OCR MEI Further Pure Core AS 2023 June — Question 6 8 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2023
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeMatrix powers by induction
DifficultyStandard +0.3 This is a straightforward matrix powers question requiring calculation of low powers, pattern recognition, and proof by induction. While it involves multiple steps, the matrix multiplication is routine, the pattern is clear from the given powers, and the induction proof follows a standard template with no conceptual obstacles. Slightly easier than average for Further Maths AS.
Spec4.01a Mathematical induction: construct proofs4.03b Matrix operations: addition, multiplication, scalar
6 The matrix \(\mathbf { M }\) is \(\left( \begin{array} { r r } 2 & 1 \\ - 1 & 0 \end{array} \right)\).
  1. Calculate \(\mathbf { M } ^ { 2 } , \mathbf { M } ^ { 3 }\) and \(\mathbf { M } ^ { 4 }\).
  2. Hence make a conjecture about the matrix \(\mathbf { M } ^ { n }\).
  3. Prove your conjecture.
Question 6:
AnswerMarks Guidance
6(a)  3 2  4 3  5 4
2= 3= 4=
M , M , M 
AnswerMarks Guidance
−2 −1 −3 −2 −4 −3B2
[2]1.1 Allow B1 if M2 correct
6(b)  n + 1 n 
M n =
AnswerMarks Guidance
− n − n + 1B1
[1]1.1 oe – allow correct unsimplified expressions
allow if correct expression is seen in part (c)
AnswerMarks Guidance
6(c)  2 1    1 + 1 1  
n = 1 : M 1 = = so true
− 1 0 − 1 − 1 + 1
k+1 k 
Assume true for n = k, so Mk = 
−k −k+1
 k + 1 k   2 1 
M k + 1 =
− k − k + 1 − 1 0
 k + 2 k + 1 
=
− k − 1 − k
 k ( + 1 ) + 1 k + 1 
= = So true for n = k + 1
− ( k + 1 ) − ( k + 1 ) + 1
As true for n = 1, and if true for n = k then true for
AnswerMarks
n = k + 1, true for all n.B1
M1
A1
A1
B1
AnswerMarks
[5]2.1
2.1
1.1
AnswerMarks
2.2aor M  Mk
or using target expression
dep first three marks gained. Must have if … then… (oe)
Question 6:
6 | (a) |  3 2  4 3  5 4
2= 3= 4=
M , M , M 
−2 −1 −3 −2 −4 −3 | B2
[2] | 1.1 | Allow B1 if M2 correct
6 | (b) |  n + 1 n 
M n =
− n − n + 1 | B1
[1] | 1.1 | oe – allow correct unsimplified expressions
allow if correct expression is seen in part (c)
6 | (c) |  2 1    1 + 1 1  
n = 1 : M 1 = = so true
− 1 0 − 1 − 1 + 1
k+1 k 
Assume true for n = k, so Mk = 
−k −k+1
 k + 1 k   2 1 
M k + 1 =
− k − k + 1 − 1 0
 k + 2 k + 1 
=
− k − 1 − k
 k ( + 1 ) + 1 k + 1 
= = So true for n = k + 1
− ( k + 1 ) − ( k + 1 ) + 1
As true for n = 1, and if true for n = k then true for
n = k + 1, true for all n. | B1
M1
A1
A1
B1
[5] | 2.1
2.1
1.1
2.2a | or M  Mk
or using target expression
dep first three marks gained. Must have if … then… (oe)
6 The matrix $\mathbf { M }$ is $\left( \begin{array} { r r } 2 & 1 \\ - 1 & 0 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item Calculate $\mathbf { M } ^ { 2 } , \mathbf { M } ^ { 3 }$ and $\mathbf { M } ^ { 4 }$.
\item Hence make a conjecture about the matrix $\mathbf { M } ^ { n }$.
\item Prove your conjecture.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2023 Q6 [8]}}
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