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Question Number 12046 by 7991 last updated on 10/Apr/17

how much matrices of integers number  A= [(a,b),(c,d) ]if A^2 +A=2I, c=0, det(A)=4

$${how}\:{much}\:{matrices}\:{of}\:{integers}\:{number} \\ $$$${A}=\begin{bmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{bmatrix}{if}\:{A}^{\mathrm{2}} +{A}=\mathrm{2}{I},\:{c}=\mathrm{0},\:{det}\left({A}\right)=\mathrm{4} \\ $$

Answered by sma3l2996 last updated on 10/Apr/17

A^2 = [(a,b),(0,d) ] [(a,b),(0,d) ]= [(a^2 ,(ab+db)),(0,d^2 ) ]  A^2 +A= [((a^2 +a),(ab+db+d)),(0,(d^2 +d)) ]= [(2,0),(0,2) ]  a^2 +a=2 ; d^2 +d=2 ; ab+db+d=0  det(A)=ad=4  (i):a^2 +a−2=0 ⇔ a=1 or a=−2  (ii):ad=4⇔d=4 or d=−2  (iii):d^2 +d−2=0⇔d=1 or d=−2  so d=a=−2  ab+db+d=0⇔ −4b−2=0 ⇔ b=−(1/2)  so A= [((−2),((−1)/2)),(0,(−2)) ]

$${A}^{\mathrm{2}} =\begin{bmatrix}{{a}}&{{b}}\\{\mathrm{0}}&{{d}}\end{bmatrix}\begin{bmatrix}{{a}}&{{b}}\\{\mathrm{0}}&{{d}}\end{bmatrix}=\begin{bmatrix}{{a}^{\mathrm{2}} }&{{ab}+{db}}\\{\mathrm{0}}&{{d}^{\mathrm{2}} }\end{bmatrix} \\ $$$${A}^{\mathrm{2}} +{A}=\begin{bmatrix}{{a}^{\mathrm{2}} +{a}}&{{ab}+{db}+{d}}\\{\mathrm{0}}&{{d}^{\mathrm{2}} +{d}}\end{bmatrix}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{2}}\end{bmatrix} \\ $$$${a}^{\mathrm{2}} +{a}=\mathrm{2}\:;\:{d}^{\mathrm{2}} +{d}=\mathrm{2}\:;\:{ab}+{db}+{d}=\mathrm{0} \\ $$$${det}\left({A}\right)={ad}=\mathrm{4} \\ $$$$\left({i}\right):{a}^{\mathrm{2}} +{a}−\mathrm{2}=\mathrm{0}\:\Leftrightarrow\:{a}=\mathrm{1}\:{or}\:{a}=−\mathrm{2} \\ $$$$\left({ii}\right):{ad}=\mathrm{4}\Leftrightarrow{d}=\mathrm{4}\:{or}\:{d}=−\mathrm{2} \\ $$$$\left({iii}\right):{d}^{\mathrm{2}} +{d}−\mathrm{2}=\mathrm{0}\Leftrightarrow{d}=\mathrm{1}\:{or}\:{d}=−\mathrm{2} \\ $$$${so}\:{d}={a}=−\mathrm{2} \\ $$$${ab}+{db}+{d}=\mathrm{0}\Leftrightarrow\:−\mathrm{4}{b}−\mathrm{2}=\mathrm{0}\:\Leftrightarrow\:{b}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${so}\:{A}=\begin{bmatrix}{−\mathrm{2}}&{\frac{−\mathrm{1}}{\mathrm{2}}}\\{\mathrm{0}}&{−\mathrm{2}}\end{bmatrix} \\ $$

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