OCR MEI Further Pure Core AS 2024 June — Question 6 9 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2024
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMatrices
TypeMatrix powers and patterns
DifficultyStandard +0.8 This is a Further Maths question requiring proof by induction for matrix powers (non-trivial as the pattern must be verified), followed by conceptual understanding of matrix definitions. The induction requires matrix multiplication and algebraic manipulation, while part (b) tests understanding of identity and inverse matrices beyond routine calculation. Moderately challenging for Further Maths students.
Spec4.01a Mathematical induction: construct proofs4.03o Inverse 3x3 matrix
6 You are given that \(\mathbf { M } = \left( \begin{array} { l l } 4 & - 9 \\ 1 & - 2 \end{array} \right)\).
  1. Prove that \(\mathbf { M } ^ { n } = \left( \begin{array} { c c } 1 + 3 n & - 9 n \\ n & 1 - 3 n \end{array} \right)\) for all positive integers \(n\).
  2. A student thinks that this formula, when \(n = 0\) and \(n = - 1\), gives the identity matrix and the inverse matrix \(\mathbf { M } ^ { - 1 }\) respectively. Determine whether the student is correct.
Question 6:
AnswerMarks Guidance
6(a) 1+3 −9  4 −9
When n = 1, M1=   =   as required
 1 1−3 1 −2
1+3k −9k 
Assume true for n = k, so Mk =  
 k 1−3k
1+3k −9k 4 −9
Mk+1=
  
 k 1−3k1 −2
 4 + 3 k − 9 − 9 k 
=
k + 1 − 2 − 3 k
 1 + 3k k (+ + 1 ) − 9 ( k + 1 ) 
= so true for n = k+1
1 1 − 3 ( k + 1 )
As true for n = 1, and if true for n = k then true for
AnswerMarks
n = k+1, true for all positive integers nB1
M1
M1
A1
A1
A1
AnswerMarks
[6]2.1
2.1
2.1
1.1
2.2a
AnswerMarks
2.4check true for n = 1
cao dep previous A1
AnswerMarks Guidance
6(b)  1 0 
n = 0 gives = I so true for n = 0
0 1
 − 2 9 
Formula with n = −1 gives
− 1 4
 − 2 9 
det M = 4  −2 − (−9)  1 = 1, so M − 1 =
− 1 4
AnswerMarks
so true for n = −1B1
B1
B1
AnswerMarks
[3]2.3
1.1
AnswerMarks
2.3[ = I may be inferred from ‘student is correct’]
Must show evidence for inverse
e.g. calculate det or show MM−1 = I
Question 6:
6 | (a) | 1+3 −9  4 −9
When n = 1, M1=   =   as required
 1 1−3 1 −2
1+3k −9k 
Assume true for n = k, so Mk =  
 k 1−3k
1+3k −9k 4 −9
Mk+1=
  
 k 1−3k1 −2
 4 + 3 k − 9 − 9 k 
=
k + 1 − 2 − 3 k
 1 + 3k k (+ + 1 ) − 9 ( k + 1 ) 
= so true for n = k+1
1 1 − 3 ( k + 1 )
As true for n = 1, and if true for n = k then true for
n = k+1, true for all positive integers n | B1
M1
M1
A1
A1
A1
[6] | 2.1
2.1
2.1
1.1
2.2a
2.4 | check true for n = 1
cao dep previous A1
6 | (b) |  1 0 
n = 0 gives = I so true for n = 0
0 1
 − 2 9 
Formula with n = −1 gives
− 1 4
 − 2 9 
det M = 4  −2 − (−9)  1 = 1, so M − 1 =
− 1 4
so true for n = −1 | B1
B1
B1
[3] | 2.3
1.1
2.3 | [ = I may be inferred from ‘student is correct’]
Must show evidence for inverse
e.g. calculate det or show MM−1 = I
6 You are given that $\mathbf { M } = \left( \begin{array} { l l } 4 & - 9 \\ 1 & - 2 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item Prove that $\mathbf { M } ^ { n } = \left( \begin{array} { c c } 1 + 3 n & - 9 n \\ n & 1 - 3 n \end{array} \right)$ for all positive integers $n$.
\item A student thinks that this formula, when $n = 0$ and $n = - 1$, gives the identity matrix and the inverse matrix $\mathbf { M } ^ { - 1 }$ respectively.

Determine whether the student is correct.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2024 Q6 [9]}}
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