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Question Number 93418 by abdomathmax last updated on 13/May/20
let A= (((1 2 3)),((3 2 1)) ) ∈M_3 (C) (1 4 2 ) 1) find A^(−1) if A inversible 2) calculate A^n 3)find cosA and sinA 4) is cos^2 A +sin^2 A =I ?
$${let}\:{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\:\:\:\in{M}_{\mathrm{3}} \left({C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\mathrm{2}\:\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{A}^{−\mathrm{1}} \:{if}\:{A}\:{inversible} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{3}\right){find}\:{cosA}\:\:{and}\:{sinA} \\ $$$$\left.\mathrm{4}\right)\:{is}\:{cos}^{\mathrm{2}} \:{A}\:+{sin}^{\mathrm{2}} \:{A}\:={I}\:? \\ $$$$ \\ $$
Commented by abdomathmax last updated on 13/May/20
sorry error of typo A = (((1 2 3)),((3 2 1)) ) ( 1 3 2 )
$${sorry}\:{error}\:{of}\:{typo}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{3}\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\mathrm{1}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\mathrm{2}\:\right) \\ $$
Commented by i jagooll last updated on 13/May/20
adj(A)= ((( 0 8 −4)),((−5 −1 8)),(( 10 −2 −4)) ) det(A) = 1(4−3)−2(4−1)+3(9−2) = 1−6+21 = 16 A^(−1) = (1/(16)) adj (A) = (1/(16)) ((( 0 8 −4)),((−5 −1 8)),(( 10 −2 −4)) )
$$\mathrm{adj}\left(\mathrm{A}\right)=\begin{pmatrix}{\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:\:\:−\mathrm{4}}\\{−\mathrm{5}\:\:\:\:−\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{8}}\\{\:\:\:\:\mathrm{10}\:\:\:\:−\mathrm{2}\:\:\:\:−\mathrm{4}}\end{pmatrix} \\ $$$$\mathrm{det}\left(\mathrm{A}\right)\:=\:\mathrm{1}\left(\mathrm{4}−\mathrm{3}\right)−\mathrm{2}\left(\mathrm{4}−\mathrm{1}\right)+\mathrm{3}\left(\mathrm{9}−\mathrm{2}\right) \\ $$$$=\:\mathrm{1}−\mathrm{6}+\mathrm{21}\:=\:\mathrm{16} \\ $$$$\mathrm{A}^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{16}}\:\mathrm{adj}\:\left(\mathrm{A}\right)\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}}\:\begin{pmatrix}{\:\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:\:\:−\mathrm{4}}\\{−\mathrm{5}\:\:\:\:−\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{8}}\\{\:\:\mathrm{10}\:\:\:\:−\mathrm{2}\:\:\:\:\:−\mathrm{4}}\end{pmatrix} \\ $$
Commented by Rio Michael last updated on 13/May/20
sir Jagoll what is the easiest method to get the Adjucate so far? please
$$\mathrm{sir}\:\mathrm{Jagoll}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{easiest}\:\mathrm{method} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{Adjucate}\:\mathrm{so}\:\mathrm{far}?\:\mathrm{please} \\ $$

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