Standard +0.8 This is a proof by induction involving 3×3 matrices requiring verification of the base case, inductive hypothesis, and inductive step with matrix multiplication. While the pattern is straightforward once M² is computed (each entry becomes 3), the algebraic manipulation of nine entries and formal proof structure elevates this above routine Further Maths questions, though it remains a standard induction exercise without requiring deep insight.
Question 13:
13 | Uses proof by induction and
investigates formula for n = 1 and
n = k (must see evidence of both
n = 1 and n = k being considered) | AO3.1a | M1 | Using induction method,
Let P(n) be the statement
3n−1 3n−1 3n−1
Mn = 3n−1 3n−1 3n−1
3n−1 3n−1 3n−1
For n =1
30 30 30
1 1 1
30 30 30 = 1 1 1 =M1
30 30 30 1 1 1
P(1) is true
∴Assume P(k) is true
Mk+1 =M×Mk
3k−1 3k−1 3k−1
1 1 1
= 1 1 1 × 3k−1 3k−1 3k−1
1 1 1 3k−1 3k−1 3k−1
(since P(k) is true)
(3k−1+3k−1+3k−1) (3k−1+..) (3k−1+..
= (3k−1+3k−1+3k−1) (3k−1+..) (3k−1+..
(3k−1+3k−1+3k−1) (3k−1+..) (3k−1+..
But
3 k−1+3 k−1+3 k−1 = 3×3 k−1
= 3 k
3k 3k 3k
Hence Mk+1= 3k 3k 3k
3k 3k 3k
3(k+1)−1 3(k+1)−1 3(k+1)−1
∴Mk+1 = 3(k+1)−1 3(k+1)−1 3(k+1)−1
3(k+1)−1 3(k+1)−1 3(k+1)−1
P(k +1) is true
Since P(1) is true
a∴nd P(k) P(k + 1) ,
hence, by induction, P(n) is true
n∈⇒�
for all
Demonstrates that formula is true
for n = 1 | AO1.1b | A1
States assumption that formula
true for n = k and uses
Mk+1 =M×Mk | AO2.1 | R1
Deduces that formula is also true
for n = k + 1 from correct working | AO2.2a | R1
Completes a rigorous argument
and explains how their argument
proves the required result. AG | AO2.4 | R1
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution