Question Number 216526 by Mingma last updated on 10/Feb/25 Answered by Wuji last updated on 10/Feb/25 $$\mathrm{let}\:\mathrm{B}=\begin{bmatrix}{\mathrm{a}\:\:\:\:\:\:\:\mathrm{b}}\\{\mathrm{c}\:\:\:\:\:\:\:\:\mathrm{d}}\\{\mathrm{e}\:\:\:\:\:\:\:\:\mathrm{f}}\end{bmatrix}\:,\:\mathrm{C}=\begin{bmatrix}{\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{4}}\\{−\mathrm{1}\:\:\:\:\mathrm{0}}\end{bmatrix}\:: \\ $$$$\mathrm{AB}=\begin{bmatrix}{\mathrm{1}\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:−\mathrm{1}\:\:\:\mathrm{1}}\end{bmatrix}\begin{bmatrix}{\mathrm{a}\:\:\:\:\mathrm{b}}\\{\mathrm{c}\:\:\:\:\:\mathrm{d}}\\{\mathrm{e}\:\:\:\:\:\mathrm{f}}\end{bmatrix}=\begin{bmatrix}{\mathrm{a}+\mathrm{3c}+\mathrm{2e}\:\:\:\:\mathrm{b}+\mathrm{3d}+\mathrm{2f}}\\{−\mathrm{c}\:\:−\mathrm{e}\:\:\:\:\:\:\:\:\:−\mathrm{d}\:\:+\mathrm{f}\:\:\:\:}\end{bmatrix} \\ $$$$\mathrm{AB}=\mathrm{C}\:\:\Rightarrow\:\mathrm{a}+\mathrm{3c}+\mathrm{2e}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{b}+\mathrm{3d}+\mathrm{2f}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:−\mathrm{c}\:+\mathrm{e}=−\mathrm{1}\:\:\Rightarrow\mathrm{e}=\mathrm{c}−\mathrm{1}…
Question Number 214713 by efronzo1 last updated on 17/Dec/24 $$\:\:\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…