Question Number 4638 by Yozzii last updated on 17/Feb/16 $${Let}\:\boldsymbol{\mathrm{P}}=\begin{pmatrix}{\mathrm{1}−{p}_{\mathrm{1}} }&{{p}_{\mathrm{2}} }\\{{p}_{\mathrm{1}} }&{\mathrm{1}−{p}_{\mathrm{2}} }\end{pmatrix}=\begin{pmatrix}{{a}_{\mathrm{1},\mathrm{1}} }&{{a}_{\mathrm{1},\mathrm{2}} }\\{{a}_{\mathrm{2},\mathrm{1}} }&{{a}_{\mathrm{2},\mathrm{2}} }\end{pmatrix} \\ $$$${and}\:{that}\:\boldsymbol{\mathrm{P}}^{{n}} \:{be}\:{the}\:{nth}\:{power}\:{of}\:\boldsymbol{\mathrm{P}}\:{evaluated} \\ $$$${as}\:\boldsymbol{\mathrm{P}}^{{n}} =\boldsymbol{\mathrm{P}}×\boldsymbol{\mathrm{P}}^{{n}−\mathrm{1}} ;{i}.{e}\:{by}\:{successive}\:…
Question Number 69954 by 20190927 last updated on 29/Sep/19 $$\mathrm{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\:\:\mathrm{find}\:\mathrm{A}^{\mathrm{n}} \\ $$ Commented by mathmax by abdo last updated on 29/Sep/19 $${A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\:={I}\:+{J} \\ $$$${we}\:{have}\:{J}^{\mathrm{2}} =\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:.\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}…
Question Number 135467 by benjo_mathlover last updated on 13/Mar/21 Answered by EDWIN88 last updated on 13/Mar/21 $$\begin{vmatrix}{\alpha\:\:\:\:\:\beta\:\:\:\:\:\:\gamma}\\{\beta\:\:\:\:\:\gamma\:\:\:\:\:\:\alpha}\\{\gamma\:\:\:\:\:\alpha\:\:\:\:\:\beta}\end{vmatrix}=\:\alpha\left(\beta\gamma−\alpha^{\mathrm{2}} \right)−\beta\left(\beta^{\mathrm{2}} −\alpha\gamma\right)+\gamma\left(\alpha\beta−\gamma^{\mathrm{2}} \right) \\ $$$$=\alpha\beta\gamma−\alpha^{\mathrm{3}} −\beta^{\mathrm{3}} +\alpha\beta\gamma+\alpha\beta\gamma−\gamma^{\mathrm{3}} \\…
Question Number 4389 by Yozzii last updated on 17/Jan/16 $$\int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} \sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{exp}\left(\frac{−\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\right){dydx}=? \\ $$ Commented by prakash jain last…
Question Number 3965 by Yozzii last updated on 25/Dec/15 $${Find}\:{det}\left({A}\right)\:{where}\:{A}\:{is}\:{an}\:{n}×{n}\:{matrix} \\ $$$${of}\:{the}\:{form}\: \\ $$$${A}=\begin{pmatrix}{\lambda}&{\mathrm{1}}&{\mathrm{1}}&{\ldots}&{\ldots}&{\ldots}&{\ldots}&{\mathrm{1}}\\{\mathrm{1}}&{\lambda}&{\mathrm{1}}&{\ldots}&{\ldots}&{\ldots}&{\ldots}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\lambda}&{\ldots}&{\ldots}&{\ldots}&{\ldots}&{\mathrm{1}}\\{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\ldots}&{\ldots}&{\ldots}&{\mathrm{1}}\\{\vdots}&{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\ldots}&{\ldots}&{\mathrm{1}}\\{\vdots}&{\vdots}&{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\ldots}&{\mathrm{1}}\\{\vdots}&{\vdots}&{\vdots}&{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\lambda}\end{pmatrix} \\ $$$$\lambda={constant}\:{for}\:{leading}\:{diagonal}\:{elements} \\ $$$$\mathrm{1}{s}\:{for}\:{all}\:{other}\:{elements}. \\ $$$$ \\ $$ Commented by 123456…
Question Number 3566 by Yozzii last updated on 15/Dec/15 $${Let}\:{A}=\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}.\:{Find}\:{a}\:{condition}\:{on} \\ $$$${a},{b},{c},{d}\:{so}\:{that}\:{A}^{{n}+\mathrm{1}} −{A}^{{n}} ={nA},\:{n}\in\mathbb{N}. \\ $$$$ \\ $$$${A}^{{n}+\mathrm{1}} −{A}^{{n}} ={nA} \\ $$$${A}^{{n}} −{A}^{{n}−\mathrm{1}} =\left({n}−\mathrm{1}\right){A} \\…
Question Number 68782 by Maclaurin Stickker last updated on 06/Oct/19 $$\mathrm{Let}\:{d}_{{n}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{determinant}\:\mathrm{of}\:\mathrm{the}\:{n}×{n} \\ $$$$\mathrm{matrix}\:\mathrm{whose}\:\mathrm{entries},\:\mathrm{from}\:\mathrm{left}\:\mathrm{to}\:\mathrm{right} \\ $$$$\mathrm{and}\:\mathrm{then}\:\mathrm{from}\:\mathrm{top}\:\mathrm{to}\:\mathrm{bottom},\:\mathrm{are} \\ $$$${cos}\:\mathrm{1},\:{cos}\:\mathrm{2},\:…,\:{cos}\:{n}^{\mathrm{2}} .\:\left(\mathrm{For}\:\mathrm{example},\right. \\ $$$${d}_{\mathrm{3}} =\begin{vmatrix}{{cos}\:\mathrm{1}\:\:{cos}\:\mathrm{2}\:\:{cos}\:\mathrm{3}}\\{{cos}\:\mathrm{4}\:\:{cos}\:\mathrm{5}\:\:{cos}\:\mathrm{6}}\\{{cos}\:\mathrm{7}\:\:{cos}\:\mathrm{8}\:\:{cos}\:\mathrm{9}}\end{vmatrix}. \\ $$$$\mathrm{The}\:\mathrm{argument}\:\mathrm{of}\:{cos}\:\mathrm{is}\:\mathrm{always}\:\mathrm{in}\:\mathrm{radians} \\…
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Question Number 2564 by Rasheed Soomro last updated on 22/Nov/15 $${A}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{2}}&{\mathrm{5}}\end{bmatrix},{B}=\begin{bmatrix}{\mathrm{3}}&{−\mathrm{1}}\\{\mathrm{4}}&{+\mathrm{2}}\end{bmatrix} \\ $$$${determine}\:{A}^{{B}} \\ $$ Answered by Yozzi last updated on 22/Nov/15 $$\mid{A}\mid=\mathrm{5}−\mathrm{4}=\mathrm{1}\neq\mathrm{0}\:\Rightarrow{A}\:{is}\:{non}−{singular}. \\ $$$${A}^{{B}}…
Question Number 2546 by Rasheed Soomro last updated on 22/Nov/15 $$\boldsymbol{\mathrm{If}}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{two}\:\mathrm{matrices}\:\mathrm{of}\:\mathrm{suitable}\:\mathrm{order} \\ $$$$\mathrm{does}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{definition}\:\mathrm{for}\:\mathrm{A}^{\mathrm{B}} \:? \\ $$ Commented by Yozzi last updated on 22/Nov/15 $${A}^{{B}} ={e}^{\left({lnA}\right){B}}…