Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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

$$\mathrm{sir}\:\mathrm{Jagoll}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{easiest}\:\mathrm{method} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{Adjucate}\:\mathrm{so}\:\mathrm{far}?\:\mathrm{please} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com