If $A = [\begin{matrix} 10 \end{matrix} \begin{matrix} 21 \end{matrix}]$ , then by the principle of Mathematical induction, prove that $A^{n} = [\begin{matrix} 10 \end{matrix} \begin{matrix} 2 n 1 \end{matrix}]$ .

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Solution

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 $n \in N$ .

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If A=[1 0 2 1 ], then by the principle of Mathematical induction, prove that An=[102n1].

Since, A=[1201]Let P(n):An=[12n01]....(i)P(1):A=[12.101]=[1201]∴P(n) is true for n=1.Let P(n) is true for n=k.∴P(k):Ak=[12k01]...(ii)Now, P(k+1):Ak+1=[12(k+1)01]....(iii)∴Ak+1=Ak×A1=[12k01][1201]=[12(k+1)01]∴P(n) is true for n=k+1.Hence, P(n) is true for all n∈N.

If $A = [\begin{matrix} 10 \end{matrix} \begin{matrix} 21 \end{matrix}]$ , then by the principle of Mathematical induction, prove that $A^{n} = [\begin{matrix} 10 \end{matrix} \begin{matrix} 2 n 1 \end{matrix}]$ .