AQA Further AS Paper 1 2019 June — Question 12 12 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2019
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind invariant lines through origin
DifficultyStandard +0.8 This is a Further Maths question combining matrix induction (routine but requiring careful algebraic manipulation), finding invariant lines (requires solving eigenvalue problems or analyzing the transformation geometrically), and invariant points (solving (A-I)v=0). Part (a) is standard induction, but parts (b) and (c) require deeper understanding of linear transformations. The 12-mark total and multi-concept nature place it above average difficulty, though the techniques are within standard Further Maths syllabus.
Spec4.01a Mathematical induction: construct proofs4.03g Invariant points and lines
The matrix \(\mathbf{A}\) is given by $$\mathbf{A} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$$
  1. Prove by induction that, for all integers \(n \geq 1\), $$\mathbf{A}^n = \begin{bmatrix} 1 & 3^n - 1 \\ 0 & 3^n \end{bmatrix}$$ [4 marks]
  2. Find all invariant lines under the transformation matrix \(\mathbf{A}\). Fully justify your answer. [6 marks]
  3. Find a line of invariant points under the transformation matrix \(\mathbf{A}\). [2 marks]
Question 12:
AnswerMarks
12(a)Demonstrates the rule is correct for
and states that it is true for (may appear at any stage).
AnswerMarks Guidance
𝑛𝑛 = 11.1b B1
1 1
1 2 1 2 1 3 βˆ’1 1 2
𝑛𝑛 = 1 οΏ½ οΏ½ = οΏ½true fοΏ½or οΏ½ 1 οΏ½ = οΏ½ οΏ½
0 3 0 3 0 3 0 3
Assume true for
∴ 𝑛𝑛 = 1
𝑛𝑛 = π‘˜π‘˜
π‘˜π‘˜
π‘˜π‘˜
1 3 βˆ’1
∴ 𝐀𝐀 = οΏ½ π‘˜π‘˜ οΏ½
0 3
π‘˜π‘˜
π‘˜π‘˜ 1 3 βˆ’1 1 2
β‡’ 𝐀𝐀 ×𝐀𝐀 = οΏ½ π‘˜π‘˜ οΏ½Γ—οΏ½ οΏ½
0 3 0 3
π‘˜π‘˜
π‘˜π‘˜+1 1 2+3οΏ½3 βˆ’1οΏ½
β‡’ 𝐀𝐀 = οΏ½ π‘˜π‘˜ οΏ½
0 3 Γ—3
π‘˜π‘˜+1
1 3 βˆ’1
= οΏ½ π‘˜π‘˜+1 οΏ½
0 3
it is also true for
True for , and true for true for
∴ 𝑛𝑛 = π‘˜π‘˜+1
Then, by induction, it is true for all integers
𝑛𝑛 = 1 𝑛𝑛 =π‘˜π‘˜ β‡’ 𝑛𝑛 = π‘˜π‘˜+1
𝑛𝑛 β‰₯ 1
𝑛𝑛 = 1
Multiplies and
π‘˜π‘˜
1 3 βˆ’1 1 2
Accept anοΏ½y letter iπ‘˜π‘˜n pοΏ½lace oοΏ½f (cοΏ½ondone ).
AnswerMarks Guidance
0 3 0 32.4 M1
π‘˜π‘˜ 𝑛𝑛
Obtains from multiplying and
π‘˜π‘˜+1 π‘˜π‘˜
1 3 βˆ’1 1 3 βˆ’1 1 2
Must inclοΏ½ude an π‘˜π‘˜in+t1ermοΏ½ediate step for thοΏ½e top rigπ‘˜π‘˜ht οΏ½elemeοΏ½nt. οΏ½
AnswerMarks Guidance
0 3 0 3 0 32.2a A1
Completes a rigorous argument and explains how their
argument proves the required result.
e.g. states β€œassume that the rule is true for ” (or equivalent)
and β€œalso true for ” (or equivalent) and β€œfor all ”
𝑛𝑛 = π‘˜π‘˜
and includes the base case with a conclusion.
𝑛𝑛 = π‘˜π‘˜+1 𝑛𝑛
Do not accept the use of in place of .
NMS scores 0/4
AnswerMarks Guidance
𝑛𝑛 π‘˜π‘˜2.1 R1
QMarking instructions AO
AnswerMarks
12(b)Sets up two equations in and its image
β€²
AnswerMarks Guidance
(π‘₯π‘₯,𝑦𝑦) (π‘₯π‘₯ ,𝑦𝑦′)3.1a M1
οΏ½ οΏ½οΏ½ aοΏ½n=d οΏ½ οΏ½
0 3 𝑦𝑦 𝑦𝑦′
β€² β€²
π‘₯π‘₯ = π‘₯π‘₯+ 2 a𝑦𝑦nd 𝑦𝑦 = 3𝑦𝑦
β€² β€²
π‘₯π‘₯ = π‘₯π‘₯+2(π‘šπ‘šπ‘₯π‘₯+𝑐𝑐) π‘šπ‘šπ‘₯π‘₯ +𝑐𝑐 = 3(π‘šπ‘šπ‘₯π‘₯+𝑐𝑐)
π‘šπ‘š(π‘₯π‘₯+2π‘šπ‘šπ‘₯π‘₯+2𝑐𝑐)+𝑐𝑐 ≑3π‘šπ‘šπ‘₯π‘₯+3𝑐𝑐
and
2
π‘šπ‘š+2π‘šπ‘š = 3π‘šπ‘š and 2π‘šπ‘šπ‘π‘+𝑐𝑐 = 3𝑐𝑐
or and or
π‘šπ‘š(π‘šπ‘šβˆ’1)= 0 𝑐𝑐(π‘šπ‘šβˆ’1)= 0
or
π‘šπ‘š = 0 π‘šπ‘š = 1 𝑐𝑐 = 0 π‘šπ‘š = 1
Invariant lines are and
𝑦𝑦 = 0π‘₯π‘₯+0 𝑦𝑦 = 1π‘₯π‘₯+𝑐𝑐
AnswerMarks Guidance
Correctly substitutes and1.1b A1
β€² β€²
Eliminates one variab𝑦𝑦le= toπ‘šπ‘š leπ‘₯π‘₯a+ve𝑐𝑐 an eq𝑦𝑦ua=tioπ‘šπ‘šnπ‘₯π‘₯ in +𝑐𝑐 and just one other
AnswerMarks Guidance
variable.1.1a M1
π‘šπ‘š,𝑐𝑐
AnswerMarks Guidance
Compares coefficients to produce two correct equations in and1.1b A1
Gives or as invariant lines. π‘šπ‘š 𝑐𝑐
AnswerMarks Guidance
Condone other incorrect invariant lines.1.1b B1
𝑦𝑦 = 0 𝑦𝑦 = π‘₯π‘₯+𝑐𝑐
Gives and as invariant lines, with no incorrect invariant
lines.
𝑦𝑦 = 0 𝑦𝑦 = π‘₯π‘₯+𝑐𝑐
AnswerMarks Guidance
NMS can score 2/62.2a B1
𝑦𝑦 = 0 𝑦𝑦 = π‘₯π‘₯+𝑐𝑐
Q12b: Alternative mark scheme for students who assume that all invariant lines pass through the origin – max 3 marks
Q12b
AnswerMarks Guidance
ALTSets up two equations in and its image
β€²3.1a M1
οΏ½ οΏ½οΏ½ aοΏ½n=d οΏ½ οΏ½
0 3 𝑦𝑦 𝑦𝑦′
β€² β€²
and
π‘₯π‘₯ = π‘₯π‘₯+2𝑦𝑦 𝑦𝑦 = 3𝑦𝑦
β€² β€²
π‘₯π‘₯ = π‘₯π‘₯+2(π‘šπ‘šπ‘₯π‘₯) π‘šπ‘šπ‘₯π‘₯ = 3(π‘šπ‘šπ‘₯π‘₯)
π‘šπ‘š(π‘₯π‘₯+2π‘šπ‘šπ‘₯π‘₯)≑ 3π‘šπ‘š π‘₯π‘₯
2
π‘šπ‘š+2π‘šπ‘š = 3π‘šπ‘š
or
π‘šπ‘š(π‘šπ‘šβˆ’1)= 0
or
π‘šπ‘š = 0 π‘šπ‘š = 1
Invariant lines are and
𝑦𝑦 = 0π‘₯π‘₯ 𝑦𝑦 = 1π‘₯π‘₯
(π‘₯π‘₯,𝑦𝑦) (π‘₯π‘₯ ,𝑦𝑦′)
Eliminates three variables to leave an equation in and just one other
AnswerMarks Guidance
variable.1.1a M1
π‘šπ‘š
Gives and as invariant lines,
with no other incorrect invariant lines.
𝑦𝑦 = 0 𝑦𝑦 = π‘₯π‘₯
AnswerMarks Guidance
NMS can score 1/32.2a B1
QMarking instructions AO
AnswerMarks Guidance
12(c)Sets up at least one correct equation in and
Accept alternative variables for this mark.1.1a M1
π‘₯π‘₯ = π‘₯π‘₯+2𝑦𝑦 𝑦𝑦 =3𝑦𝑦
π‘₯π‘₯ 𝑦𝑦
Gives as the only line of invariant points.
AnswerMarks Guidance
NMS can score 2/21.1b A1
𝑦𝑦 = 0
AnswerMarks Guidance
Total12 𝑦𝑦 = 0
QMarking instructions AO 
Question 12:
--- 12(a) ---
12(a) | Demonstrates the rule is correct for
and states that it is true for (may appear at any stage).
𝑛𝑛 = 1 | 1.1b | B1 | Try : and
1 1
1 2 1 2 1 3 βˆ’1 1 2
𝑛𝑛 = 1 οΏ½ οΏ½ = οΏ½true fοΏ½or οΏ½ 1 οΏ½ = οΏ½ οΏ½
0 3 0 3 0 3 0 3
Assume true for
∴ 𝑛𝑛 = 1
𝑛𝑛 = π‘˜π‘˜
π‘˜π‘˜
π‘˜π‘˜
1 3 βˆ’1
∴ 𝐀𝐀 = οΏ½ π‘˜π‘˜ οΏ½
0 3
π‘˜π‘˜
π‘˜π‘˜ 1 3 βˆ’1 1 2
β‡’ 𝐀𝐀 ×𝐀𝐀 = οΏ½ π‘˜π‘˜ οΏ½Γ—οΏ½ οΏ½
0 3 0 3
π‘˜π‘˜
π‘˜π‘˜+1 1 2+3οΏ½3 βˆ’1οΏ½
β‡’ 𝐀𝐀 = οΏ½ π‘˜π‘˜ οΏ½
0 3 Γ—3
π‘˜π‘˜+1
1 3 βˆ’1
= οΏ½ π‘˜π‘˜+1 οΏ½
0 3
it is also true for
True for , and true for true for
∴ 𝑛𝑛 = π‘˜π‘˜+1
Then, by induction, it is true for all integers
𝑛𝑛 = 1 𝑛𝑛 =π‘˜π‘˜ β‡’ 𝑛𝑛 = π‘˜π‘˜+1
𝑛𝑛 β‰₯ 1
𝑛𝑛 = 1
Multiplies and
π‘˜π‘˜
1 3 βˆ’1 1 2
Accept anοΏ½y letter iπ‘˜π‘˜n pοΏ½lace oοΏ½f (cοΏ½ondone ).
0 3 0 3 | 2.4 | M1
π‘˜π‘˜ 𝑛𝑛
Obtains from multiplying and
π‘˜π‘˜+1 π‘˜π‘˜
1 3 βˆ’1 1 3 βˆ’1 1 2
Must inclοΏ½ude an π‘˜π‘˜in+t1ermοΏ½ediate step for thοΏ½e top rigπ‘˜π‘˜ht οΏ½elemeοΏ½nt. οΏ½
0 3 0 3 0 3 | 2.2a | A1
Completes a rigorous argument and explains how their
argument proves the required result.
e.g. states β€œassume that the rule is true for ” (or equivalent)
and β€œalso true for ” (or equivalent) and β€œfor all ”
𝑛𝑛 = π‘˜π‘˜
and includes the base case with a conclusion.
𝑛𝑛 = π‘˜π‘˜+1 𝑛𝑛
Do not accept the use of in place of .
NMS scores 0/4
𝑛𝑛 π‘˜π‘˜ | 2.1 | R1
Q | Marking instructions | AO | Marks | Typical solution
--- 12(b) ---
12(b) | Sets up two equations in and its image
β€²
(π‘₯π‘₯,𝑦𝑦) (π‘₯π‘₯ ,𝑦𝑦′) | 3.1a | M1 | 1 2 π‘₯π‘₯ π‘₯π‘₯β€²
οΏ½ οΏ½οΏ½ aοΏ½n=d οΏ½ οΏ½
0 3 𝑦𝑦 𝑦𝑦′
β€² β€²
π‘₯π‘₯ = π‘₯π‘₯+ 2 a𝑦𝑦nd 𝑦𝑦 = 3𝑦𝑦
β€² β€²
π‘₯π‘₯ = π‘₯π‘₯+2(π‘šπ‘šπ‘₯π‘₯+𝑐𝑐) π‘šπ‘šπ‘₯π‘₯ +𝑐𝑐 = 3(π‘šπ‘šπ‘₯π‘₯+𝑐𝑐)
π‘šπ‘š(π‘₯π‘₯+2π‘šπ‘šπ‘₯π‘₯+2𝑐𝑐)+𝑐𝑐 ≑3π‘šπ‘šπ‘₯π‘₯+3𝑐𝑐
and
2
π‘šπ‘š+2π‘šπ‘š = 3π‘šπ‘š and 2π‘šπ‘šπ‘π‘+𝑐𝑐 = 3𝑐𝑐
or and or
π‘šπ‘š(π‘šπ‘šβˆ’1)= 0 𝑐𝑐(π‘šπ‘šβˆ’1)= 0
or
π‘šπ‘š = 0 π‘šπ‘š = 1 𝑐𝑐 = 0 π‘šπ‘š = 1
Invariant lines are and
𝑦𝑦 = 0π‘₯π‘₯+0 𝑦𝑦 = 1π‘₯π‘₯+𝑐𝑐
Correctly substitutes and | 1.1b | A1
β€² β€²
Eliminates one variab𝑦𝑦le= toπ‘šπ‘š leπ‘₯π‘₯a+ve𝑐𝑐 an eq𝑦𝑦ua=tioπ‘šπ‘šnπ‘₯π‘₯ in +𝑐𝑐 and just one other
variable. | 1.1a | M1
π‘šπ‘š,𝑐𝑐
Compares coefficients to produce two correct equations in and | 1.1b | A1
Gives or as invariant lines. π‘šπ‘š 𝑐𝑐
Condone other incorrect invariant lines. | 1.1b | B1
𝑦𝑦 = 0 𝑦𝑦 = π‘₯π‘₯+𝑐𝑐
Gives and as invariant lines, with no incorrect invariant
lines.
𝑦𝑦 = 0 𝑦𝑦 = π‘₯π‘₯+𝑐𝑐
NMS can score 2/6 | 2.2a | B1
𝑦𝑦 = 0 𝑦𝑦 = π‘₯π‘₯+𝑐𝑐
Q12b: Alternative mark scheme for students who assume that all invariant lines pass through the origin – max 3 marks
Q12b
ALT | Sets up two equations in and its image
β€² | 3.1a | M1 | 1 2 π‘₯π‘₯ π‘₯π‘₯β€²
οΏ½ οΏ½οΏ½ aοΏ½n=d οΏ½ οΏ½
0 3 𝑦𝑦 𝑦𝑦′
β€² β€²
and
π‘₯π‘₯ = π‘₯π‘₯+2𝑦𝑦 𝑦𝑦 = 3𝑦𝑦
β€² β€²
π‘₯π‘₯ = π‘₯π‘₯+2(π‘šπ‘šπ‘₯π‘₯) π‘šπ‘šπ‘₯π‘₯ = 3(π‘šπ‘šπ‘₯π‘₯)
π‘šπ‘š(π‘₯π‘₯+2π‘šπ‘šπ‘₯π‘₯)≑ 3π‘šπ‘š π‘₯π‘₯
2
π‘šπ‘š+2π‘šπ‘š = 3π‘šπ‘š
or
π‘šπ‘š(π‘šπ‘šβˆ’1)= 0
or
π‘šπ‘š = 0 π‘šπ‘š = 1
Invariant lines are and
𝑦𝑦 = 0π‘₯π‘₯ 𝑦𝑦 = 1π‘₯π‘₯
(π‘₯π‘₯,𝑦𝑦) (π‘₯π‘₯ ,𝑦𝑦′)
Eliminates three variables to leave an equation in and just one other
variable. | 1.1a | M1
π‘šπ‘š
Gives and as invariant lines,
with no other incorrect invariant lines.
𝑦𝑦 = 0 𝑦𝑦 = π‘₯π‘₯
NMS can score 1/3 | 2.2a | B1
Q | Marking instructions | AO | Marks | Typical solution
--- 12(c) ---
12(c) | Sets up at least one correct equation in and
Accept alternative variables for this mark. | 1.1a | M1 | and
π‘₯π‘₯ = π‘₯π‘₯+2𝑦𝑦 𝑦𝑦 =3𝑦𝑦
π‘₯π‘₯ 𝑦𝑦
Gives as the only line of invariant points.
NMS can score 2/2 | 1.1b | A1
𝑦𝑦 = 0
Total | 12 | 𝑦𝑦 = 0
Q | Marking instructions | AO | Marks | Typical solution
The matrix $\mathbf{A}$ is given by
$$\mathbf{A} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$$

\begin{enumerate}[label=(\alph*)]
\item Prove by induction that, for all integers $n \geq 1$,
$$\mathbf{A}^n = \begin{bmatrix} 1 & 3^n - 1 \\ 0 & 3^n \end{bmatrix}$$ [4 marks]

\item Find all invariant lines under the transformation matrix $\mathbf{A}$.

Fully justify your answer. [6 marks]

\item Find a line of invariant points under the transformation matrix $\mathbf{A}$. [2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2019 Q12 [12]}}
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