Powers of M: Direct computation gives $M^2=\left[\begin{array}{c}3-2\\ 2-1\end{array}\right]$, $M^3=\left[\begin{array}{c}4-3\\ 3-2\end{array}\right]$. By induction,

$M^k=\left[\begin{array}{c}k+1-k\\ k1-k\end{array}\right]$

Hence $M^{26}=\left[\begin{array}{c}27-26\\ 26-25\end{array}\right]$, giving $p=27,q=-26,r=26,s=-25$.

Using $\sum _{k=1}^{26}k=\frac{26\cdot 27}{2}=351$ and $\sum _{k=1}^{26}(k+1)=377$:

$\sum _{k=1}^{26}{M}^k=\left[\begin{array}{c}377-351\\ 351-325\end{array}\right]$

so $a=377,b=-351,c=351,d=-325$.

(A) Writing $N=\left[\begin{array}{c}\alpha \beta \\ \gamma \delta \end{array}\right]$ and equating $MN=N\left[\begin{array}{c}11\\ 01\end{array}\right]$ yields $\alpha =\gamma$ and $\beta =\alpha +\delta$. Choosing for example $\alpha =1,\delta =0$ gives $N=\left[\begin{array}{c}11\\ 10\end{array}\right]$ with $\det N=-1\ne 0$. TRUE.

(B) $a=377$, not $378$. FALSE.

(C) The determinant of the coefficient matrix is $\left|\begin{array}{c}27-26\\ 26-25\end{array}\right|=-675+676=1$. Since the determinant is $\pm 1$, Cramer's rule produces integer solutions for any integer right-hand side. TRUE.

(D) The coefficient determinant is

$\Delta =(377+t)(t-325)-(-351)(351)=t^2+52t+351^2-377\cdot 325$

Using $377\cdot 325=(351+26)(351-26)=351^2-26^2$:

$\Delta =t^2+52t+26^2=(t+26{)}^2$

For $t>0$, $\Delta =(t+26{)}^2>0$, so the system has a unique solution. TRUE.

$\therefore$ The correct options are (A), (C) and (D).