Question Number 93534 by naka3546 last updated on 13/May/20
$${A}^{\mathrm{2}} \:\:=\:\:\begin{pmatrix}{\mathrm{7}}&{\mathrm{3}}\\{\mathrm{9}}&{\mathrm{4}}\end{pmatrix}\:\:\:\Rightarrow\:\:{A}\:=\:\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix} \\ $$$${Find}\:\:{the}\:\:{all}\:\:{of}\:\:\:{different}\:\:{matrices}\:\:{A}\:\: \\ $$$$\left({i}\right)\:.\:{If}\:\:{a},\:{b},\:{c},\:{d}\:\in\:\mathbb{Z}\:\:\: \\ $$$$\left({ii}\right)\:.\:{If}\:\:{a},\:{b},\:{c},\:{d}\:\in\:\mathbb{R}^{+} \: \\ $$
Answered by prakash jain last updated on 13/May/20
![A^2 = [((a^2 +bc),(ad+bd)),((ac+cd),(bc+d^2 )) ] a^2 +bc=7 ad+bd=3 ac+cd=9 bc+d^2 =4 Z^+ a^2 −d^2 =3⇒a+d=3,a−d=1 a=2,d=1 b=1 c=3 A= [(2,1),(3,1) ]](Q93595.png)
$${A}^{\mathrm{2}} =\begin{bmatrix}{{a}^{\mathrm{2}} +{bc}}&{{ad}+{bd}}\\{{ac}+{cd}}&{{bc}+{d}^{\mathrm{2}} }\end{bmatrix} \\ $$$${a}^{\mathrm{2}} +{bc}=\mathrm{7} \\ $$$${ad}+{bd}=\mathrm{3} \\ $$$${ac}+{cd}=\mathrm{9} \\ $$$${bc}+{d}^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathbb{Z}^{+} \\ $$$${a}^{\mathrm{2}} −{d}^{\mathrm{2}} =\mathrm{3}\Rightarrow{a}+{d}=\mathrm{3},{a}−{d}=\mathrm{1} \\ $$$${a}=\mathrm{2},{d}=\mathrm{1} \\ $$$${b}=\mathrm{1} \\ $$$${c}=\mathrm{3} \\ $$$${A}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{1}}\end{bmatrix} \\ $$
Commented by naka3546 last updated on 13/May/20
Commented by naka3546 last updated on 13/May/20
$${I}\:{have}\:\:{to}\:\:{find}\:\:{the}\:\:{different}\:\:\:{two}\:\:{matrices}\:\:{more}\:. \\ $$
Commented by prakash jain last updated on 13/May/20
One more solution exists for Real numbers case. i am still trying to solve real number cae
Commented by prakash jain last updated on 13/May/20
in z there are two soltion, one that i gave for z+. another one with sign changed for all numbers