Question 8

Given $M = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$. The characteristic equation is $\det(M - \lambda I) = 0 \Rightarrow \lambda^2 - 2\lambda + 1 = 0 \Rightarrow (\lambda - 1)^2 = 0$. The eigenvalue is $\lambda = 1$ (repeated).

(A) Since $M$ has a single eigenvalue $\lambda = 1$ and is not a scalar matrix, its Jordan Canonical Form is $J = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. Thus, there exists an invertible matrix $N$ such that $N^{-1}MN = J$, which implies $MN = NJ$. Option (A) is correct.

(B) Let $M = I + A$, where $A = M - I = \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}$. Note that $A^2 = 0$.

Then $M^k = (I + A)^k = I + kA = \begin{bmatrix} 1+k & -k \\ k & 1-k \end{bmatrix}$.

$\sum_{k=1}^{26} M^k = \sum_{k=1}^{26} \begin{bmatrix} 1+k & -k \\ k & 1-k \end{bmatrix} = \begin{bmatrix} 26 + \frac{26 \times 27}{2} & -\frac{26 \times 27}{2} \\ \frac{26 \times 27}{2} & 26 - \frac{26 \times 27}{2} \end{bmatrix} = \begin{bmatrix} 26 + 351 & -351 \\ 351 & 26 - 351 \end{bmatrix} = \begin{bmatrix} 377 & -351 \\ 351 & -325 \end{bmatrix}$.

Thus $a = 377$. Option (B) is incorrect.

(C) The system is $M^{26} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} m \\ n \end{bmatrix}$. Since $\det(M) = 1$, $\det(M^{26}) = 1^{26} = 1$. As the determinant is 1 and all entries of $M^{26}$ are integers, $(M^{26})^{-1}$ exists and has integer entries. Thus, $x$ and $y$ will be unique integers for any integer $m, n$. Option (C) is correct.

(D) Let $S = \sum_{k=1}^{26} M^k = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = 26I + 351A$.

The system is $(S + tI) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$.

$\det(S + tI) = \det((26+t)I + 351A)$. Since $A$ is nilpotent ($A^2=0$), $\det(\lambda I + kA) = \lambda^2$.

Here $\det(S + tI) = (26+t)^2$. For $t > 0$, $(26+t)^2 \neq 0$, so the system always has a unique solution. Option (D) is correct.