Let .A=1 a a0 1 b0 0 1 If for some n∈N, An=14821600196001 then n + a + b is equal to ____________

$$A=1$$ a $$a0$$ $$1$$ $$b0$$ $$0$$ $$1$$⇒$$An=1nana1+n-1b201nb001$$ ,$$An=14821600196001$$⇒$$na=48$$,$$nb=96$$ ⇒$$na1+n-1b2=2160$$ ⇒$$481+96-b2=2160$$⇒$$b=8$$ ⇒$$n=12$$,$$a=4$$⇒$$n+a+b=24$$

Q.

Let . $A = [\begin{array}{lll} 1 a a \\ 0 1 b \\ 0 0 1 \end{array}]$ If for some $n \in N$ , $A^{n} = [\begin{array}{ccc} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{array}]$ then n + a + b is equal to ____________

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By Expert Faculty of Sri Chaitanya

answer is 24.

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Detailed Solution

$A = [\begin{array}{lll} 1 a a \\ 0 1 b \\ 0 0 1 \end{array}] \Rightarrow A^{n} = [\begin{matrix} 1 & n a & n a (1 + \frac{(n - 1) b}{2}) \\ 0 & 1 & n b \\ 0 & 0 & 1 \end{matrix}]$ ,

$A^{n} = [\begin{array}{ccc} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{array}] \Rightarrow n a = 48, n b = 96 \Rightarrow n a (1 + \frac{(n - 1) b}{2}) = 2160 \Rightarrow 48 (1 + \frac{96 - b}{2}) = 2160 \Rightarrow b = 8 \Rightarrow n = 12, a = 4 \Rightarrow n + a + b = 24$

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