Question Number 34355 by srikarkailash9@gmail.com last updated on 05/May/18 Commented by math khazana by abdo last updated on 05/May/18 $${let}\:{prove}\:{by}\:{recurrence}\:{that}\: \\ $$$${A}^{{n}} \:=\:\begin{pmatrix}{{p}^{{n}} \:\:\:\:\:\:\:\:{q}\frac{{p}^{{n}} −\mathrm{1}}{{p}−\mathrm{1}}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}…
Question Number 98661 by bemath last updated on 15/Jun/20 $$\mathrm{using}\:\mathrm{cayley}\:−\:\mathrm{hamilton} \\ $$$$\mathrm{theorem}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of} \\ $$$$\mathrm{matrix}\:\mathrm{A}=\:\begin{bmatrix}{\mathrm{0}\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\\{\mathrm{1}\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\end{bmatrix}\: \\ $$ Commented by john santu last updated on 15/Jun/20 $$\mathrm{we}\:\mathrm{first}\:\mathrm{compute}\:\mathrm{the}\:\mathrm{characteristic}…
Question Number 164133 by Avijit007 last updated on 14/Jan/22 Answered by cortano1 last updated on 15/Jan/22 $$\:{A}=\begin{bmatrix}{{a}\:\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{bmatrix};\:{B}=\begin{bmatrix}{{e}\:\:\:\:{f}}\\{{g}\:\:\:\:{h}}\end{bmatrix} \\ $$$$\:\begin{bmatrix}{\mathrm{2}{a}+\mathrm{3}{e}\:\:\:\:\mathrm{2}{b}+\mathrm{3}{f}}\\{\mathrm{2}{c}+\mathrm{3}{g}\:\:\:\:\:\mathrm{2}{d}+\mathrm{3}{h}}\end{bmatrix}=\:\begin{bmatrix}{\mathrm{8}\:\:\:\:\:\mathrm{3}}\\{\mathrm{7}\:\:\:\:\:\mathrm{6}}\end{bmatrix} \\ $$$$\:\begin{bmatrix}{{a}+{e}\:\:\:\:{b}+{g}}\\{{c}+{f}\:\:\:\:{d}+{h}}\end{bmatrix}=\:\begin{bmatrix}{\mathrm{3}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{3}\:\:\:\:\:\:\mathrm{3}}\end{bmatrix} \\ $$$$\:\mathrm{2}{a}+\mathrm{3}{e}=\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{d}+\mathrm{3}{h}=\mathrm{6} \\ $$$$\:\mathrm{2}{a}+\mathrm{2}{e}=\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{d}+\mathrm{2}{h}=\mathrm{6}…
Question Number 164134 by Avijit007 last updated on 14/Jan/22 Answered by Rasheed.Sindhi last updated on 14/Jan/22 $${A}=\begin{bmatrix}{{a}_{\mathrm{11}} }&{{a}_{\mathrm{12}} }\\{{a}_{\mathrm{21}} }&{{a}_{\mathrm{22}} }\end{bmatrix},{B}=\begin{bmatrix}{{b}_{\mathrm{11}} }&{{b}_{\mathrm{12}} }\\{{b}_{\mathrm{21}} }&{{b}_{\mathrm{22}} }\end{bmatrix}\left({say}\right)…
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 $${A}\:=\:\begin{pmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}\end{pmatrix} \\ $$$${A}^{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{4}}&{\mathrm{4}}\\{\mathrm{0}}&{\mathrm{4}}\end{pmatrix} \\…
Question Number 32878 by 7991 last updated on 05/Apr/18 $${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 $$\mathrm{det}\left({A}\right)=\alpha\mathrm{det}\begin{bmatrix}{\alpha}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\beta}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\end{bmatrix}−\mathrm{det}\begin{bmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\beta}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\end{bmatrix}+\mathrm{det}\begin{bmatrix}{\mathrm{1}}&{\alpha}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\end{bmatrix}−\mathrm{det}\begin{bmatrix}{\mathrm{1}}&{\alpha}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\end{bmatrix}= \\ $$$$\:\:\:\:\:\left[\mathrm{the}\:\mathrm{last}\:\mathrm{2}\:\mathrm{are}\:\mathrm{the}\:\mathrm{same},\:\mathrm{so}\:\mathrm{they}\right.…
Question Number 98270 by M±th+et+s last updated on 12/Jun/20 $${prove} \\ $$$${Fg}={G}\frac{{m}_{\mathrm{1}} {m}_{\mathrm{2}} }{{r}^{\mathrm{2}} } \\ $$ Answered by Rio Michael last updated on 12/Jun/20…
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 $$\mathrm{find}\:\mathrm{eigen}\:\mathrm{vector}\:\mathrm{det}\left(\mathrm{A}−\lambda\mathrm{I}\right)=\mathrm{0} \\…
Question Number 98003 by M±th+et+s last updated on 11/Jun/20 $${prove}\:{that}\: \\ $$$${E}={mc}^{\mathrm{2}} \:\: \\ $$ Answered by Rio Michael last updated on 11/Jun/20 $$\mathrm{Let}\:\mathrm{me}\:\mathrm{begin}\:\mathrm{with}\:\mathrm{Plank}'\mathrm{s}\:\mathrm{equation} \\…
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