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Principle of Mathematical Induction

Short Answer Type

11.
If A = $[\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}]$ , then by the principle of Mathematical induction, prove that Aⁿ = $[\begin{matrix} 1 & 2 n \\ 0 & 1 \end{matrix}]$

$Since, A = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] Let P (n) : A^{n} = [\begin{matrix} 1 & 2 n \\ 0 & 1 \end{matrix}] . . . (i) P (1) : A = [\begin{matrix} 1 & 2 . 1 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] ∴ P (n) is true for n = 1 . Let P (n) is true for n = k . ∴ P (k) : A^{k} = [\begin{matrix} 1 & 2 k \\ 0 & 1 \end{matrix}] . . . (ii) Now, P (k + 1) : A^{k + 1} = [\begin{matrix} 1 & 2 (k + 1) \\ 0 & 1 \end{matrix}] . . . (iii) ∴ A^{k + 1} = A^{k} \times A^{1} = [\begin{matrix} 1 & 2 k \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 2 (k + 1) \\ 0 & 1 \end{matrix}] ∴ P (n) is true for n = k + 1 . Hence, P (n) is true for all n \in N$