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 \(n\),
$$\mathbf { A } ^ { n } = \left( \begin{array} { c c }
2 ^ { n } & 3 \left( 2 ^ { n } - 1 \right) \\
0 & 1
\end{array} \right)$$
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Question 2:
Answer Marks
Guidance
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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# 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]}}