Matrix Power Sum and Invertibility Analysis

To determine the correct statements, we first analyze the matrix $M = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$.

1. Powers of $M$

Calculating the first few powers of $M$: $M^1 = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$ $M^2 = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ 2 & -1 \end{bmatrix}$ $M^3 = \begin{bmatrix} 3 & -2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 3 & -2 \end{bmatrix}$

By mathematical induction, for any positive integer $k$: $M^k = \begin{bmatrix} k+1 & -k \\ k & -(k-1) \end{bmatrix}$

For $k = 26$: $M^{26} = \begin{bmatrix} 27 & -26 \\ 26 & -25 \end{bmatrix} = \begin{bmatrix} p & q \\ r & s \end{bmatrix}$ Thus, $p = 27$, $q = -26$, $r = 26$, and $s = -25$.

Now, let $S = \sum_{k=1}^{26} M^k = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.

• $a = \sum_{k=1}^{26} (k+1) = \frac{26 \times 27}{2} + 26 = 351 + 26 = 377$
• $b = \sum_{k=1}^{26} (-k) = -351$
• $c = \sum_{k=1}^{26} k = 351$
• $d = \sum_{k=1}^{26} (1-k) = 26 - 351 = -325$

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2. Analysis of Options

Option (A): Existence of Invertible Matrix $N$

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

For $\lambda = 1$, the eigenvector equation is: $(M - I)v = 0 \implies \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies v_1 = v_2$ Since the geometric multiplicity is $1$, $M$ is non-diagonalizable, and its Jordan Canonical Form is $J = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.

Thus, there exists an invertible $2 \times 2$ real matrix $N$ such that: $N^{-1}MN = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \implies MN = N \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ Therefore, Option (A) is TRUE.

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Option (B): Value of $a$

As derived above: $a = \sum_{k=1}^{26} (k+1) = 377$ Hence, the statement $a = 378$ is incorrect. Therefore, Option (B) is FALSE.

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Option (C): Unique Integer Solutions for $x, y$

The system of equations can be written in matrix form as: $\begin{bmatrix} p & q \\ r & s \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} m \\ n \end{bmatrix} \implies M^{26} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} m \\ n \end{bmatrix}$

The determinant of $M^{26}$ is: $\det(M^{26}) = (\det M)^{26} = (0 - (-1))^{26} = 1^{26} = 1$

Since $\det(M^{26}) = 1 \neq 0$, $(M^{26})^{-1}$ exists and is given by: $(M^{26})^{-1} = \begin{bmatrix} s & -q \\ -r & p \end{bmatrix} = \begin{bmatrix} -25 & 26 \\ -26 & 27 \end{bmatrix}$ Since all entries of $(M^{26})^{-1}$ are integers, the solution vector: $\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -25 & 26 \\ -26 & 27 \end{bmatrix} \begin{bmatrix} m \\ n \end{bmatrix} = \begin{bmatrix} -25m + 26n \\ -26m + 27n \end{bmatrix}$ yields a unique pair of integers $(x, y)$ for any given integers $m$ and $n$. Therefore, Option (C) is TRUE.

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Option (D): System of Equations for Positive Real $t$

The system of linear equations is: $\begin{bmatrix} a+t & b \\ c & d+t \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \implies (S + tI) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

The determinant of the coefficient matrix is: $\det(S + tI) = t^2 + \text{tr}(S)t + \det(S)$

Calculating trace and determinant of $S$:

• $\text{tr}(S) = a + d = 377 - 325 = 52$
• $\det(S) = ad - bc = (377)(-325) - (-351)(351) = -122525 + 123201 = 676 = 26^2$

Thus, $\det(S + tI) = t^2 + 52t + 676 = (t + 26)^2$

For any positive real number $t > 0$, $(t + 26)^2 > 26^2 > 0$, which means $\det(S + tI) \neq 0$. Hence, the system has a unique solution for each $t > 0$. Therefore, Option (D) is TRUE.

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Conclusion

The correct options are A, C, and D.