OCR MEI Further Pure Core 2019 June — Question 3 7 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
Topic3x3 Matrices
TypeMatrix properties verification
DifficultyModerate -0.8 This is a straightforward verification question requiring routine matrix operations (multiplication, finding inverses, checking commutativity). All techniques are standard A-level procedures with no problem-solving insight needed. The presence of parameter k adds minimal complexity. Easier than average as it's purely mechanical computation.
Spec4.03c Matrix multiplication: properties (associative, not commutative)4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1)
3 Matrices \(\mathbf { A }\) and \(\mathbf { B }\) are defined by \(\mathbf { A } = \left( \begin{array} { l l } 3 & 1 \\ 2 & 1 \end{array} \right)\) and \(\mathbf { B } = \left( \begin{array} { l l } k & 1 \\ 2 & 0 \end{array} \right)\), where \(k\) is a constant.
  1. Verify the result \(( \mathbf { A B } ) ^ { - 1 } = \mathbf { B } ^ { - 1 } \mathbf { A } ^ { - 1 }\) in this case.
  2. Investigate whether \(\mathbf { A }\) and \(\mathbf { B }\) are commutative under matrix multiplication.
Question 3:
AnswerMarks Guidance
3(a) 3 1 k 1  3k2 3
AB 
    
2 12 0 2k2 2
1 2 3 
(AB)1 
 
22k2 3k2
 1 1
A 1   
2 3
10 1 
B 1   
22 k
1 2 3 
B 1A 1   
22k2 3k2
AnswerMarks
[so (AB)–1 = B–1A–1]B1
B1ft
B1
B1
B1
AnswerMarks
[5]1.1b
1.1b
1.1b
1.1b
AnswerMarks Guidance
2.2aft their AB provided det ≠ 0 [isw]
3(b) 3k2 k1
BA
 
 6 2
AnswerMarks
AB = BA when k = 2 [and not otherwise]B1
B1
AnswerMarks
[2]1.1b
2.3
Question 3:
3 | (a) | 3 1 k 1  3k2 3
AB 
    
2 12 0 2k2 2
1 2 3 
(AB)1 
 
22k2 3k2
 1 1
A 1   
2 3
10 1 
B 1   
22 k
1 2 3 
B 1A 1   
22k2 3k2
[so (AB)–1 = B–1A–1] | B1
B1ft
B1
B1
B1
[5] | 1.1b
1.1b
1.1b
1.1b
2.2a | ft their AB provided det ≠ 0 | [isw]
3 | (b) | 3k2 k1
BA
 
 6 2
AB = BA when k = 2 [and not otherwise] | B1
B1
[2] | 1.1b
2.3
3 Matrices $\mathbf { A }$ and $\mathbf { B }$ are defined by $\mathbf { A } = \left( \begin{array} { l l } 3 & 1 \\ 2 & 1 \end{array} \right)$ and $\mathbf { B } = \left( \begin{array} { l l } k & 1 \\ 2 & 0 \end{array} \right)$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Verify the result $( \mathbf { A B } ) ^ { - 1 } = \mathbf { B } ^ { - 1 } \mathbf { A } ^ { - 1 }$ in this case.
\item Investigate whether $\mathbf { A }$ and $\mathbf { B }$ are commutative under matrix multiplication.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2019 Q3 [7]}}
This paper (17 questions)
View full paper
Q1 4 Q2 3 Q3 7 Q4 3 Q5 5 Q6 4 Q7 8 Q8 8 Q9 7 Q10 8 Q11 12 Q12 9 Q13 11 Q14 13 Q15 8 Q16 12 Q17 22