Matrices and Determinants

Concept:The matrix $M$ equals $I + B$ where $B$ is nilpotent ($B^2 = 0$). This allows binomial expansion for powers and sums. Determinants decide existence of unique solutions.Explanation:Write $M = \begin{bmatrix}2 & -1 \\ 1 & 0\end{bmatrix} = I + B$ with $I$ identity and $B = \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix}$.Check $B^2 = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$, so $B$ is nilpotent.Compute $M^{26} = (I+B)^{26} = I + 26B$ (since higher powers vanish).Thus $M^{26} = \begin{bmatrix}27 & -26 \\ 26 & -25\end{bmatrix}$, so $p=27, q=-26, r=26, s=-25$.Sum $S = \sum_{k=1}^{26} M^k = \sum_{k=1}^{26} (I + kB) = 26I + (1+2+...+26)B = 26I + 351B$.This gives $S = \begin{bmatrix}377 & -351 \\ 351 & -325\end{bmatrix}$, so $a=377, b=-351, c=351, d=-325$.Hence statement B ($a=378$) is false.Check option C: The system $27x - 26y = m$, $26x - 25y = n$ has coefficient matrix with determinant $27(-25) - (-26)26 = -675+676 = 1 \neq 0$.Since the determinant is non‑zero and the coefficients are integers, for any integers $m,n$ the unique solution $x,y$ are integers. So C is true.Check option D: The determinant of the matrix $\begin{bmatrix}a+t & b \\ c & d+t\end{bmatrix}$ is $(a+t)(d+t)-bc$.Substituting the values: $(377+t)(-325+t) - (-351)(351) = t^2 + 52t + 676 = (t+26)^2$.For any $t>0$, $(t+26)^2 \neq 0$, so the system has a unique solution. Thus D is true.Check option A: We need an invertible real $N = \begin{bmatrix}j & k \\ l & m\end{bmatrix}$ such that $MN = N\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$.Equating entries gives $j = l$ and $2k-m = j+k \Rightarrow m-k = j$.Then $\det(N) = jm - lk = j(m-k) = j^2$.Choosing $j \neq 0$ (e.g., $j=1, k=0, l=1, m=1$) makes $\det(N)=1 \neq 0$, so $N$ is invertible. Hence A is true.Answer:Statements A, C, and D are true.