OCR MEI Further Pure Core AS Specimen — Question 9 14 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
SessionSpecimen
Marks14
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TopicInvariant lines and eigenvalues and vectors
TypeMatrix powers by induction
DifficultyChallenging +1.2 This is a Further Maths AS question on matrix powers and invariant points. Part (i) requires proof by induction (standard technique), part (ii) is routine eigenvalue work, and parts (iii)-(iv) connect these ideas but with clear guidance. While it's multi-step and requires several techniques, the question structure heavily scaffolds the solution, making it moderately above average difficulty but not requiring significant novel insight.
Spec4.01a Mathematical induction: construct proofs4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.03g Invariant points and lines
You are given that matrix \(\mathbf{M} = \begin{pmatrix} -3 & 8 \\ -2 & 5 \end{pmatrix}\).
  1. Prove that, for all positive integers \(n\), \(\mathbf{M}^n = \begin{pmatrix} 1-4n & 8n \\ -2n & 1+4n \end{pmatrix}\). [6]
  2. Determine the equation of the line of invariant points of the transformation represented by the matrix \(\mathbf{M}\). [3]
It is claimed that the answer to part (ii) is also a line of invariant points of the transformation represented by the matrix \(\mathbf{M}^n\), for any positive integer \(n\).
  1. Explain geometrically why this claim is true. [2]
  2. Verify algebraically that this claim is true. [3]
Question 9:
AnswerMarks Guidance
9(i) (cid:167)1(cid:16)4 8 (cid:183) (cid:167)(cid:16)3 8(cid:183)
When n = 1, M1 = (cid:168) (cid:184)(cid:32)(cid:168) (cid:184)
(cid:169) (cid:16)2 1(cid:14)4(cid:185) (cid:169)(cid:16)2 5(cid:185)
(cid:167)1(cid:16)4k 8k (cid:183)
Assume true for n = k: Mk (cid:32)(cid:168) (cid:184)
(cid:169) (cid:16)2k 1(cid:14)4k(cid:185)
(cid:167)1(cid:16)4k 8k (cid:183)(cid:167)(cid:16)3 8(cid:183)
Mk(cid:14)1 (cid:32)Mk (cid:117)M(cid:32)(cid:168) (cid:184)(cid:168) (cid:184)
(cid:169) (cid:16)2k 1(cid:14)4k(cid:185)(cid:169)(cid:16)2 5(cid:185)
(cid:167)(cid:16)3(cid:16)4k 8(cid:14)8k(cid:183)
=
(cid:168) (cid:184)
(cid:169)(cid:16)2(cid:16)2k 5(cid:14)4k(cid:185)
(cid:167)1(cid:16)4(cid:11)k(cid:14)1(cid:12) 8(cid:11)k(cid:14)1(cid:12) (cid:183)
= (cid:168) (cid:184)
(cid:16)2(cid:11)k(cid:14)1(cid:12) 1(cid:14)4(cid:11)k(cid:14)1(cid:12)
(cid:169) (cid:185)
which is the formula with n = k + 1.
Therefore if true for n, true for n + 1.
E
AnswerMarks
True for 1 (cid:159) true for 2, 3,… and all positive integersB1
E1
M1
A1
I
C
A1
E1
AnswerMarks
[6]1.1
2.1
1.1
M
2.1
2.4
AnswerMarks
2.2aN
(cid:167)(cid:16)3 8(cid:183)(cid:167)1(cid:16)4k 8k (cid:183)
OR Mk(cid:14)1(cid:32)M(cid:117)Mk (cid:32)(cid:168) (cid:184)(cid:168) (cid:184)
E (cid:169)(cid:16)2 5(cid:185)(cid:169) (cid:16)2k 1(cid:14)4k(cid:185)
Not required to check both ways round.
Convincingly expressed in
terms of
k + 1
Completion of proof by
induction. Dependent on B1
and previous E1
AnswerMarks Guidance
9(ii) P
(cid:167)(cid:16)3 8(cid:183)(cid:167)x(cid:183) (cid:167)x(cid:183)
(cid:168) (cid:169)(cid:16)2 5 (cid:184) (cid:185) (cid:168) (cid:169)y (cid:184) (cid:185) (cid:32)(cid:168) (cid:169)y (cid:184) (cid:185) S
(cid:159) −3x + 8y = x, −2x + 5y = y
1
(cid:159) y = x
AnswerMarks
2M1
M1
A1
AnswerMarks
[3]1.1a
1.1
AnswerMarks
2.2aUsing Mx(cid:32)x
Obtaining an equation
Any correct form; must state
both equations lead to this
answer.
AnswerMarks Guidance
9(iii) Mn represents the transformation repeated n times.
1
Each repeat leaves points on y = x unchanged.
AnswerMarks
2E1
E1
AnswerMarks
[2]2.4
2.1
AnswerMarks Guidance
9(iv) (cid:167)1(cid:16)4n 8n (cid:183) (cid:167) x (cid:183)
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:169) (cid:16)2n 1(cid:14)4n(cid:185) (cid:169) 1 x (cid:185)
2
(cid:167) x(cid:16)4nx(cid:14)8n(1 x) (cid:183)
2
=(cid:168) (cid:184)
(cid:16)2nx(cid:14) 1 x(cid:14)4n(1 x)
(cid:169) (cid:185)
2 2
(cid:167) x (cid:183)
=(cid:168) (cid:184)
1 x
(cid:169) (cid:185)
2
AnswerMarks
so same invariant line for all n.M1
A1
I
E1
AnswerMarks
[3]1.1
M
1.1
AnswerMarks
2.3N
E
Multiplication with some
progress
cao
AnswerMarks Guidance
QuestionAO1 AO2
1i2 0
1ii2 0
2i2 0
2ii1 1
3i2 0
3ii1 0
4i3 1
4ii0 1
54 0
6i1 0
6ii3 0
7i2 1
7ii2 0
8i0 2
8ii4 1
7
AnswerMarks Guidance
9i2 4
9ii2 1
03
9iii1 1
9iv2 1
Totals36 14
PPMMTT
Y410 Mark Scheme June 20XX
N
E
M
I
C
E
P
S
16
Question 9:
9 | (i) | (cid:167)1(cid:16)4 8 (cid:183) (cid:167)(cid:16)3 8(cid:183)
When n = 1, M1 = (cid:168) (cid:184)(cid:32)(cid:168) (cid:184)
(cid:169) (cid:16)2 1(cid:14)4(cid:185) (cid:169)(cid:16)2 5(cid:185)
(cid:167)1(cid:16)4k 8k (cid:183)
Assume true for n = k: Mk (cid:32)(cid:168) (cid:184)
(cid:169) (cid:16)2k 1(cid:14)4k(cid:185)
(cid:167)1(cid:16)4k 8k (cid:183)(cid:167)(cid:16)3 8(cid:183)
Mk(cid:14)1 (cid:32)Mk (cid:117)M(cid:32)(cid:168) (cid:184)(cid:168) (cid:184)
(cid:169) (cid:16)2k 1(cid:14)4k(cid:185)(cid:169)(cid:16)2 5(cid:185)
(cid:167)(cid:16)3(cid:16)4k 8(cid:14)8k(cid:183)
=
(cid:168) (cid:184)
(cid:169)(cid:16)2(cid:16)2k 5(cid:14)4k(cid:185)
(cid:167)1(cid:16)4(cid:11)k(cid:14)1(cid:12) 8(cid:11)k(cid:14)1(cid:12) (cid:183)
= (cid:168) (cid:184)
(cid:16)2(cid:11)k(cid:14)1(cid:12) 1(cid:14)4(cid:11)k(cid:14)1(cid:12)
(cid:169) (cid:185)
which is the formula with n = k + 1.
Therefore if true for n, true for n + 1.
E
True for 1 (cid:159) true for 2, 3,… and all positive integers | B1
E1
M1
A1
I
C
A1
E1
[6] | 1.1
2.1
1.1
M
2.1
2.4
2.2a | N
(cid:167)(cid:16)3 8(cid:183)(cid:167)1(cid:16)4k 8k (cid:183)
OR Mk(cid:14)1(cid:32)M(cid:117)Mk (cid:32)(cid:168) (cid:184)(cid:168) (cid:184)
E (cid:169)(cid:16)2 5(cid:185)(cid:169) (cid:16)2k 1(cid:14)4k(cid:185)
Not required to check both ways round.
Convincingly expressed in
terms of
k + 1
Completion of proof by
induction. Dependent on B1
and previous E1
9 | (ii) | P
(cid:167)(cid:16)3 8(cid:183)(cid:167)x(cid:183) (cid:167)x(cid:183)
(cid:168) (cid:169)(cid:16)2 5 (cid:184) (cid:185) (cid:168) (cid:169)y (cid:184) (cid:185) (cid:32)(cid:168) (cid:169)y (cid:184) (cid:185) S
(cid:159) −3x + 8y = x, −2x + 5y = y
1
(cid:159) y = x
2 | M1
M1
A1
[3] | 1.1a
1.1
2.2a | Using Mx(cid:32)x
Obtaining an equation
Any correct form; must state
both equations lead to this
answer.
9 | (iii) | Mn represents the transformation repeated n times.
1
Each repeat leaves points on y = x unchanged.
2 | E1
E1
[2] | 2.4
2.1
9 | (iv) | (cid:167)1(cid:16)4n 8n (cid:183) (cid:167) x (cid:183)
(cid:168) (cid:184) (cid:168) (cid:184)
(cid:169) (cid:16)2n 1(cid:14)4n(cid:185) (cid:169) 1 x (cid:185)
2
(cid:167) x(cid:16)4nx(cid:14)8n(1 x) (cid:183)
2
=(cid:168) (cid:184)
(cid:16)2nx(cid:14) 1 x(cid:14)4n(1 x)
(cid:169) (cid:185)
2 2
(cid:167) x (cid:183)
=(cid:168) (cid:184)
1 x
(cid:169) (cid:185)
2
so same invariant line for all n. | M1
A1
I
E1
[3] | 1.1
M
1.1
2.3 | N
E
Multiplication with some
progress
cao
Question | AO1 | AO2 | AO3(PS) | AO3(M) | Total
1i | 2 | 0 | 0 | 0 | 2
1ii | 2 | 0 | 0 | 0 | 2
2i | 2 | 0 | 0 | 0 | 2
2ii | 1 | 1 | 0 | 0 | 2
3i | 2 | 0 | 0 | 0 | 2
3ii | 1 | 0 | 1 | 0 | 2
4i | 3 | 1 | 0 | 0 | 4
4ii | 0 | 1 | 1 | 0 | 2
5 | 4 | 0 | 3 | 0 | 7
6i | 1 | 0 | 0 | 0 | 1
6ii | 3 | 0 | 1 | 0 | 4
7i | 2 | 1 | 1 | 0 | 4
7ii | 2 | 0 | 1 | 0 | 3
8i | 0 | 2 | 0 | 0 | 2
8ii | 4 | 1 | 2 | 0 | N
7
9i | 2 | 4 | 0 | 0 | 6
9ii | 2 | 1 | 0 | E
0 | 3
9iii | 1 | 1 | 0 | 0 | 2
9iv | 2 | 1 | 0 | 0 | 3
Totals | 36 | 14 | 10 | M | 0 | 60
PPMMTT
Y410 Mark Scheme June 20XX
N
E
M
I
C
E
P
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16
You are given that matrix $\mathbf{M} = \begin{pmatrix} -3 & 8 \\ -2 & 5 \end{pmatrix}$.

\begin{enumerate}[label=(\roman*)]
\item Prove that, for all positive integers $n$, $\mathbf{M}^n = \begin{pmatrix} 1-4n & 8n \\ -2n & 1+4n \end{pmatrix}$. [6]

\item Determine the equation of the line of invariant points of the transformation represented by the matrix $\mathbf{M}$. [3]
\end{enumerate}

It is claimed that the answer to part (ii) is also a line of invariant points of the transformation represented by the matrix $\mathbf{M}^n$, for any positive integer $n$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{2}
\item Explain geometrically why this claim is true. [2]

\item Verify algebraically that this claim is true. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core AS  Q9 [14]}}
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