Raising The Stakes

By Cayley Hamilton theorem , we can show that $A^2- 6A + 9 \; \mathbb I_2 =\mathbb O_2$ , where $\mathbb I_2$ denotes the identity matrix of order 2 and $\mathbb O_2$ denotes the zero matrix of order 2. Then $(A-3\; \mathbb I_2 )^2 = \mathbb O_2$ .

Let $A - 3\; \mathbb I_2 = B$ , we have $B^2 = \mathbb O_2$ , thus $B^m = \mathbb O_2$ for $m= 2, 3,4,\ldots$ .

By binomial theorem ,

$\Large{ \begin{aligned} A^n &=& (3\; \mathbb I_2 + B)^n \\ &=& \dbinom n0 (3\; \mathbb I_2)^n + \dbinom n1 (3\; \mathbb I_2)^{n-1} B + \underbrace{\dbinom n2 (3\;\mathbb I_2)^{n-2} B^2 + \cdots + \dbinom n2 (3 \; \mathbb I_2)^{n-2} B^n}_{= \; \mathbb O_2 \text{ because } B^m =\; \mathbb O_2} \\ &=& 3^n \; \mathbb I_2 + n\cdot 3^{n-1} B \\ &=& 3^n \; \mathbb I_2 + n\cdot 3^{n-1} (A-3\; \mathbb I_2 ) \\ &=& n \cdot 3^{n-1} A - (n-1) 3^n \; \mathbb I_2 \\ &=& n \cdot 3^{n-1} \begin{bmatrix}2 & 1 \\-1 & 4 \end{bmatrix} - (n-1) 3^n \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \\ &=& \begin{bmatrix} (3-n) 3^{n-1} & n \cdot 3^{n-1} \\ -n \cdot 3^{n-1} & (n+3) \cdot 3^{n-1} \end{bmatrix} \; . \end{aligned}}$

Hence, $b_{22}$ is equal to $(2016+3) \cdot 3^{2016 - 1} = 2019 \cdot 3^{2015}$ .

By Fermat's little theorem , we can see that $b_{22} \pmod 7 \equiv 3\times 3^5 = 3^{7-1} \pmod 7 = \boxed1$ .

It is trivial to note that $A = 3I + \begin{bmatrix}-1 & 1 \\-1 & 1 \end{bmatrix}$ where $I$ is Identity matrix of order $2$ .

Now let $C = \begin{bmatrix}-1 & 1 \\-1 & 1 \end{bmatrix}$ . Note that C is nilpotent with $n=2$ .

Therefore,

$A^{2} = 9I + 6C$

$A^{3} = 27I + 27C$

$A^{4} = 81I + 108C$

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$A^{n} = 3^{n}I + n.3^{n-1}C$

So, $B = A^{2016} = 3^{2016}I + 2016.3^{2015}C$ and hence, $b_{22} = 3^{2016} + 2016.3^{2015} \equiv 1 (mod 7)$ .