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

Questions

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 ____________

Unlock the full solution & master the concept.

Get a detailed solution and exclusive access to our masterclass to ensure you never miss a concept

By Expert Faculty of Sri Chaitanya

NEW

Ready to Test Your Skills?

Check Your Performance Today with our Free Mock Tests used by Toppers!

detailed solution

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

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