CAIE FP1 2011 June — Question 2 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve matrix power formula
DifficultyStandard +0.3 This is a straightforward proof by induction with matrix multiplication. The base case is trivial, and the inductive step requires multiplying two 2×2 matrices and simplifying using standard index laws (2^k × 2 = 2^(k+1)). The matrix structure is simple (upper triangular with a zero row), making the algebra routine. This is a standard textbook exercise testing the mechanical application of induction to matrices, slightly easier than average due to its predictable structure.
Spec4.01a Mathematical induction: construct proofs4.03b Matrix operations: addition, multiplication, scalar

2 Let \(\mathbf { A } = \left( \begin{array} { l l } 2 & 3 \\ 0 & 1 \end{array} \right)\). Prove by mathematical induction that, for every positive integer \(n\), $$\mathbf { A } ^ { n } = \left( \begin{array} { c c } 2 ^ { n } & 3 \left( 2 ^ { n } - 1 \right) \\ 0 & 1 \end{array} \right)$$

Question 2:
AnswerMarks Guidance
Working/AnswerMarks Guidance
Let \(P_n\): \(\mathbf{A} = \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix} \Rightarrow \mathbf{A}^n = \begin{pmatrix} 2^n & 3(2^n-1) \\ 0 & 1 \end{pmatrix}\) States proposition
\(\mathbf{A}^1 = \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2^1 & 3(2-1) \\ 0 & 1 \end{pmatrix} \Rightarrow P_1\) is trueB1 Shows base case
Assume \(P_k\) is true for some integer \(k\)B1
\(\mathbf{A}^{k+1} = \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2^k & 3(2^k-1) \\ 0 & 1 \end{pmatrix}\)M1 Proves inductive step
\(= \begin{pmatrix} 2^{k+1} & 3.2(2^k-1)+3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2^{k+1} & 3(2^{k+1}-1) \\ 0 & 1 \end{pmatrix}\)A1
Since \(P_1\) true and \(P_k \Rightarrow P_{k+1}\), hence by PMI \(P_n\) is true \(\forall\) positive integers \(n\)A1 States conclusion
# Question 2:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Let $P_n$: $\mathbf{A} = \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix} \Rightarrow \mathbf{A}^n = \begin{pmatrix} 2^n & 3(2^n-1) \\ 0 & 1 \end{pmatrix}$ | — | States proposition |
| $\mathbf{A}^1 = \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2^1 & 3(2-1) \\ 0 & 1 \end{pmatrix} \Rightarrow P_1$ is true | B1 | Shows base case |
| Assume $P_k$ is true for some integer $k$ | B1 | |
| $\mathbf{A}^{k+1} = \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2^k & 3(2^k-1) \\ 0 & 1 \end{pmatrix}$ | M1 | Proves inductive step |
| $= \begin{pmatrix} 2^{k+1} & 3.2(2^k-1)+3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2^{k+1} & 3(2^{k+1}-1) \\ 0 & 1 \end{pmatrix}$ | A1 | |
| Since $P_1$ true and $P_k \Rightarrow P_{k+1}$, hence by PMI $P_n$ is true $\forall$ positive integers $n$ | A1 | States conclusion |

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2 Let $\mathbf { A } = \left( \begin{array} { l l } 2 & 3 \\ 0 & 1 \end{array} \right)$. Prove by mathematical induction that, for every positive integer $n$,

$$\mathbf { A } ^ { n } = \left( \begin{array} { c c } 
2 ^ { n } & 3 \left( 2 ^ { n } - 1 \right) \\
0 & 1
\end{array} \right)$$

\hfill \mbox{\textit{CAIE FP1 2011 Q2 [5]}}