Question Number 12045 by 7991 last updated on 10/Apr/17
![how much matrices of integers number A= [(a,b),(c,d) ]if A^2 =I and b=c](Q12045.png)
$${how}\:{much}\:{matrices}\:{of}\:{integers}\:{number} \\ $$$${A}=\begin{bmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{bmatrix}{if}\:{A}^{\mathrm{2}} ={I}\:{and}\:{b}={c} \\ $$
Answered by sma3l2996 last updated on 10/Apr/17
![A^2 = [(a,b),(c,d) ] [(a,b),(c,d) ]= [((a^2 +bc),(ab+db)),((ac+dc),(bc+d^2 )) ]= [(1,0),(0,1) ] a^2 +bc=1 ; b(a+d)=0 ; c(a+d)=0 ; bc+d^2 =1 b(a+d)=0⇔b=0 or a=−d if b=0 so a^2 =1 ⇒a=1 or a=−1 and d=1 or d=−1 if b≠0 so a=−d a^2 +b^2 =1 ⇔ b=(√(1−a^2 )) or b=−(√(1−a^2 )) A= [((−1),0),(0,1) ] or A= [(1,0),(0,(−1)) ] or A= [((−1),0),(0,(−1)) ] or A=I or A= [(a,(√(1−a^2 ))),((√(1−a^2 )),(−a)) ] or A= [(a,(−(√(1−a^2 )))),((−(√(1−a^2 ))),(−a)) ]](Q12051.png)
$${A}^{\mathrm{2}} =\begin{bmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{bmatrix}\begin{bmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{bmatrix}=\begin{bmatrix}{{a}^{\mathrm{2}} +{bc}}&{{ab}+{db}}\\{{ac}+{dc}}&{{bc}+{d}^{\mathrm{2}} }\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$${a}^{\mathrm{2}} +{bc}=\mathrm{1}\:;\:{b}\left({a}+{d}\right)=\mathrm{0}\:;\:{c}\left({a}+{d}\right)=\mathrm{0}\:;\:{bc}+{d}^{\mathrm{2}} =\mathrm{1} \\ $$$${b}\left({a}+{d}\right)=\mathrm{0}\Leftrightarrow{b}=\mathrm{0}\:{or}\:{a}=−{d} \\ $$$${if}\:{b}=\mathrm{0}\:{so}\:{a}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{a}=\mathrm{1}\:{or}\:{a}=−\mathrm{1}\:{and}\:{d}=\mathrm{1}\:{or}\:{d}=−\mathrm{1} \\ $$$${if}\:{b}\neq\mathrm{0}\:{so}\:{a}=−{d} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1}\:\Leftrightarrow\:{b}=\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\:{or}\:{b}=−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$${A}=\begin{bmatrix}{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\:{or}\:{A}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{−\mathrm{1}}\end{bmatrix}\:{or}\:{A}=\begin{bmatrix}{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{−\mathrm{1}}\end{bmatrix}\:{or}\:{A}={I} \\ $$$${or}\:{A}=\begin{bmatrix}{{a}}&{\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\\{\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}&{−{a}}\end{bmatrix}\:{or}\:{A}=\begin{bmatrix}{{a}}&{−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\\{−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}&{−{a}}\end{bmatrix} \\ $$