Question Number 142204 by mnjuly1970 last updated on 27/May/21
Answered by MJS_new last updated on 28/May/21
![what are you waiting for? A^2 = [(7,(2m−5),5),(2,(m^2 −1),(m+2)),(2,(2−m),9) ] A^4 = [((4m+49),(2m^3 −5m^2 +7m−20),(2m^2 −m+70)),((2m^3 +2m+16),(m^4 −3m^2 +4m−5),(m^3 +2m^2 +8m+26)),((36−2m),(−m^3 +2m^2 −4m+6),(95−m^2 )) ] m^3 +2m^2 +8m+26=24 ⇒ m=−((2+((−37+3(√(1041))))^(1/3) −((37+3(√(1041))))^(1/3) )/3)≈−.265257381969](Q142223.png)
$$\mathrm{what}\:\mathrm{are}\:\mathrm{you}\:\mathrm{waiting}\:\mathrm{for}? \\ $$$${A}^{\mathrm{2}} =\begin{bmatrix}{\mathrm{7}}&{\mathrm{2}{m}−\mathrm{5}}&{\mathrm{5}}\\{\mathrm{2}}&{{m}^{\mathrm{2}} −\mathrm{1}}&{{m}+\mathrm{2}}\\{\mathrm{2}}&{\mathrm{2}−{m}}&{\mathrm{9}}\end{bmatrix} \\ $$$${A}^{\mathrm{4}} =\begin{bmatrix}{\mathrm{4}{m}+\mathrm{49}}&{\mathrm{2}{m}^{\mathrm{3}} −\mathrm{5}{m}^{\mathrm{2}} +\mathrm{7}{m}−\mathrm{20}}&{\mathrm{2}{m}^{\mathrm{2}} −{m}+\mathrm{70}}\\{\mathrm{2}{m}^{\mathrm{3}} +\mathrm{2}{m}+\mathrm{16}}&{{m}^{\mathrm{4}} −\mathrm{3}{m}^{\mathrm{2}} +\mathrm{4}{m}−\mathrm{5}}&{{m}^{\mathrm{3}} +\mathrm{2}{m}^{\mathrm{2}} +\mathrm{8}{m}+\mathrm{26}}\\{\mathrm{36}−\mathrm{2}{m}}&{−{m}^{\mathrm{3}} +\mathrm{2}{m}^{\mathrm{2}} −\mathrm{4}{m}+\mathrm{6}}&{\mathrm{95}−{m}^{\mathrm{2}} }\end{bmatrix} \\ $$$${m}^{\mathrm{3}} +\mathrm{2}{m}^{\mathrm{2}} +\mathrm{8}{m}+\mathrm{26}=\mathrm{24} \\ $$$$\Rightarrow \\ $$$${m}=−\frac{\mathrm{2}+\sqrt[{\mathrm{3}}]{−\mathrm{37}+\mathrm{3}\sqrt{\mathrm{1041}}}−\sqrt[{\mathrm{3}}]{\mathrm{37}+\mathrm{3}\sqrt{\mathrm{1041}}}}{\mathrm{3}}\approx−.\mathrm{265257381969} \\ $$