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 $${A}^{{B}} ={e}^{\left({lnA}\right){B}}…
Question Number 1771 by hareem ali last updated on 19/Sep/15 $$\mathrm{2}{x}−{y}+\mathrm{2}{z}=\mathrm{4} \\ $$$${x}+\mathrm{10}{y}−\mathrm{3}{z}=\mathrm{10} \\ $$ Answered by 123456 last updated on 19/Sep/15 $$\begin{bmatrix}{\mathrm{2}}&{−\mathrm{1}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{10}}&{−\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix}=\begin{bmatrix}{\mathrm{4}}\\{\mathrm{10}}\\{\mathrm{0}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{2}}&{−\mathrm{1}}&{\mathrm{2}}&{\mathrm{4}}\\{\mathrm{1}}&{\mathrm{10}}&{−\mathrm{3}}&{\mathrm{10}}\end{bmatrix}…
Question Number 605 by magmarsenpai last updated on 10/Feb/15 $${Given}\:{a}\:{matrix}\:{A}\:\in\:\mathbb{M}_{{n}\:×\:{n}\:} \:\forall\:\:{k}\:\in\:\mathbb{N}\:{define}: \\ $$$${A}=\begin{cases}{\varnothing_{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:{A}=\varnothing\:\:{and}\:\:{k}\geqslant\mathrm{1}}\\{{I}_{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:{A}\neq\varnothing_{{n}} \:{and}\:{k}\neq\mathrm{0}}\\{{A}^{{k}−\mathrm{1}} {A}\:\:\:\:{if}\:{A}\neq\varnothing_{{n}} \:{and}\:{k}\geqslant\mathrm{1}}\end{cases} \\ $$$${Prove}\:{that}\:{A}^{{k}} {A}^{{r}} ={A}^{{k}+{r}} \:\forall\:{k},{r}\:\in\:\mathbb{N}. \\ $$…
Question Number 529 by 8905571695 last updated on 25/Jan/15 $$ \\ $$ Answered by prakash jain last updated on 25/Jan/15 $$\mathrm{blank} \\ $$ Terms of…
Question Number 131509 by snipers237 last updated on 05/Feb/21 $${Let}\:{a},{b},{c}\:\:{be}\:\:{no}\:{null}\:{integers}.\:{A}\:{ballot}\:{box}\:{contains}\:\:“{a}''\:{black}\:{bowls}\:{and}\:“{b}''{white}\:{bowls}. \\ $$$${After}\:{a}\:{print}\:{we}\:{put}\:{the}\:{bowl}\:{back}\:{in}\:{the}\:{ballot}\:{box}\:{with}\:“{c}''\:{another}\:{bowls}\:{of}\:{the}\:{same}\:{color}. \\ $$$${Prove}\:{that}\:{the}\:{probability}\:{to}\:{extract}\:{a}\:{green}\:{bowl}\:{at}\:{any}\:\:\:{print}\:{is}\:{always} \\ $$$${p}\:=\:\frac{{a}}{{a}+{b}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com