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 $${we}\:{know}\:{that}\:{if}\:{A}\:{is}\:{m}×{n}\:{matrix}\:{then} \\…
Question Number 54903 by Meritguide1234 last updated on 14/Feb/19 Answered by kaivan.ahmadi last updated on 14/Feb/19 $$\mathrm{det}\left(\mathrm{ABA}^{\mathrm{T}} \right)=\mathrm{det}\left(\mathrm{A}^{\mathrm{2}} \mathrm{B}\right)=\mathrm{detA}^{\mathrm{2}} \mathrm{detB}=\mathrm{8}\:\:\left(\mathrm{i}\right) \\ $$$$\mathrm{det}\left(\mathrm{AB}^{−\mathrm{1}} \right)=\frac{\mathrm{detA}}{\mathrm{detB}}=\mathrm{8}\Rightarrow\mathrm{detA}=\mathrm{8detB}\:\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{replace}\:\left(\mathrm{ii}\right)\:\mathrm{in}\:\left(\mathrm{i}\right)\:\mathrm{deduce}…
Question Number 54741 by gunawan last updated on 10/Feb/19 $$\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{1}.\begin{pmatrix}{\mathrm{n}}\\{\mathrm{0}}\end{pmatrix}^{\mathrm{2}} +\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}^{\mathrm{2}} +…+\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}^{\mathrm{2}} =\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}.\:\begin{pmatrix}{\mathrm{n}}\\{\mathrm{0}}\end{pmatrix}+\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{1}}\end{pmatrix}+…+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{n}}\end{pmatrix}^{\mathrm{2}} =\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}} \\ $$ Commented by Abdo…
Question Number 54644 by gunawan last updated on 08/Feb/19 $$\mathrm{A}=\begin{bmatrix}{\mathrm{3}}&{\mathrm{7}}\\{−\mathrm{1}}&{−\mathrm{2}}\end{bmatrix} \\ $$$${A}^{\mathrm{27}} +{A}^{\mathrm{31}} +{A}^{\mathrm{40}} =… \\ $$ Answered by kaivan.ahmadi last updated on 10/Feb/19 $$\mathrm{A}^{\mathrm{2}}…
Question Number 119124 by abdullahquwatan last updated on 22/Oct/20 $$\mathrm{if}\:\mathrm{matrix}\:\mathrm{A}^{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{1}\:\:\mathrm{3}}\\{\mathrm{0}\:\:\mathrm{1}}\end{pmatrix}\:\:\mathrm{then}\:\mathrm{matrix}\:\mathrm{A}=… \\ $$ Answered by benjo_mathlover last updated on 22/Oct/20 $$\Leftrightarrow\:{A}=\:\begin{pmatrix}{\mathrm{1}\:\:\:\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${A}^{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\frac{\mathrm{6}}{\mathrm{2}}}\\{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\…