Question Number 54969 by gunawan last updated on 15/Feb/19 $$\mathrm{Let}\:\lambda\:\mathrm{is}\:\mathrm{value}\:\mathrm{of}\:\mathrm{characteristic} \\ $$$$\mathrm{matrices}\:{P}\:\:\mathrm{the}\:\mathrm{fill}\:{P}^{{t}} ={P}^{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{all}\:\lambda\:\mathrm{the}\:\mathrm{posible} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 54968 by gunawan last updated on 15/Feb/19 $$\mathrm{Let}\:{A}\:\mathrm{matrices}\:\mathrm{order}\:\mathrm{2}×\mathrm{2}\:\mathrm{the}\:\mathrm{fill} \\ $$$$\mathrm{tr}\left({A}^{\mathrm{2}} \right)=\left[{tr}\left({A}\right)\right]^{\mathrm{2}} \\ $$$$\mathrm{a}.\:\mathrm{Find}\:\mathrm{det}\left(\mathrm{A}\right) \\ $$$$\mathrm{b}.\:\mathrm{If}\:\mathrm{A}\:\mathrm{can}'\mathrm{t}\:\mathrm{diagonalizing},\:\mathrm{find}\:\mathrm{tr}\left({A}\right) \\ $$$$ \\ $$ Answered by kaivan.ahmadi last…
Question Number 54967 by gunawan last updated on 15/Feb/19 $$\mathrm{Let}\:{x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} ,\:{x}_{\mathrm{3}} \:\mathrm{the}\:\mathrm{number}\:\mathrm{real} \\ $$$${x}_{\mathrm{1}} <{x}_{\mathrm{2}} <{x}_{\mathrm{3}} .\:{T}\::\:{P}_{\mathrm{2}} \rightarrow{R}^{\mathrm{3}} \:\mathrm{defined} \\ $$$$\mathrm{with}\:\mathrm{rule}\:{T}=\begin{bmatrix}{{P}\left({x}_{\mathrm{1}} \right)}\\{{P}\left({x}_{\mathrm{2}} \right)}\\{{P}\left({x}_{\mathrm{3}} \right)}\end{bmatrix}…
Question Number 54962 by gunawan last updated on 15/Feb/19 $$\mathrm{Let}\:{X}=\left\{\left(−\mathrm{1},\:\mathrm{0},\:\mathrm{0}\right),\:\left(\mathrm{1},\:\mathrm{1},\:\mathrm{0}\right),\:\left(\mathrm{0},\:\mathrm{1},\:\mathrm{1}\right)\right. \\ $$$$\mathrm{and}\:\mathrm{ortogonal}\:\mathrm{projection}\:\mathrm{at}\:{X}. \\ $$$$\mathrm{Matrices}\:\mathrm{representation}\:\mathrm{P}\:\mathrm{to} \\ $$$$\mathrm{basic}\:\mathrm{basis}\:\mathrm{in}\:\mathrm{space}\:\mathrm{Euclid}\:\mathrm{R}^{\mathrm{3}} \:\mathrm{is}.. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 54963 by gunawan last updated on 15/Feb/19 $$\mathrm{Let}\:{a}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{so}\:\mathrm{matrices} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{{a}}&{{a}}\\{{a}}&{\mathrm{1}}&{{a}}\\{{a}}&{{a}}&{\mathrm{1}}\end{bmatrix}\mathrm{having}\:\mathrm{three}\:\mathrm{value} \\ $$$$\mathrm{characteristic}\:\mathrm{real}\:\lambda_{\mathrm{1}} \geqslant\lambda_{\mathrm{2}} \geqslant\lambda_{\mathrm{3}} >\mathrm{0} \\ $$$$\mathrm{then}\:{a}\:\mathrm{have}\:\mathrm{to}\:\mathrm{lies}\:\mathrm{in}\:\mathrm{interval}… \\ $$ Terms of Service Privacy…
Question Number 54953 by gunawan last updated on 15/Feb/19 $$\mathrm{The}\:\mathrm{characteristic}\:\mathrm{polynomial} \\ $$$$\mathrm{matrices}\:\begin{bmatrix}{\mathrm{1}}&{−\mathrm{2}}&{\mathrm{3}}\\{\mathrm{4}}&{\mathrm{5}}&{−\mathrm{6}}\\{−\mathrm{7}}&{\mathrm{8}}&{\mathrm{9}}\end{bmatrix}\mathrm{is}… \\ $$ Commented by maxmathsup by imad last updated on 16/Feb/19 $${det}\left({A}−{xI}\right)\:=\begin{vmatrix}{\mathrm{1}−{x}\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}−{x}\:\:\:\:\:\:\:\:−\mathrm{6}}\end{vmatrix} \\…
Question Number 54952 by gunawan last updated on 15/Feb/19 $$\mathrm{If}\:\:{T}\::\:\mathbb{C}\rightarrow\mathbb{C}\:\:\mathrm{is}\:\mathrm{linear}\:\mathrm{transformation} \\ $$$$\mathrm{and}\:{x}\:\in\:\mathbb{C},\:\mathrm{then}\:{T}\left({x}\right)=… \\ $$ Commented by kaivan.ahmadi last updated on 15/Feb/19 $${let}\:{x}={ai}+{b}\:{then} \\ $$$${T}\left({x}\right)={T}\left({ai}+{b}\right)={T}\left({ai}\right)+{T}\left({b}\right)= \\…
Question Number 54939 by gunawan last updated on 15/Feb/19 $$\mathrm{coordinate}\:{x}^{\mathrm{2}} \:\mathrm{to}\:\mathrm{basis}\:\left\{{x}^{\mathrm{2}} +{x},\:{x}+\mathrm{1},\:{x}^{\mathrm{2}} +\mathrm{1}\right\} \\ $$$$\mathrm{at}\:\mathrm{P}_{\mathrm{2}} \:\mathrm{is}… \\ $$ Commented by Abdo msup. last updated on…
Question Number 54938 by gunawan last updated on 14/Feb/19 $$\mathrm{If}\:{A}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{2}}\end{bmatrix},\mathrm{then}\:{A}^{\mathrm{2006}} =… \\ $$ Commented by Abdo msup. last updated on 15/Feb/19 $${we}\:{have}\:{A}\:=\begin{pmatrix}{\mathrm{2}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{2}}\end{pmatrix}\:\:+\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\:=\mathrm{2}{I}\:+{J} \\ $$$${J}^{\mathrm{2}} =\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:.\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:={I}\:\Rightarrow…
Question Number 54937 by gunawan last updated on 14/Feb/19 $$\mathrm{If}\:{A}\:\mathrm{matrices}\:\mathrm{order}\:\mathrm{1999}×\mathrm{2006}, \\ $$$$\mathrm{then}\:\mathrm{minimal}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{rank}\left({A}\right)+{null}\left({A}\right) \\ $$$$ \\ $$ Answered by kaivan.ahmadi last updated on 15/Feb/19 $${we}\:{know}\:{that}\:{if}\:{A}\:{is}\:{m}×{n}\:{matrix}\:{then} \\…