Fun with Matrix

Let $I=\begin{bmatrix}1&0\\0&1\end{bmatrix},B=\begin{bmatrix}1&3\\0&0\end{bmatrix}$
These little properties allow us to find $A^n$ easily: $A=I+B$ , $IB=BI=B$ and $B^n=B$ for positive integers $n$ .
$\;\;\;\;A^n$
$=(I+B)^n$
$=\displaystyle\sum_{i=0}^n C_i^n I^i B^{n-i}$ (usually this is not true, but this time it is, becuase $IB=BI$ )
$=\displaystyle I+\sum_{i=1}^n C_i^n I^i B^{n-i}$
$=\displaystyle I+\sum_{i=1}^n C_i^n B$ ( $I$ is identity, $B$ is an idempotent element)
$=I+(2^n-1)B$
$=\begin{bmatrix}2^n&3\times 2^n-3\\0&1\end{bmatrix}$
$b=(3\times 2^{20}-3)-2(3\times 2^{19}-3)+3=6$

Let $A=\begin{bmatrix} 2 & 3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix} = I + B$ , where $B = \begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix}$ . We note that $B^2 = \begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix} = B$ $\implies B^n = B$ , where $n \in \mathbb N$ . Therefore,

$\begin{aligned} A^n & = (I+B)^n \\ & = I + \binom n1 B + \binom n2 B^2 + \binom n3 B^3 + \cdots + \binom nn B^n \\ & = I + \sum_{k=1}^n \binom nk \color{#3D99F6} B^k & \small \color{#3D99F6} \text{Note that }B^k = B \\ & = I + {\color{#D61F06}\sum_{k=0}^n \binom nk}\color{#3D99F6}B - B &\small \color{#D61F06} \text{By binomial theorem} \\ & = I + {\color{#D61F06}2^n}B - B \\ & = I + (2^n-1)B \end{aligned}$

Then we have:

$\begin{aligned} X & = A^{20} - 2A^{19} + A \\ & = I + (2^{20}-1)B - 2\left(I + (2^{19}-1)B\right) + I + B \\ & = 2^{20}B - B - 2^{20}B + 2B + B \\ & = 2B \\ & = \begin{bmatrix} 2 & 6 \\ 0 & 0 \end{bmatrix} \end{aligned}$

Therefore, $b = \boxed{6}$ .

We can establish a recurrence $A(n + 1) = \begin{bmatrix} 2T_{11}(n) & 2T_{12}(n) + 3 \\ 0 & 1 \end{bmatrix} \\$

Where $A(1) = \begin{bmatrix} 2 & 3 \\ 0 & 1 \end{bmatrix}$ , $A(2) = \begin{bmatrix} 4 & 9 \\ 0 & 1 \end{bmatrix}$ , $A(3) = \begin{bmatrix} 8 & 21 \\ 0 & 1 \end{bmatrix}, A(4) = \begin{bmatrix} 16 & 45 \\ 0 & 1 \end{bmatrix}, \cdots \\$

Thus, you can view the above equation as,

$A^{20} - 2A^{19} + A = \begin{bmatrix} T_{11}(n + 1) - 2T_{11}(n) + 2 & 2T_{12}(n + 1) - 2T_{12}(n) + 3 \\ 0 & 1 -2 +1 \end{bmatrix} = \begin{bmatrix} 2 & b \\ 0 & 0 \end{bmatrix} \\$

We can see that $A_{21} and A_{22}$ are pretty obvious.

$A_{11} = T_{11}(n + 1) - 2T_{11}(n) + 2 = 2T_{11}(n) - 2T_{11}(n) + 2 = 2$ and,

$A_{12} = T_{12}(n+1) - 2T_{12}(n) + 3 = 2T_{12}(n) + 3 - 2T_{12}(n) + 3 = 6 \\$

Thus, $b = 6$ .

It's actually not necessary to calculate $A^{19}$ . Instead, assume that

$A^{19} = \begin{bmatrix} p & q \\ r & s\end{bmatrix}$

Then we can write out

$\begin{bmatrix}0 & 3 \\ 0 & -1\end{bmatrix} \begin{bmatrix}p & q \\ r & s\end{bmatrix}= \begin{bmatrix}0 & b-3 \\ 0 & -1\end{bmatrix}$

Therefore

$\left\{\begin{aligned}3s &= b-3\\ -s &= -1 \end{aligned}\right. \Rightarrow b= 6$