Consider the $2 \times 2$ matrix $A = \begin{bmatrix} x & y \\ 0 & 1 \end{bmatrix}$
Step-by-Step Solution
Key Concept: Testing special matrix properties by computing A² and comparing with identity or original matrix. For idempotent matrices (A² = A), equating corresponding elements gives constraints on x and y; for involutory matrices (A² = I), the constraint xy + y = 0 factors as y(x+1) = 0, forcing y = 0 when x = ±1.
Testing properties of matrix $A$: if $A^2 = A$ (idempotent), then $x^2 = x$ and $xy + y = 0$, giving either $x = 0, y \in \mathbb{R}$ or $x = 1, y = 0$. If $A^2 = I$ (involuntary), then $x^2 = 1$ and $xy + y = 0$, yielding $x = \pm 1, y = 0$. If $AA^T = I$ (orthogonal), then $x^2 + y^2 = 1$ and $y = 0$, so $x = \pm 1$. If $\det(A) = 0$ (singular), then $x = 0$ for all $y \in \mathbb{R}$.
Correct Answer: [A-q, r] [B-p, s] [C-s] [D-r]