AQA Further Paper 1 2021 June — Question 5 5 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2021
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve matrix power formula
DifficultyStandard +0.8 This is a standard Further Maths proof by induction for matrix powers requiring verification of the base case, assumption of P(k), and proof of P(k+1) through matrix multiplication. While the matrix multiplication involves multiple terms and careful algebraic manipulation (particularly showing 3^{k+1} - 1 = 3(3^k - 1) + 2), it follows a well-established template with no novel insight required. The 5-mark allocation and routine nature place it moderately above average difficulty.
Spec4.01a Mathematical induction: construct proofs4.03b Matrix operations: addition, multiplication, scalar

The matrix M is defined by \(\mathbf{M} = \begin{pmatrix} 3 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) Prove by induction that \(\mathbf{M}^n = \begin{pmatrix} 3^n & 3^n - 1 & -3^n + 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) for all integers \(n \geq 1\) [5 marks]

Question 5:
AnswerMarks
5Demonstrates the result for
n =1 and states that it is true
AnswerMarks Guidance
for n =11.1b B1
 
Let n =1then M1= 0 1 0 =Mso
 
0 0 1
the result is true for n =1
Assume the result if true for n =k:
3 2 −23k 3k −1 −3k +1
  
Mk+1= 0 1 0 0 1 0
  
0 0 1  0 0 1 
 
3k+1 3k+1−1 −3k+1+1
 
= 0 1 0
 
 0 0 1 
 
Hence true for n =k +1
It is true for n =1. If it is true for
then it is true for . Hence true
by induction for all integers n ≥1𝑛𝑛 = 𝑘𝑘
𝑛𝑛 = 𝑘𝑘+1
States the assumption that the
result true for n =k
AnswerMarks Guidance
Condone use of instead of2.4 B1
Writes Mk+1 as M 𝑛𝑛 Mk or MkM 𝑘𝑘
AnswerMarks Guidance
Condone use of instead of3.1a M1
Calculates Mk+1 c 𝑛𝑛 orrectly (full 𝑘𝑘 y
simplified)
AnswerMarks Guidance
Condone use of instead of1.1b A1
Completes a rigo𝑛𝑛rous 𝑘𝑘
argument by stating that
It is true for n =1;
that if it is true for then it
is true for
And hence (by ind𝑛𝑛uc=tio𝑘𝑘n) true
for all integ𝑛𝑛e=rs 𝑘𝑘n+≥11
Do not condone use of
instead of in the inductive
AnswerMarks Guidance
step 𝑛𝑛2.1 R1
𝑘𝑘 Total5
QMarking Instructions AO
Question 5:
5 | Demonstrates the result for
n =1 and states that it is true
for n =1 | 1.1b | B1 | 3 2 −2
 
Let n =1then M1= 0 1 0 =Mso
 
0 0 1
the result is true for n =1
Assume the result if true for n =k:
3 2 −23k 3k −1 −3k +1
  
Mk+1= 0 1 0 0 1 0
  
0 0 1  0 0 1 
 
3k+1 3k+1−1 −3k+1+1
 
= 0 1 0
 
 0 0 1 
 
Hence true for n =k +1
It is true for n =1. If it is true for
then it is true for . Hence true
by induction for all integers n ≥1𝑛𝑛 = 𝑘𝑘
𝑛𝑛 = 𝑘𝑘+1
States the assumption that the
result true for n =k
Condone use of instead of | 2.4 | B1
Writes Mk+1 as M 𝑛𝑛 Mk or MkM 𝑘𝑘
Condone use of instead of | 3.1a | M1
Calculates Mk+1 c 𝑛𝑛 orrectly (full 𝑘𝑘 y
simplified)
Condone use of instead of | 1.1b | A1
Completes a rigo𝑛𝑛rous 𝑘𝑘
argument by stating that
It is true for n =1;
that if it is true for then it
is true for
And hence (by ind𝑛𝑛uc=tio𝑘𝑘n) true
for all integ𝑛𝑛e=rs 𝑘𝑘n+≥11
Do not condone use of
instead of in the inductive
step 𝑛𝑛 | 2.1 | R1
𝑘𝑘 Total | 5
Q | Marking Instructions | AO | Marks | Typical solution
The matrix M is defined by $\mathbf{M} = \begin{pmatrix} 3 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

Prove by induction that $\mathbf{M}^n = \begin{pmatrix} 3^n & 3^n - 1 & -3^n + 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ for all integers $n \geq 1$
[5 marks]

\hfill \mbox{\textit{AQA Further Paper 1 2021 Q5 [5]}}