Consider the matrix
$M=\left[\begin{array}{cc}2& -1\\ 1& 0\end{array}\right]$
Let $p,q,r,s,a,b,c,d$ be integers such that
$M^{26}=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]$
and
$\sum _{k=1}^{26}{M}^k=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Then which of the following statements is (are) TRUE?

The correct statements are:
1. There exists a 2×2 invertible matrix $N$ with real entries such that $MN=N\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$
2. For any two given integers $m$ and $n$, there exist unique integers $x$ and $y$ such that $px+qy=m$ and $rx+sy=n$
3. For each positive real number $t$, the system of linear equations $(a+t)x+by=1$ and $cx+(d+t)y=-1$ has a unique solution

Step 1: Finding the properties and powers of matrix $M$

We are given the matrix:

$M=\left[\begin{array}{cc}2& -1\\ 1& 0\end{array}\right]$

Let us split matrix $M$ into the identity matrix $I$ and a nilpotent matrix $A$:

$M=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& -1\\ 1& -1\end{array}\right]=I+A$

where $A=\left[\begin{array}{cc}1& -1\\ 1& -1\end{array}\right]$.

Notice that:

$A^2=\left[\begin{array}{cc}1& -1\\ 1& -1\end{array}\right]\left[\begin{array}{cc}1& -1\\ 1& -1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Thus, $A^k=0$ for all integers $k\ge 2$.

Using the Binomial Theorem for commuting matrices:

$M^k={(I+A)}^k=I+kA=\left[\begin{array}{cc}1+k& -k\\ k& 1-k\end{array}\right]$

Step 2: Checking Statement 1

The characteristic equation of matrix $M$ is:

$det(M-\lambda I)={\lambda }^2-2\lambda +1={(\lambda -1)}^2=0$

The Jordan Canonical Form of $M$ is $J=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$. Since $M$ is similar to its Jordan Form, there exists an invertible matrix $N$ such that $N^{-1}MN=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$, which implies:

$MN=N\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

Thus, Statement 1 is TRUE.

Step 3: Calculating $M^{26}$ and checking Statement 3

For $k=26$, we have:

$M^{26}=\left[\begin{array}{cc}1+26& -26\\ 26& 1-26\end{array}\right]=\left[\begin{array}{cc}27& -26\\ 26& -25\end{array}\right]=\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]$

The given system of equations is:

$\left[\begin{array}{cc}p& q\\ r& s\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}m\\ n\end{array}\right]$

The determinant of this matrix is $det\left(M^{26}\right)={(detM)}^{26}=1^{26}=1$.

The inverse matrix is given by:

${\left(M^{26}\right)}^{-1}=\left[\begin{array}{cc}s& -q\\ -r& p\end{array}\right]=\left[\begin{array}{cc}-25& 26\\ -26& 27\end{array}\right]$

Since all entries of ${\left(M^{26}\right)}^{-1}$ are integers, for any given integers $m$ and $n$, the unique solution vector is:

$\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-25m+26n\\ -26m+27n\end{array}\right]$

These values for $x$ and $y$ are guaranteed to be unique integers. Thus, Statement 3 is TRUE.

Step 4: Calculating $\sum _{k=1}^{26}{M}^k$ and checking Statement 2 & Statement 4

Summing $M^k$ from $k=1$ to $26$:

$\sum _{k=1}^{26}{M}^k=\sum _{k=1}^{26}\left[\begin{array}{cc}1+k& -k\\ k& 1-k\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Since $\sum _{k=1}^{26}k=\frac{26\times 27}{2}=351$:

$a=26+351=377$
$b=-351$
$c=351$
$d=26-351=-325$

Statement 2 claims that $a=378$, which is FALSE because $a=377$.

Now, for Statement 4, the determinant of the matrix of coefficients of the system $(a+t)x+by=1$ and $cx+(d+t)y=-1$ is:

$\Delta =\left|\begin{array}{cc}a+t& b\\ c& d+t\end{array}\right|=(a+t)(d+t)-bc=t^2+(a+d)t+(ad-bc)$

Calculating the values:

$a+d=377-325=52$
$ad-bc=\left(377\right)(-325)-(-351)\left(351\right)=-122525+123201=676={26}^2$

Substituting these into $\Delta$:

$\Delta =t^2+52t+676={(t+26)}^2$

For any positive real number $t>0$, $t+26>26>0$, which means $\Delta ={(t+26)}^2\ne 0$.
Since the determinant is non-zero, the system of linear equations always has a unique solution for each positive real number $t$. Thus, Statement 4 is TRUE.

Conclusion:
The TRUE statements are: Statement 1, Statement 3, and Statement 4.