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Category: Matrices and Determinants

5-3-3-2-A-2-5-5-1-4-7-6-2-Find-4A-1-

Question Number 22432 by Joel577 last updated on 18/Oct/17 $$\begin{pmatrix}{\mathrm{5}\:\:\:\mathrm{3}}\\{\mathrm{3}\:\:\:\mathrm{2}}\end{pmatrix}{A}\:+\:\begin{pmatrix}{\mathrm{2}\:\:\:\mathrm{5}}\\{\mathrm{5}\:\:\:\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{4}\:\:\:\:\mathrm{7}}\\{\mathrm{6}\:\:\:\:\mathrm{2}}\end{pmatrix} \\ $$$$\mathrm{Find}\:\mid\mathrm{4}{A}^{−\mathrm{1}} \mid \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-153486

Question Number 153486 by roxceefocus last updated on 07/Sep/21 Answered by liberty last updated on 08/Sep/21 $$\:{det}\left({F}\right)=\mathrm{3}\left(\mathrm{0}\right)−\mathrm{4}\left(\mathrm{0}−\mathrm{4}\right)+\mathrm{2}\left(\mathrm{0}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{16}+\mathrm{4}=\mathrm{20} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:=\frac{\mathrm{1}}{\mathrm{20}}\:{adj}\begin{pmatrix}{\mathrm{3}\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:−\mathrm{1}\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{2}\:\:\:\:\:\:\mathrm{1}\:\:−\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{−\mathrm{1}}\\{−\mathrm{8}}\\{\:\:\:\mathrm{6}}\end{pmatrix} \\ $$ Terms of…

let-D-v-5-1-3-m-find-number-v-and-m-such-that-D-2-5I-I-identity-matrix-

Question Number 153227 by Eric002 last updated on 05/Sep/21 $${let}\:{D}=\begin{bmatrix}{{v}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}}\\{\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:{m}}\end{bmatrix}\:{find}\:{number}\:\left({v}\right)\:{and} \\ $$$$\left({m}\right)\:{such}\:{that}\:{D}^{\mathrm{2}} =\mathrm{5}{I}\:\:\:\:\:\left({I}={identity}\:{matrix}\right) \\ $$ Answered by puissant last updated on 05/Sep/21 $${D}^{\mathrm{2}} =\begin{bmatrix}{{v}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\mathrm{5}{v}+\mathrm{5}{m}}\\{\frac{{v}}{\mathrm{3}}+\frac{{m}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}}{\mathrm{3}}+{m}^{\mathrm{2}}…

Question-152545

Question Number 152545 by liberty last updated on 29/Aug/21 Answered by Olaf_Thorendsen last updated on 31/Aug/21 $$\: \\ $$$$\mathrm{cos}\frac{\pi}{\mathrm{12}}\:=\:\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{6}}\right) \\ $$$$\mathrm{cos}\frac{\pi}{\mathrm{12}}\:=\:\mathrm{cos}\frac{\pi}{\mathrm{4}}\mathrm{cos}\frac{\pi}{\mathrm{6}}+\mathrm{sin}\frac{\pi}{\mathrm{4}}\mathrm{sin}\frac{\pi}{\mathrm{6}} \\ $$$$\mathrm{cos}\frac{\pi}{\mathrm{12}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}\frac{\pi}{\mathrm{12}}\:=\:\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\:\left(\mathrm{1}\right)…