Consider the matrix M = bmatrix 2 & -1 1 & 0 bmatrix . Let p, q, r,...

The characteristic equation of $M$ is given by $\det(M - \lambda I) = 0$:
$\begin{vmatrix} 2 - \lambda & -1 \\ 1 & -\lambda \end{vmatrix} = \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0$
The only eigenvalue is $\lambda = 1$ with algebraic multiplicity $2$.

For the eigenvector, $(M - I)X = 0 \Rightarrow \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \Rightarrow x = y$.
Since there is only one linearly independent eigenvector, $M$ is not diagonalizable but can be reduced to its Jordan canonical form $J = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.
Thus, there exists an invertible matrix $N$ such that $M = N J N^{-1} \Rightarrow MN = N \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.
Statement (A) is TRUE.

We can write $M = I + A$, where $A = \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}$.
Notice that $A^2 = \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Using the binomial theorem (since $I$ and $A$ commute), $M^k = (I + A)^k = I + kA = \begin{bmatrix} 1+k & -k \\ k & 1-k \end{bmatrix}$.

For $k=26$, $M^{26} = \begin{bmatrix} 27 & -26 \\ 26 & -25 \end{bmatrix} = \begin{bmatrix} p & q \\ r & s \end{bmatrix}$.
The determinant of $M^{26}$ is $ps - qr = 27(-25) - (-26)(26) = 1$.
Since $\det(M^{26}) = 1$, its inverse has integer entries. Thus, for any integers $m$ and $n$, the system $\begin{bmatrix} p & q \\ r & s \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} m \\ n \end{bmatrix}$ has a unique integer solution.
Statement (C) is TRUE.

Now, let's find $\sum_{k=1}^{26} M^k = \sum_{k=1}^{26} \begin{bmatrix} 1+k & -k \\ k & 1-k \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
$a = \sum_{k=1}^{26} (1+k) = 2 + 3 + \dots + 27 = \dfrac{26}{2}(2 + 27) = 13 \times 29 = 377$.
Statement (B) is FALSE.

We also have:
$b = \sum_{k=1}^{26} (-k) = -351$
$c = \sum_{k=1}^{26} k = 351$
$d = \sum_{k=1}^{26} (1-k) = 26 - 351 = -325$

For statement (D), the system of equations is $\begin{bmatrix} a+t & b \\ c & d+t \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$.
The determinant of the coefficient matrix is:
$\Delta = (a+t)(d+t) - bc = (377+t)(-325+t) - (-351)(351)$
$\Delta = t^2 + 52t - 122525 + 123201 = t^2 + 52t + 676 = (t+26)^2$
For any positive real number $t > 0$, $\Delta = (t+26)^2 > 0$, meaning the determinant is non-zero. Hence, the system has a unique solution.
Statement (D) is TRUE.

Answer: There exists a $2 \times 2$ invertible matrix $N$ with real entries such that $MN = N \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$; For any two given integers $m$ and $n$, there exist unique integers $x$ and $y$ such that $px + qy = m$ and $rx + sy = n$; 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