Question Number 214713 by efronzo1 last updated on 17/Dec/24 $$\:\:\mathrm{Find}\:\mathrm{matrix}\:\mathrm{B}\:\mathrm{if}\:\mathrm{given}\:\mathrm{AB}=\mathrm{BA}=\begin{pmatrix}{\mathrm{0}\:\:\mathrm{0}}\\{\mathrm{0}\:\:\mathrm{0}}\end{pmatrix} \\ $$$$\:\:\mathrm{where}\:\mathrm{A}=\:\begin{pmatrix}{\mathrm{5}\:\:\:\mathrm{3}}\\{\mathrm{5}\:\:\:\mathrm{3}}\end{pmatrix}\:\mathrm{and}\:\mathrm{B}\:\neq\:\begin{pmatrix}{\mathrm{0}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$$ \\ $$ Answered by golsendro last updated on 18/Dec/24 $$\:\mathrm{det}\left(\mathrm{A}\right)=\:\mathrm{15}−\mathrm{15}=\mathrm{0} \\…
Question Number 214495 by mdshorifulislam last updated on 10/Dec/24 Commented by TonyCWX08 last updated on 10/Dec/24 $${What}\:{is}\:{this}???? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 214447 by 2universe456 last updated on 08/Dec/24 Commented by Frix last updated on 08/Dec/24 $$={abc} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 213435 by mnjuly1970 last updated on 14/Nov/24 $$ \\ $$$$\:\:\:\:\:\:\:{A}\:\in\:\mathrm{M}_{\mathrm{2}×\mathrm{2}} \:\:,{and}\:,{det}\:\left({A}\right)\neq\mathrm{0}\::\:\:\:{A}^{\mathrm{3}} \:=\:{A}^{\mathrm{2}} \:+\:{A} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\:\:{det}\:\left(\:{A}\:−\mathrm{2}{I}\:\right)=? \\ $$$$\:\:\:\:\:\: \\ $$ Terms of Service Privacy…
Question Number 212844 by efronzo1 last updated on 25/Oct/24 $$\:\:\:\cancel{\downharpoonleft}\underline{\:} \\ $$ Commented by AlagaIbile last updated on 29/Oct/24 $$\:{det}\left({S}^{\mathrm{2}} \right)\:=\:{det}^{\mathrm{2}} \left({S}\right)\:=\:\mathrm{36} \\ $$$$\:{det}\left({S}\right)\:=\:\pm\:\mathrm{6} \\…
Question Number 211902 by Gangadhar last updated on 23/Sep/24 Commented by BHOOPENDRA last updated on 24/Sep/24 $$\mathrm{3} \\ $$ Answered by BHOOPENDRA last updated on…
Question Number 208624 by vipin last updated on 19/Jun/24 Answered by Berbere last updated on 19/Jun/24 $$\frac{\mathrm{1}}{{cosec}^{−} \left(−\sqrt{\mathrm{2}}\right)}=\mathrm{sin}^{−\mathrm{1}} \left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=−\frac{\pi}{\mathrm{4}} \\ $$$${f}\left({x}\right).\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}{\mathrm{1}+\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}\right)…{E} \\ $$$${g}\left({y}\right)=\mathrm{tan}^{−\mathrm{1}}…
Question Number 208359 by alcohol last updated on 13/Jun/24 Commented by alcohol last updated on 13/Jun/24 $${please}\:{help} \\ $$ Answered by Berbere last updated on…
Question Number 207931 by efronzo1 last updated on 31/May/24 Answered by Lochinbek last updated on 31/May/24 Answered by mr W last updated on 31/May/24 Commented…
Question Number 205586 by necx122 last updated on 25/Mar/24 $${Givem}\:{that}\:{the}\:{matrix}\:{A}\:=\:\begin{pmatrix}{\mathrm{3}}&{\mathrm{1}}&{\mathrm{5}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{5}}\\{\mathrm{5}}&{\mathrm{1}}&{\mathrm{6}}\end{pmatrix}.\: \\ $$$${If}\:{Adj}.\:{A}\:=\:\begin{pmatrix}{\mathrm{13}}&{-\mathrm{1}}&{-\mathrm{10}}\\{\mathrm{13}}&{-\mathrm{7}}&{-\mathrm{5}}\\{-\mathrm{13}}&{\mathrm{2}}&{\mathrm{7}}\end{pmatrix} \\ $$$$\left({i}\right)\:{find}\:{A}^{−\mathrm{1}} \\ $$$$\left({ii}\right)\:{Use}\:{the}\:{result}\:{in}\:\left({i}\right)\:{to}\:{find}\:{the} \\ $$$${values}\:{of}\:{x},\:{y}\:{and}\:{z}\:{that}\:{will}\:{satisfy}\:{the} \\ $$$${equations}: \\ $$$$\mathrm{3}{x}\:+\:{y}\:+\:\mathrm{5}{z}\:=\:\mathrm{8} \\ $$$$\mathrm{2}{x}\:+\mathrm{3}{y}\:+\:\mathrm{5}{z}\:=\:\mathrm{0} \\…