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Question Number 93388 by niely_special last updated on 12/May/20 Commented by prakash jain last updated on 12/May/20 $$\mathrm{English}\:\mathrm{Please}. \\ $$ Commented by turbo msup by…
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Question Number 27343 by abdo imad last updated on 05/Jan/18 $${let}\:{give}\:{A}=\left(_{\mathrm{2}\:\:\:\:\:\:\:\mathrm{2}} ^{\mathrm{1}\:\:\:\:\:\:\mathrm{2}} \right)\:\:\:{find}\:\:{A}^{{n}} \:\:\:{and}\:\:{e}^{{A}} \\ $$$${and}\:\:{e}^{{tA}} \:\:\:\:\:\:.\:{we}\:{remind}\:{that}\:\:{e}^{{A}} =\:\sum_{} \:\frac{{A}^{{n}} }{{n}!} \\ $$ Commented by Rasheed.Sindhi…
Question Number 91977 by Power last updated on 04/May/20 Commented by jagoll last updated on 04/May/20 $$\mathrm{1}−\frac{\mathrm{5y}}{\mathrm{x}}\:=\:\frac{\mathrm{16}}{\mathrm{x}^{\mathrm{2}} }\:,\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:\mathrm{u}\:,\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{v} \\ $$$$\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\mathrm{2}} +\frac{\mathrm{2y}}{\mathrm{x}}\:=\:\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{u}^{\mathrm{2}} \:+\:\mathrm{2u}\:=\:\mathrm{3v} \\…
Question Number 26264 by Karan last updated on 23/Dec/17 $$\frac{\mathrm{x}}{\mathrm{3}}+\mathrm{2x}=\mathrm{14} \\ $$ Answered by Rasheed.Sindhi last updated on 23/Dec/17 $$\frac{\mathrm{x}}{\mathrm{3}}+\mathrm{2x}=\mathrm{14} \\ $$$$\mathrm{x}+\mathrm{6x}=\mathrm{42}\:\:\left[\mathrm{Multiplying}\:\mathrm{by}\:\mathrm{3}\right] \\ $$$$\mathrm{7x}=\mathrm{42} \\…
Question Number 25975 by mariahameed97@gmail.com last updated on 17/Dec/17 Answered by Rasheed.Sindhi last updated on 17/Dec/17 $$\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{6} \\ $$$$\mathrm{x}−\mathrm{y}+\mathrm{z}=\mathrm{2} \\ $$$$\mathrm{2x}−\mathrm{y}+\mathrm{3z}=\mathrm{6} \\ $$$$\:\:\mathrm{D}=\begin{vmatrix}{\mathrm{1}}&{\:\:\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{-\mathrm{1}}&{\mathrm{1}}\\{\mathrm{2}}&{-\mathrm{1}}&{\mathrm{3}}\end{vmatrix} \\ $$$$\:\:\:\:\:=\begin{vmatrix}{\mathrm{1}}&{\:\:\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{-\mathrm{2}}&{\mathrm{0}}\\{\mathrm{2}}&{-\mathrm{3}}&{\mathrm{1}}\end{vmatrix}\:\mathrm{C}_{\mathrm{2}}…
Question Number 25623 by Sunilll last updated on 12/Dec/17 $${solve}\:{for}\:{A}\:{and}\:{B}\:{if}\:\mathrm{2}{A}+{B}\:\begin{bmatrix}{\mathrm{6}\:\:\:\mathrm{3}}\\{\mathrm{6}\:\:−\mathrm{2}}\end{bmatrix} \\ $$$${and}\:\mathrm{3}{A}+\mathrm{2}{B}\:\begin{bmatrix}{\mathrm{1}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{5}}\end{bmatrix} \\ $$ Answered by mrW1 last updated on 12/Dec/17 $$\mathrm{2}{A}+{B}={C}\:\:…\left({i}\right) \\ $$$$\mathrm{3}{A}+\mathrm{2}{B}={D}\:\:…\left({ii}\right) \\…
Question Number 156610 by jlewis last updated on 13/Oct/21 $$\mathrm{If}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{invertible}\:\mathrm{matrices},\mathrm{then}: \\ $$$$\left(\mathrm{AB}\right)^{−\mathrm{1}} =\mathrm{B}^{−\mathrm{1}} \mathrm{A}^{−\mathrm{1}} \neq\:\mathrm{A}^{−\mathrm{1}} \mathrm{B}^{−\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{proove}}. \\ $$ Answered by physicstutes last updated…
Question Number 156248 by alcohol last updated on 09/Oct/21 Answered by physicstutes last updated on 09/Oct/21 $${e}^{{i}\theta} =\:\mathrm{cos}\:\theta\:+\:{i}\:\mathrm{sin}\theta…….\left({i}\right) \\ $$$${e}^{−{i}\theta} \:=\:\mathrm{cos}\:\theta\:−\:{i}\:\mathrm{sin}\:\theta…..\left({ii}\right) \\ $$$$\mathrm{equation}\:\left({i}\right)\:\mathrm{and}\:\left({ii}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{and}\:\mathrm{index}\:\left(\mathrm{euler}\right)\:\mathrm{form} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{general}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{with}\:\mathrm{modulus}\:\mathrm{1}.…