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Category: Matrices and Determinants

2-1-0-2-2018-

Question Number 32877 by 7991 last updated on 05/Apr/18 $$\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}\end{bmatrix}^{\mathrm{2018}} =…..??? \\ $$ Answered by Joel578 last updated on 05/Apr/18 $${A}\:=\:\begin{pmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}\end{pmatrix} \\ $$$${A}^{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{4}}&{\mathrm{4}}\\{\mathrm{0}}&{\mathrm{4}}\end{pmatrix} \\…

A-1-1-1-1-1-1-1-1-1-1-1-1-with-2-1-2-det-A-

Question Number 32878 by 7991 last updated on 05/Apr/18 $${A}=\begin{bmatrix}{\alpha}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\alpha}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\beta}\end{bmatrix}{with}\:\alpha^{\mathrm{2}} \neq\mathrm{1}\neq\beta^{\mathrm{2}} \\ $$$${det}\left({A}\right)=….??? \\ $$ Answered by MJS last updated on 05/Apr/18 $$\mathrm{det}\left({A}\right)=\alpha\mathrm{det}\begin{bmatrix}{\alpha}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\beta}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\end{bmatrix}−\mathrm{det}\begin{bmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\beta}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\end{bmatrix}+\mathrm{det}\begin{bmatrix}{\mathrm{1}}&{\alpha}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\end{bmatrix}−\mathrm{det}\begin{bmatrix}{\mathrm{1}}&{\alpha}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\end{bmatrix}= \\ $$$$\:\:\:\:\:\left[\mathrm{the}\:\mathrm{last}\:\mathrm{2}\:\mathrm{are}\:\mathrm{the}\:\mathrm{same},\:\mathrm{so}\:\mathrm{they}\right.…

Let-A-2-2-1-3-Find-a-non-singular-matrix-P-such-that-P-1-AP-is-a-diagonal-matrix-

Question Number 98250 by bobhans last updated on 12/Jun/20 $$\mathrm{Let}\:\mathrm{A}=\begin{pmatrix}{\mathrm{2}\:\:\:\:\mathrm{2}}\\{\mathrm{1}\:\:\:\:\mathrm{3}}\end{pmatrix}\:.\:\mathrm{Find}\:\mathrm{a}\:\mathrm{non}\:\mathrm{singular}\:\mathrm{matrix} \\ $$$$\mathrm{P}\:\mathrm{such}\:\mathrm{that}\:\mathrm{P}^{−\mathrm{1}} \mathrm{AP}\:\mathrm{is}\:\mathrm{a}\:\mathrm{diagonal}\:\mathrm{matrix}. \\ $$ Commented by john santu last updated on 12/Jun/20 $$\mathrm{find}\:\mathrm{eigen}\:\mathrm{vector}\:\mathrm{det}\left(\mathrm{A}−\lambda\mathrm{I}\right)=\mathrm{0} \\…

hello-every-one-how-do-they-calculated-the-universe-old-wich-is-13-8-billion-years-

Question Number 97885 by  M±th+et+s last updated on 10/Jun/20 $${hello}\:{every}\:{one} \\ $$$${how}\:{do}\:{they}\:{calculated}\:{the}\:{universe}\:{old} \\ $$$${wich}\:{is}\:\mathrm{13}.\mathrm{8}\:{billion}\:{years} \\ $$ Commented by EmericGent last updated on 10/Jun/20 One way is use the density in function of time and see when the density reach Planck's limit Commented…

A-1-5-6-7-find-A-k-

Question Number 31785 by yesasitya22@gmail.com last updated on 14/Mar/18 $${A}=\:\begin{bmatrix}{\mathrm{1}\:\mathrm{5}}\\{\mathrm{6}\:\mathrm{7}}\end{bmatrix}{find}\:{A}^{{k}} \\ $$ Answered by kaivan.ahmadi last updated on 16/Dec/18 $$\mathrm{sice}\:\mathrm{A}\:\mathrm{is}\:\mathrm{2}×\mathrm{1}\:\mathrm{so}\:\mathrm{we}\:\mathrm{cant}\:\mathrm{calcute}\:\mathrm{A}^{\mathrm{2}} \\ $$$$\mathrm{also}\:\mathrm{A}^{\mathrm{k}} \\ $$ Terms…