1
Total6
QMarking Instructions AO 
Question 9:
--- 9(a) ---
9(a) | Explains that the claim is incorrect
as singular square matrices do not
have inverses. | AO2.3 | E1 | Statement is incorrect if either
matrix is singular/has determinant
equal to zero as the inverse will not
exist
2 2
Eg is singular
 
3 3
Correctly gives an example of a
singular matrix. | AO1.1b | B1
(b) | Correctly refines the statement
using ‘non-singular’ or equivalent
wording | AO2.3 | B1 | Given any two non-singular
square matrices, A and B, then
(AB)–1 = B–1A–1
(c) | Correctly recalls the inverse
property for matrices A and B
(seen at least once) | AO1.2 | B1 | A and B are non-singular so
inverses exist hence
A and B are non-singular so
inverses exist hence
(AB)(B1A1)A(BB1)A1
AIA1
AA1
 I
Since
(AB)(B1A1) I
Then
(AB)1 (B1A1)
Correctly uses associativity by
regrouping (seen at least once) | AO2.5 | B1
Correctly applies the identity
property throughout and concludes
their rigorous mathematical
argument with no errors or
omissions | AO2.1 | R1
Total | 6
Q | Marking Instructions | AO | Marks | Typical Solution
A student claims:

"Given any two non-zero square matrices, A and B, then $(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$"

\begin{enumerate}[label=(\alph*)]
\item Explain why the student's claim is incorrect giving a counter example.
[2 marks]

\item Refine the student's claim to make it fully correct.
[1 mark]

\item Prove that your answer to part (b) is correct.
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2  Q9 [6]}}
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