Consider the $2 imes 2$ matrix $A=\left[\begin{array}{ll}x & y \ 0 & 1\end{array} ight]$

Question

Consider the $2 	imes 2$ matrix $A=\left[\begin{array}{ll}x & y \ 0 & 1\end{array}ight]$

Professor PC Thomas Classes · Answer

A is idempotent $\Rightarrow \mathrm{A}^2=\mathrm{A}$ $$ \left[\begin{array}{ll} x & y \ 0 & 1 \end{array}ight]\left[\begin{array}{ll} x & y \ 0 & 1 \end{array}ight]=\left[\begin{array}{ll} x & y \ 0 & 1 \end{array}ight] ;\left[\begin{array}{cc} x^2 & x y+y \ 0 & 1 \end{array}ight]=\left[\begin{array}{ll} x & y \ 0 & 1 \end{array}ight] $$ $\Rightarrow x^2=x \& x y+y=y \quad \Rightarrow x=0$ or $1 \& x y=0 \Rightarrow x=0, y \in R$ or $x=1, y=0 \quad(A) ightarrow Q, R$ A is involuntary $$ \begin{aligned} & \Rightarrow A^2=I \ & {\left[\begin{array}{ll} x & y \ 0 & 1 \end{array}ight]\left[\begin{array}{ll} x & y \ 0 & 1 \end{array}ight]=\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}ight] ;\left[\begin{array}{cc} x^2 & x y+y \ 0 & 1 \end{array}ight]=\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}ight]} \ & \Rightarrow x^2=18 x x y+y=0 \Rightarrow x=1, y=0 	ext { or } x=-1, y \in R \quad 	ext { (B) } ightarrow P, S \end{aligned} $$ $A$ is orthogonal $\Rightarrow A A=I$ $$ \begin{align*} & \Rightarrow\left[\begin{array}{ll} x & y \ 0 & 1 \end{array}ight]\left[\begin{array}{ll} x & 0 \ y & 1 \end{array}ight]=\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}ight] \ & \Rightarrow\left[\begin{array}{cc} x^2+y^2 & y \ y & 1 \end{array}ight]=\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}ight] \Rightarrow x^2+y^2=1, y=0 \quad \Rightarrow x= \pm 1, y=0 \quad(\mathrm{C}) ightarrow S 	ag{C} \end{align*} $$ A is singular $\Rightarrow|A|=0 \Rightarrow x=0, y \in R$ $(D) ightarrow R$