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Category: Matrices and Determinants

Question-208624

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}}…

Givem-that-the-matrix-A-3-1-5-2-3-5-5-1-6-If-Adj-A-13-1-10-13-7-5-13-2-7-i-find-A-1-ii-Use-the-result-in-i-to-find-the-values-of-x-y-and-z-that

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} \\…

Question-205306

Question Number 205306 by peter frank last updated on 15/Mar/24 Answered by A5T last updated on 15/Mar/24 $$\mathrm{2}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{xy}\right)=\mathrm{450}\Rightarrow\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{xy}=\mathrm{225} \\ $$$$\Rightarrow{y}=\frac{\mathrm{225}−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}{x}}\Rightarrow{v}=\mathrm{3}{x}^{\mathrm{2}} {y}=\frac{\mathrm{3}{x}\left(\mathrm{225}−\mathrm{3}{x}^{\mathrm{2}} \right)}{\mathrm{4}}…