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Author: Tinku Tara

let-T-be-a-n-n-matrix-with-integral-entries-and-Q-T-1-2-I-where-I-denote-the-n-n-identity-matrix-then-prove-that-matrix-Q-is-invertible-

Question Number 208292 by universe last updated on 10/Jun/24 $$\:\mathrm{let}\:\mathrm{T}\:\mathrm{be}\:\mathrm{a}\:{n}×{n}\:\mathrm{matrix}\:\mathrm{with}\:\mathrm{integral}\: \\ $$$$\:\mathrm{entries}\:\mathrm{and}\:\:\mathrm{Q}\:=\:\mathrm{T}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{I}\:\:\:\mathrm{where}\:\mathrm{I}\:\mathrm{denote} \\ $$$$\:\:\mathrm{the}\:\mathrm{n}×\mathrm{n}\:\mathrm{identity}\:\mathrm{matrix}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\:\:\mathrm{that}\:\mathrm{matrix}\:\mathrm{Q}\:\mathrm{is}\:\mathrm{invertible} \\ $$ Answered by Berbere last updated on 10/Jun/24…

Question-208277

Question Number 208277 by Mastermind last updated on 10/Jun/24 Answered by Rasheed.Sindhi last updated on 10/Jun/24 $$\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{5}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{5}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}\end{vmatrix}\: \\ $$$${R}_{\mathrm{2}} ={R}_{\mathrm{2}} −\mathrm{2}{R}_{\mathrm{1}} \\ $$$$=\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\mathrm{1}}&{\:\:\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{-\mathrm{5}}\\{\mathrm{1}}&{\mathrm{5}}&{\mathrm{2}}&{\mathrm{0}}&{\:\:\mathrm{0}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{0}}&{\mathrm{2}}&{\:\:\mathrm{0}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\:\:\mathrm{2}}\end{vmatrix}\: \\ $$$${R}_{\mathrm{4}}…

Question-208288

Question Number 208288 by efronzo1 last updated on 10/Jun/24 Answered by A5T last updated on 10/Jun/24 $$\mathrm{8}\left(\mathrm{8}+{x}\right)=\left(\mathrm{2}{r}\right)\left(\mathrm{2}{r}+\mathrm{2}{R}\right)=\mathrm{4}{r}^{\mathrm{2}} +\mathrm{4}{rR} \\ $$$$\Rightarrow{x}=\frac{{r}^{\mathrm{2}} +{rR}−\mathrm{16}}{\mathrm{2}}…\left({i}\right) \\ $$$$\left(\mathrm{8}+{x}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +\left(\mathrm{2}{r}+{R}\right)^{\mathrm{2}}…

cos-2pi-21-cos-4pi-21-cos-8pi-21-cos-10pi-22-cos-16pi-21-cos-20pi-21-

Question Number 208252 by efronzo1 last updated on 09/Jun/24 $$\:\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{10}\pi}{\mathrm{22}}\right)\mathrm{cos}\:\left(\frac{\mathrm{16}\pi}{\mathrm{21}}\right)\mathrm{cos}\:\left(\frac{\mathrm{20}\pi}{\mathrm{21}}\right)=? \\ $$ Answered by som(math1967) last updated on 09/Jun/24 $$\:{let}\:\frac{\pi}{\mathrm{21}}={x} \\ $$$${cos}\mathrm{2}{xcos}\mathrm{4}{xcos}\mathrm{8}{xcos}\mathrm{10}{xcos}\mathrm{16}{xcos}\mathrm{20}{x} \\ $$$${cosxcos}\mathrm{2}{xcos}\mathrm{4}{xcos}\mathrm{8}{xcos}\mathrm{10}{xcos}\mathrm{5}{x} \\…